Internal problem ID [9395]
Internal file name [OUTPUT/8332_Monday_June_06_2022_02_45_14_AM_39273400/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1063.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_x_method_1", "second_order_change_of_variable_on_x_method_2", "second_order_change_of_variable_on_y_method_1"
Maple gives the following as the ode type
[[_2nd_order, _linear, _nonhomogeneous]]
\[ \boxed {y^{\prime \prime }-\left (2 \,{\mathrm e}^{x}+1\right ) y^{\prime }+{\mathrm e}^{2 x} y={\mathrm e}^{3 x}} \]
This is second order non-homogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the solution to \[ y^{\prime \prime }+\left (-2 \,{\mathrm e}^{x}-1\right ) y^{\prime }+{\mathrm e}^{2 x} y = 0 \] In normal form the ode \begin {align*} y^{\prime \prime }+\left (-2 \,{\mathrm e}^{x}-1\right ) y^{\prime }+{\mathrm e}^{2 x} y&=0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=-2 \,{\mathrm e}^{x}-1\\ q \left (x \right )&={\mathrm e}^{2 x} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) gives \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(p_{1} = 0\). Eq (4) simplifies to \begin {align*} \tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )&=0 \end {align*}
This ode is solved resulting in \begin {align*} \tau &= \int {\mathrm e}^{-\left (\int p \left (x \right )d x \right )}d x\\ &= \int {\mathrm e}^{-\left (\int \left (-2 \,{\mathrm e}^{x}-1\right )d x \right )}d x\\ &= \int e^{x +2 \,{\mathrm e}^{x}} \,dx\\ &= \int {\mathrm e}^{x +2 \,{\mathrm e}^{x}}d x\\ &= \frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{2}\tag {6} \end {align*}
Using (6) to evaluate \(q_{1}\) from (5) gives \begin {align*} q_{1} \left (\tau \right ) &= \frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &= \frac {{\mathrm e}^{2 x}}{{\mathrm e}^{2 x +4 \,{\mathrm e}^{x}}}\\ &= {\mathrm e}^{-4 \,{\mathrm e}^{x}}\tag {7} \end {align*}
Substituting the above in (3) and noting that now \(p_{1} = 0\) results in \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+q_{1} y \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+{\mathrm e}^{-4 \,{\mathrm e}^{x}} y \left (\tau \right )&=0 \\ \end {align*}
But in terms of \(\tau \) \begin {align*} {\mathrm e}^{-4 \,{\mathrm e}^{x}}&=\frac {1}{4 \tau ^{2}} \end {align*}
Hence the above ode becomes \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+\frac {y \left (\tau \right )}{4 \tau ^{2}}&=0 \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). The ode can be written as \[ 4 \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right ) \tau ^{2}+y \left (\tau \right ) = 0 \] Which shows it is a Euler ODE. This is Euler second order ODE. Let the solution be \(y \left (\tau \right ) = \tau ^r\), then \(y'=r \tau ^{r-1}\) and \(y''=r(r-1) \tau ^{r-2}\). Substituting these back into the given ODE gives \[ 4 \tau ^{2}(r(r-1))\tau ^{r-2}+0 r \tau ^{r-1}+\tau ^{r} = 0 \] Simplifying gives \[ 4 r \left (r -1\right )\tau ^{r}+0\,\tau ^{r}+\tau ^{r} = 0 \] Since \(\tau ^{r}\neq 0\) then dividing throughout by \(\tau ^{r}\) gives \[ 4 r \left (r -1\right )+0+1 = 0 \] Or \[ 4 r^{2}-4 r +1 = 0 \tag {1} \] Equation (1) is the characteristic equation. Its roots determine the form of the general solution. Using the quadratic equation the roots are \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= {\frac {1}{2}} \end {align*}
Since the roots are equal, then the general solution is \[ y \left (\tau \right )= c_{1} y_1 + c_{2} y_2 \] Where \(y_1 = \tau ^{r}\) and \(y_2 = \tau ^{r} \ln \left (\tau \right )\). Hence \[ y \left (\tau \right ) = c_{1} \sqrt {\tau }+c_{2} \sqrt {\tau }\, \ln \left (\tau \right ) \] The above solution is now transformed back to \(y\) using (6) which results in \begin {align*} y &= \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \left (c_{1} +c_{2} \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )-c_{2} \ln \left (2\right )\right )}{2} \end {align*}
Therefore the homogeneous solution \(y_h\) is \[ y_h = \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \left (c_{1} +c_{2} \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )-c_{2} \ln \left (2\right )\right )}{2} \] The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let \begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}} \\ y_2 &= \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2} \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*} Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}} & \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2} \\ \frac {d}{dx}\left (\sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\right ) & \frac {d}{dx}\left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}} & \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2} \\ \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{x} & \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right ) {\mathrm e}^{x}}{2}+\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{x}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right ) {\mathrm e}^{x}}{2} \end {vmatrix} \] Therefore \[ W = \left (\sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\right )\left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right ) {\mathrm e}^{x}}{2}+\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{x}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right ) {\mathrm e}^{x}}{2}\right ) - \left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2}\right )\left (\sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{x}\right ) \] Which simplifies to \[ W = \sqrt {2}\, {\mathrm e}^{2 \,{\mathrm e}^{x}} {\mathrm e}^{x} \] Which simplifies to \[ W = \sqrt {2}\, {\mathrm e}^{x +2 \,{\mathrm e}^{x}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2}\right ) {\mathrm e}^{3 x}}{\sqrt {2}\, {\mathrm e}^{x +2 \,{\mathrm e}^{x}}}\,dx \] Which simplifies to \[ u_1 = - \int \frac {{\mathrm e}^{2 x -2 \,{\mathrm e}^{x}} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )}{2}d x \] Hence \[ u_1 = -\left (\int _{0}^{x}\frac {{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )}{2}d \alpha \right ) \] And Eq. (3) becomes \[ u_2 = \int \frac {\sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{3 x}}{\sqrt {2}\, {\mathrm e}^{x +2 \,{\mathrm e}^{x}}}\,dx \] Which simplifies to \[ u_2 = \int \frac {\sqrt {2}\, {\mathrm e}^{2 x -2 \,{\mathrm e}^{x}} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}d x \] Hence \[ u_2 = \int _{0}^{x}\frac {\sqrt {2}\, {\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}}{2}d \alpha \] Which simplifies to \begin{align*} u_1 &= -\frac {\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )}{2} \\ u_2 &= \frac {\sqrt {2}\, \left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right )}{2} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\frac {\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}+\frac {\sqrt {2}\, \left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2}\right )}{2} \] Which simplifies to \[ y_p(x) = \frac {\left (-\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )+\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \left (c_{1} +c_{2} \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )-c_{2} \ln \left (2\right )\right )}{2}\right ) + \left (\frac {\left (-\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )+\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \left (c_{1} +c_{2} \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )-c_{2} \ln \left (2\right )\right )}{2}+\frac {\left (-\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )+\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \left (c_{1} +c_{2} \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )-c_{2} \ln \left (2\right )\right )}{2}+\frac {\left (-\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )+\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2} \] Verified OK.
This is second order non-homogeneous ODE. In standard form the ODE is \[ A y''(x) + B y'(x) + C y(x) = f(x) \] Where \(A=1, B=-2 \,{\mathrm e}^{x}-1, C={\mathrm e}^{2 x}, f(x)={\mathrm e}^{3 x}\). Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). Solving for \(y_h\) from \[ y^{\prime \prime }+\left (-2 \,{\mathrm e}^{x}-1\right ) y^{\prime }+{\mathrm e}^{2 x} y = 0 \] In normal form the ode \begin {align*} y^{\prime \prime }+\left (-2 \,{\mathrm e}^{x}-1\right ) y^{\prime }+{\mathrm e}^{2 x} y&=0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=-2 \,{\mathrm e}^{x}-1\\ q \left (x \right )&={\mathrm e}^{2 x} \end {align*}
Applying change of variables \(\tau = g \left (x \right )\) to (2) results \begin {align*} \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }y \left (\tau \right )\right )+q_{1} y \left (\tau \right )&=0 \tag {3} \end {align*}
Where \(\tau \) is the new independent variable, and \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\tag {5} \end {align*}
Let \(q_1=c^2\) where \(c\) is some constant. Therefore from (5) \begin {align*} \tau ' &= \frac {1}{c}\sqrt {q}\\ &= \frac {\sqrt {{\mathrm e}^{2 x}}}{c}\tag {6} \\ \tau '' &= \frac {\sqrt {{\mathrm e}^{2 x}}}{c} \end {align*}
Substituting the above into (4) results in \begin {align*} p_{1} \left (\tau \right ) &=\frac {\tau ^{\prime \prime }\left (x \right )+p \left (x \right ) \tau ^{\prime }\left (x \right )}{{\tau ^{\prime }\left (x \right )}^{2}}\\ &=\frac {\frac {\sqrt {{\mathrm e}^{2 x}}}{c}-2 \,{\mathrm e}^{x}-1\frac {\sqrt {{\mathrm e}^{2 x}}}{c}}{\left (\frac {\sqrt {{\mathrm e}^{2 x}}}{c}\right )^2} \\ &=-2 c \end {align*}
