Internal problem ID [11104]
Internal file name [OUTPUT/10361_Wednesday_January_24_2024_10_18_15_PM_64620218/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.3-1. Equations with
exponential functions
Problem number: 17.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "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, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }-\left (a +2 b \,{\mathrm e}^{a x}\right ) y^{\prime }+b^{2} {\mathrm e}^{2 a x} y=0} \]
In normal form the ode \begin {align*} y^{\prime \prime }+\left (-2 b \,{\mathrm e}^{a x}-a \right ) y^{\prime }+b^{2} {\mathrm e}^{2 a 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 b \,{\mathrm e}^{a x}-a\\ q \left (x \right )&=b^{2} {\mathrm e}^{2 a 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 b \,{\mathrm e}^{a x}-a \right )d x \right )}d x\\ &= \int e^{\frac {a^{2} x +2 b \,{\mathrm e}^{a x}}{a}} \,dx\\ &= \int {\mathrm e}^{\frac {a^{2} x +2 b \,{\mathrm e}^{a x}}{a}}d x\\ &= \frac {{\mathrm e}^{\frac {2 b \,{\mathrm e}^{a x}}{a}}}{2 b}\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 {b^{2} {\mathrm e}^{2 a x}}{{\mathrm e}^{\frac {2 a^{2} x +4 b \,{\mathrm e}^{a x}}{a}}}\\ &= b^{2} {\mathrm e}^{-\frac {4 b \,{\mathrm e}^{a x}}{a}}\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 )+b^{2} {\mathrm e}^{-\frac {4 b \,{\mathrm e}^{a x}}{a}} y \left (\tau \right )&=0 \\ \end {align*}
But in terms of \(\tau \) \begin {align*} b^{2} {\mathrm e}^{-\frac {4 b \,{\mathrm e}^{a x}}{a}}&=\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 {\frac {{\mathrm e}^{\frac {2 b \,{\mathrm e}^{a x}}{a}}}{b}}\, \left (c_{1} +c_{2} \ln \left (\frac {{\mathrm e}^{\frac {2 b \,{\mathrm e}^{a x}}{a}}}{b}\right )-c_{2} \ln \left (2\right )\right )}{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sqrt {2}\, \sqrt {\frac {{\mathrm e}^{\frac {2 b \,{\mathrm e}^{a x}}{a}}}{b}}\, \left (c_{1} +c_{2} \ln \left (\frac {{\mathrm e}^{\frac {2 b \,{\mathrm e}^{a x}}{a}}}{b}\right )-c_{2} \ln \left (2\right )\right )}{2} \\ \end{align*}
Verification of solutions
\[ y = \frac {\sqrt {2}\, \sqrt {\frac {{\mathrm e}^{\frac {2 b \,{\mathrm e}^{a x}}{a}}}{b}}\, \left (c_{1} +c_{2} \ln \left (\frac {{\mathrm e}^{\frac {2 b \,{\mathrm e}^{a x}}{a}}}{b}\right )-c_{2} \ln \left (2\right )\right )}{2} \] Verified OK.
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 b \,{\mathrm e}^{a x}-a\\ q \left (x \right )&=b^{2} {\mathrm e}^{2 a x} \end {align*}
Calculating the Liouville ode invariant \(Q\) given by \begin {align*} Q &= q - \frac {p'}{2}- \frac {p^2}{4} \\ &= b^{2} {\mathrm e}^{2 a x} - \frac {\left (-2 b \,{\mathrm e}^{a x}-a\right )'}{2}- \frac {\left (-2 b \,{\mathrm e}^{a x}-a\right )^2}{4} \\ &= b^{2} {\mathrm e}^{2 a x} - \frac {\left (-2 b a \,{\mathrm e}^{a x}\right )}{2}- \frac {\left (\left (-2 b \,{\mathrm e}^{a x}-a \right )^{2}\right )}{4} \\ &= b^{2} {\mathrm e}^{2 a x} - \left (-b a \,{\mathrm e}^{a x}\right )-\frac {\left (-2 b \,{\mathrm e}^{a x}-a \right )^{2}}{4}\\ &= -\frac {a^{2}}{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 b \,{\mathrm e}^{a x}-a}{2} }\\ &= {\mathrm e}^{\frac {a^{2} x +2 b \,{\mathrm e}^{a x}}{2 a}}\tag {5} \end {align*}
Hence (3) becomes \begin {align*} y = v \left (x \right ) {\mathrm e}^{\frac {a^{2} x +2 b \,{\mathrm e}^{a x}}{2 a}}\tag {4} \end {align*}
Applying this change of variable to the original ode results in \begin {align*} {\mathrm e}^{\frac {a^{2} x +2 b \,{\mathrm e}^{a x}}{2 a}} \left (-a^{2} v \left (x \right )+4 v^{\prime \prime }\left (x \right )\right ) = 0 \end {align*}
Which is now solved for \(v \left (x \right )\) 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 \(A=4, B=0, C=-a^{2}\). Let the solution be \(v \left (x \right )=e^{\lambda x}\). Substituting this into the ODE gives \[ 4 \lambda ^{2} {\mathrm e}^{\lambda x}-a^{2} {\mathrm e}^{\lambda x} = 0 \tag {1} \] Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda x}\) gives \[ -a^{2}+4 \lambda ^{2} = 0 \tag {2} \] 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 \(A=4, B=0, C=-a^{2}\) into the above gives \begin {align*} \lambda _{1,2} &= \frac {0}{(2) \left (4\right )} \pm \frac {1}{(2) \left (4\right )} \sqrt {0^2 - (4) \left (4\right )\left (-a^{2}\right )}\\ &= \pm \frac {\sqrt {a^{2}}}{2} \end {align*}
Hence \begin{align*} \lambda _1 &= + \frac {\sqrt {a^{2}}}{2} \\ \lambda _2 &= - \frac {\sqrt {a^{2}}}{2} \\ \end{align*} Which simplifies to \begin{align*} \lambda _1 &= \frac {\sqrt {a^{2}}}{2} \\ \lambda _2 &= -\frac {\sqrt {a^{2}}}{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 {\sqrt {a^{2}}}{2}\right )x} +c_{2} e^{\left (-\frac {\sqrt {a^{2}}}{2}\right )x} \\ \end{align*} Or \[ v \left (x \right ) =c_{1} {\mathrm e}^{\frac {\sqrt {a^{2}}\, x}{2}}+c_{2} {\mathrm e}^{-\frac {\sqrt {a^{2}}\, x}{2}} \] 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 {\sqrt {a^{2}}\, x}{2}}+c_{2} {\mathrm e}^{-\frac {\sqrt {a^{2}}\, x}{2}}\right ) \left (z \left (x \right )\right )\tag {7} \end {align*}
But from (5) \begin {align*} z \left (x \right )&= {\mathrm e}^{\frac {a^{2} x +2 b \,{\mathrm e}^{a x}}{2 a}} \end {align*}
Hence (7) becomes \begin {align*} y = \left (c_{1} {\mathrm e}^{\frac {\sqrt {a^{2}}\, x}{2}}+c_{2} {\mathrm e}^{-\frac {\sqrt {a^{2}}\, x}{2}}\right ) {\mathrm e}^{\frac {a^{2} x +2 b \,{\mathrm e}^{a x}}{2 a}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (c_{1} {\mathrm e}^{\frac {\sqrt {a^{2}}\, x}{2}}+c_{2} {\mathrm e}^{-\frac {\sqrt {a^{2}}\, x}{2}}\right ) {\mathrm e}^{\frac {a^{2} x +