Therefore ode (3) now becomes \begin {align*} y \left (\tau \right )'' + p_1 y \left (\tau \right )' + q_1 y \left (\tau \right ) &= 0 \\ \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )-2 c \left (\frac {d}{d \tau }y \left (\tau \right )\right )+c^{2} y \left (\tau \right ) &= 0 \tag {7} \end {align*}
The above ode is now solved for \(y \left (\tau \right )\). Since the ode is now constant coefficients, it can be easily solved to give \begin {align*} y \left (\tau \right ) &= {\mathrm e}^{c \tau } c_{1} \end {align*}
Now from (6) \begin {align*} \tau &= \int \frac {1}{c} \sqrt q \,dx \\ &= \frac {\int \sqrt {{\mathrm e}^{2 x}}d x}{c}\\ &= \frac {\sqrt {{\mathrm e}^{2 x}}}{c} \end {align*}
Substituting the above into the solution obtained gives \[ y = {\mathrm e}^{{\mathrm e}^{x}} c_{1} \] Now the particular solution to this ODE is found \[ y^{\prime \prime }+\left (-2 \,{\mathrm e}^{x}-1\right ) y^{\prime }+{\mathrm e}^{2 x} y = {\mathrm e}^{3 x} \] The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let \begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}} \\ y_2 &= \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2} \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*} Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}} & \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2} \\ \frac {d}{dx}\left (\sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\right ) & \frac {d}{dx}\left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}} & \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2} \\ \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{x} & \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right ) {\mathrm e}^{x}}{2}+\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{x}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right ) {\mathrm e}^{x}}{2} \end {vmatrix} \] Therefore \[ W = \left (\sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\right )\left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right ) {\mathrm e}^{x}}{2}+\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{x}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right ) {\mathrm e}^{x}}{2}\right ) - \left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2}\right )\left (\sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{x}\right ) \] Which simplifies to \[ W = \sqrt {2}\, {\mathrm e}^{2 \,{\mathrm e}^{x}} {\mathrm e}^{x} \] Which simplifies to \[ W = \sqrt {2}\, {\mathrm e}^{x +2 \,{\mathrm e}^{x}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2}\right ) {\mathrm e}^{3 x}}{\sqrt {2}\, {\mathrm e}^{x +2 \,{\mathrm e}^{x}}}\,dx \] Which simplifies to \[ u_1 = - \int \frac {{\mathrm e}^{2 x -2 \,{\mathrm e}^{x}} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )}{2}d x \] Hence \[ u_1 = -\left (\int _{0}^{x}\frac {{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )}{2}d \alpha \right ) \] And Eq. (3) becomes \[ u_2 = \int \frac {\sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{3 x}}{\sqrt {2}\, {\mathrm e}^{x +2 \,{\mathrm e}^{x}}}\,dx \] Which simplifies to \[ u_2 = \int \frac {\sqrt {2}\, {\mathrm e}^{2 x -2 \,{\mathrm e}^{x}} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}d x \] Hence \[ u_2 = \int _{0}^{x}\frac {\sqrt {2}\, {\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}}{2}d \alpha \] Which simplifies to \begin{align*} u_1 &= -\frac {\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )}{2} \\ u_2 &= \frac {\sqrt {2}\, \left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right )}{2} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\frac {\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}+\frac {\sqrt {2}\, \left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2}\right )}{2} \] Which simplifies to \[ y_p(x) = \frac {\left (-\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )+\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{{\mathrm e}^{x}} c_{1}\right ) + \left (\frac {\left (-\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )+\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}\right ) \\ &= \frac {\left (-\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )+\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}+{\mathrm e}^{{\mathrm e}^{x}} c_{1} \\ \end{align*} Which simplifies to \[ y = \frac {\left (-\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )+\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}+{\mathrm e}^{{\mathrm e}^{x}} c_{1} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )+\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}+{\mathrm e}^{{\mathrm e}^{x}} c_{1} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )+\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}+{\mathrm e}^{{\mathrm e}^{x}} c_{1} \] Verified OK.
This is second order non-homogeneous ODE. Let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ODE \( A y''(x) + B y'(x) + C y(x) = 0\), and \(y_p\) is a particular solution to the non-homogeneous ODE \(A y''(x) + B y'(x) + C y(x) = f(x)\). \(y_h\) is the solution to \[ y^{\prime \prime }+\left (-2 \,{\mathrm e}^{x}-1\right ) y^{\prime }+{\mathrm e}^{2 x} y = 0 \] In normal form the given ode is written as \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=-2 \,{\mathrm e}^{x}-1\\ q \left (x \right )&={\mathrm e}^{2 x} \end {align*}
Calculating the Liouville ode invariant \(Q\) given by \begin {align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= {\mathrm e}^{2 x} - \frac {\left (-2 \,{\mathrm e}^{x}-1\right )'}{2}- \frac {\left (-2 \,{\mathrm e}^{x}-1\right )^2}{4} \\ &= {\mathrm e}^{2 x} - \frac {\left (-2 \,{\mathrm e}^{x}\right )}{2}- \frac {\left (\left (-2 \,{\mathrm e}^{x}-1\right )^{2}\right )}{4} \\ &= {\mathrm e}^{2 x} - \left (-{\mathrm e}^{x}\right )-\frac {\left (-2 \,{\mathrm e}^{x}-1\right )^{2}}{4}\\ &= -{\frac {1}{4}} \end {align*}
Since the Liouville ode invariant does not depend on the independent variable \(x\) then the transformation \begin {align*} y = v \left (x \right ) z \left (x \right )\tag {3} \end {align*}
is used to change the original ode to a constant coefficients ode in \(v\). In (3) the term \(z \left (x \right )\) is given by \begin {align*} z \left (x \right )&={\mathrm e}^{-\left (\int \frac {p \left (x \right )}{2}d x \right )}\\ &= e^{-\int \frac {-2 \,{\mathrm e}^{x}-1}{2} }\\ &= {\mathrm e}^{\frac {x}{2}+{\mathrm e}^{x}}\tag {5} \end {align*}
Hence (3) becomes \begin {align*} y = v \left (x \right ) {\mathrm e}^{\frac {x}{2}+{\mathrm e}^{x}}\tag {4} \end {align*}
Applying this change of variable to the original ode results in \begin {align*} {\mathrm e}^{\frac {x}{2}+{\mathrm e}^{x}} \left (4 v^{\prime \prime }\left (x \right )-v \left (x \right )\right ) = 4 \,{\mathrm e}^{3 x} \end {align*}
Which is now solved for \(v \left (x \right )\) Simplyfing the ode gives \[ v^{\prime \prime }\left (x \right )-\frac {v \left (x \right )}{4} = {\mathrm e}^{\frac {5 x}{2}-{\mathrm e}^{x}} \] This is second order non-homogeneous ODE. In standard form the ODE is \[ A v''(x) + B v'(x) + C v(x) = f(x) \] Where \(A=1, B=0, C=-{\frac {1}{4}}, f(x)={\mathrm e}^{\frac {5 x}{2}-{\mathrm e}^{x}}\). Let the solution be \[ v \left (x \right ) = v_h + v_p \] Where \(v_h\) is the solution to the homogeneous ODE \( A v''(x) + B v'(x) + C v(x) = 0\), and \(v_p\) is a particular solution to the non-homogeneous ODE \(A v''(x) + B v'(x) + C v(x) = f(x)\). \(v_h\) is the solution to \[ v^{\prime \prime }\left (x \right )-\frac {v \left (x \right )}{4} = 0 \] This is second order with constant coefficients homogeneous ODE. In standard form the ODE is \[ A v''(x) + B v'(x) + C v(x) = 0 \] Where in the above \(A=1, B=0, C=-{\frac {1}{4}}\). Let the solution be \(v \left (x \right )=e^{\lambda x}\). Substituting this into the ODE gives \[ \lambda ^{2} {\mathrm e}^{\lambda x}-\frac {{\mathrm e}^{\lambda x}}{4} = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda x}\) gives \[ \lambda ^{2}-\frac {1}{4} = 0 \tag {2} \] Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula \[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \] Substituting \(A=1, B=0, C=-{\frac {1}{4}}\) into the above gives \begin {align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (-{\frac {1}{4}}\right )}\\ &= \pm {\frac {1}{2}} \end {align*}
Hence \begin{align*} \lambda _1 &= + {\frac {1}{2}} \\ \lambda _2 &= - {\frac {1}{2}} \\ \end{align*} Which simplifies to \begin{align*} \lambda _1 &= {\frac {1}{2}} \\ \lambda _2 &= -{\frac {1}{2}} \\ \end{align*} Since roots are real and distinct, then the solution is \begin{align*} v \left (x \right ) &= c_{1} e^{\lambda _1 x} + c_{2} e^{\lambda _2 x} \\ v \left (x \right ) &= c_{1} e^{\left ({\frac {1}{2}}\right )x} +c_{2} e^{\left (-{\frac {1}{2}}\right )x} \\ \end{align*} Or \[ v \left (x \right ) =c_{1} {\mathrm e}^{\frac {x}{2}}+c_{2} {\mathrm e}^{-\frac {x}{2}} \] Therefore the homogeneous solution \(v_h\) is \[ v_h = c_{1} {\mathrm e}^{\frac {x}{2}}+c_{2} {\mathrm e}^{-\frac {x}{2}} \] The particular solution \(v_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let \begin{equation} \tag{1} v_p(x) = u_1 v_1 + u_2 v_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(v_1,v_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} v_1 &= {\mathrm e}^{\frac {x}{2}} \\ v_2 &= {\mathrm e}^{-\frac {x}{2}} \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {v_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {v_1 f(x)}{a W(x)} \\ \end{align*} Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(v''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} v_1 & v_{2} \\ v_{1}^{\prime } & v_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} {\mathrm e}^{\frac {x}{2}} & {\mathrm e}^{-\frac {x}{2}} \\ \frac {d}{dx}\left ({\mathrm e}^{\frac {x}{2}}\right ) & \frac {d}{dx}\left ({\mathrm e}^{-\frac {x}{2}}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} {\mathrm e}^{\frac {x}{2}} & {\mathrm e}^{-\frac {x}{2}} \\ \frac {{\mathrm e}^{\frac {x}{2}}}{2} & -\frac {{\mathrm e}^{-\frac {x}{2}}}{2} \end {vmatrix} \] Therefore \[ W = \left ({\mathrm e}^{\frac {x}{2}}\right )\left (-\frac {{\mathrm e}^{-\frac {x}{2}}}{2}\right ) - \left ({\mathrm e}^{-\frac {x}{2}}\right )\left (\frac {{\mathrm e}^{\frac {x}{2}}}{2}\right ) \] Which simplifies to \[ W = -{\mathrm e}^{\frac {x}{2}} {\mathrm e}^{-\frac {x}{2}} \] Which simplifies to \[ W = -1 \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {{\mathrm e}^{-\frac {x}{2}} {\mathrm e}^{\frac {5 x}{2}-{\mathrm e}^{x}}}{-1}\,dx \] Which simplifies to \[ u_1 = - \int -{\mathrm e}^{2 x -{\mathrm e}^{x}}d x \] Hence \[ u_1 = -{\mathrm e}^{x} {\mathrm e}^{-{\mathrm e}^{x}}-{\mathrm e}^{-{\mathrm e}^{x}} \] And Eq. (3) becomes \[ u_2 = \int \frac {{\mathrm e}^{\frac {x}{2}} {\mathrm e}^{\frac {5 x}{2}-{\mathrm e}^{x}}}{-1}\,dx \] Which simplifies to \[ u_2 = \int -{\mathrm e}^{3 x -{\mathrm e}^{x}}d x \] Hence \[ u_2 = {\mathrm e}^{2 x} {\mathrm e}^{-{\mathrm e}^{x}}+2 \,{\mathrm e}^{x} {\mathrm e}^{-{\mathrm e}^{x}}+2 \,{\mathrm e}^{-{\mathrm e}^{x}} \] Which simplifies to \begin{align*} u_1 &= -{\mathrm e}^{x -{\mathrm e}^{x}}-{\mathrm e}^{-{\mathrm e}^{x}} \\ u_2 &= {\mathrm e}^{2 x -{\mathrm e}^{x}}+2 \,{\mathrm e}^{x -{\mathrm e}^{x}}+2 \,{\mathrm e}^{-{\mathrm e}^{x}} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ v_p(x) = \left (-{\mathrm e}^{x -{\mathrm e}^{x}}-{\mathrm e}^{-{\mathrm e}^{x}}\right ) {\mathrm e}^{\frac {x}{2}}+\left ({\mathrm e}^{2 x -{\mathrm e}^{x}}+2 \,{\mathrm e}^{x -{\mathrm e}^{x}}+2 \,{\mathrm e}^{-{\mathrm e}^{x}}\right ) {\mathrm e}^{-\frac {x}{2}} \] Which simplifies to \[ v_p(x) = {\mathrm e}^{\frac {x}{2}-{\mathrm e}^{x}}+2 \,{\mathrm e}^{-\frac {x}{2}-{\mathrm e}^{x}} \] Therefore the general solution is \begin{align*} v &= v_h + v_p \\ &= \left (c_{1} {\mathrm e}^{\frac {x}{2}}+c_{2} {\mathrm e}^{-\frac {x}{2}}\right ) + \left ({\mathrm e}^{\frac {x}{2}-{\mathrm e}^{x}}+2 \,{\mathrm e}^{-\frac {x}{2}-{\mathrm e}^{x}}\right ) \\ \end{align*} Now that \(v \left (x \right )\) is known, then \begin {align*} y&= v \left (x \right ) z \left (x \right )\\ &= \left (c_{1} {\mathrm e}^{\frac {x}{2}}+c_{2} {\mathrm e}^{-\frac {x}{2}}+{\mathrm e}^{\frac {x}{2}-{\mathrm e}^{x}}+2 \,{\mathrm e}^{-\frac {x}{2}-{\mathrm e}^{x}}\right ) \left (z \left (x \right )\right )\tag {7} \end {align*}
But from (5) \begin {align*} z \left (x \right )&= {\mathrm e}^{\frac {x}{2}+{\mathrm e}^{x}} \end {align*}
Hence (7) becomes \begin {align*} y = \left (c_{1} {\mathrm e}^{\frac {x}{2}}+c_{2} {\mathrm e}^{-\frac {x}{2}}+{\mathrm e}^{\frac {x}{2}-{\mathrm e}^{x}}+2 \,{\mathrm e}^{-\frac {x}{2}-{\mathrm e}^{x}}\right ) {\mathrm e}^{\frac {x}{2}+{\mathrm e}^{x}} \end {align*}
Therefore the homogeneous solution \(y_h\) is \[ y_h = \left (c_{1} {\mathrm e}^{\frac {x}{2}}+c_{2} {\mathrm e}^{-\frac {x}{2}}+{\mathrm e}^{\frac {x}{2}-{\mathrm e}^{x}}+2 \,{\mathrm e}^{-\frac {x}{2}-{\mathrm e}^{x}}\right ) {\mathrm e}^{\frac {x}{2}+{\mathrm e}^{x}} \] The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let \begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation} Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as \begin{align*} y_1 &= \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}} \\ y_2 &= \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2} \\ \end{align*} In the Variation of parameters \(u_1,u_2\) are found using \begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*} Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence \[ W = \begin {vmatrix} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}} & \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2} \\ \frac {d}{dx}\left (\sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\right ) & \frac {d}{dx}\left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2}\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}} & \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2} \\ \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{x} & \frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right ) {\mathrm e}^{x}}{2}+\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{x}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right ) {\mathrm e}^{x}}{2} \end {vmatrix} \] Therefore \[ W = \left (\sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\right )\left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right ) {\mathrm e}^{x}}{2}+\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{x}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right ) {\mathrm e}^{x}}{2}\right ) - \left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2}\right )\left (\sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{x}\right ) \] Which simplifies to \[ W = \sqrt {2}\, {\mathrm e}^{2 \,{\mathrm e}^{x}} {\mathrm e}^{x} \] Which simplifies to \[ W = \sqrt {2}\, {\mathrm e}^{x +2 \,{\mathrm e}^{x}} \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2}\right ) {\mathrm e}^{3 x}}{\sqrt {2}\, {\mathrm e}^{x +2 \,{\mathrm e}^{x}}}\,dx \] Which simplifies to \[ u_1 = - \int \frac {{\mathrm e}^{2 x -2 \,{\mathrm e}^{x}} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )}{2}d x \] Hence \[ u_1 = -\left (\int _{0}^{x}\frac {{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )}{2}d \alpha \right ) \] And Eq. (3) becomes \[ u_2 = \int \frac {\sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, {\mathrm e}^{3 x}}{\sqrt {2}\, {\mathrm e}^{x +2 \,{\mathrm e}^{x}}}\,dx \] Which simplifies to \[ u_2 = \int \frac {\sqrt {2}\, {\mathrm e}^{2 x -2 \,{\mathrm e}^{x}} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}d x \] Hence \[ u_2 = \int _{0}^{x}\frac {\sqrt {2}\, {\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}}{2}d \alpha \] Which simplifies to \begin{align*} u_1 &= \frac {\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (\ln \left (2\right )-\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right )}{2} \\ u_2 &= \frac {\sqrt {2}\, \left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right )}{2} \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = \frac {\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (\ln \left (2\right )-\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha \right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}+\frac {\sqrt {2}\, \left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )}{2}-\frac {\sqrt {2}\, \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}\, \ln \left (2\right )}{2}\right )}{2} \] Which simplifies to \[ y_p(x) = \frac {\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (\ln \left (2\right )-\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha +\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2} \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left (\left (c_{1} {\mathrm e}^{\frac {x}{2}}+c_{2} {\mathrm e}^{-\frac {x}{2}}+{\mathrm e}^{\frac {x}{2}-{\mathrm e}^{x}}+2 \,{\mathrm e}^{-\frac {x}{2}-{\mathrm e}^{x}}\right ) {\mathrm e}^{\frac {x}{2}+{\mathrm e}^{x}}\right ) + \left (\frac {\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (\ln \left (2\right )-\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha +\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2}\right ) \\ \end{align*} Which simplifies to \[ y = c_{1} {\mathrm e}^{x +{\mathrm e}^{x}}+{\mathrm e}^{x}+c_{2} {\mathrm e}^{{\mathrm e}^{x}}+2+\frac {\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (\ln \left (2\right )-\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha +\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{x +{\mathrm e}^{x}}+{\mathrm e}^{x}+c_{2} {\mathrm e}^{{\mathrm e}^{x}}+2+\frac {\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (\ln \left (2\right )-\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha +\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{x +{\mathrm e}^{x}}+{\mathrm e}^{x}+c_{2} {\mathrm e}^{{\mathrm e}^{x}}+2+\frac {\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}\, \left (\ln \left (2\right )-\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{\alpha }}\right )\right )d \alpha +\left (\int _{0}^{x}{\mathrm e}^{2 \alpha -2 \,{\mathrm e}^{\alpha }} \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{\alpha }}}d \alpha \right ) \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{x}}\right )\right )\right ) \sqrt {{\mathrm e}^{2 \,{\mathrm e}^{x}}}}{2} \] Verified OK.
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature trying high order exact linear fully integrable trying differential order: 2; linear nonhomogeneous with symmetry [0,1] trying a double symmetry of the form [xi=0, eta=F(x)] trying symmetries linear in x and y(x) -> Try solving first the homogeneous part of the ODE trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Reducible group (found another exponential solution) <- Kovacics algorithm successful <- solving first the homogeneous part of the ODE successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 34
dsolve(diff(diff(y(x),x),x)-(2*exp(x)+1)*diff(y(x),x)+exp(2*x)*y(x)-exp(3*x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{\frac {x}{2}+{\mathrm e}^{x}} \sinh \left (\frac {x}{2}\right ) c_{2} +{\mathrm e}^{\frac {x}{2}+{\mathrm e}^{x}} \cosh \left (\frac {x}{2}\right ) c_{1} +{\mathrm e}^{x}+2 \]
✓ Solution by Mathematica
Time used: 0.078 (sec). Leaf size: 28
DSolve[-E^(3*x) + E^(2*x)*y[x] - (1 + 2*E^x)*y'[x] + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^x+c_1 e^{e^x}+c_2 e^{x+e^x}+2 \]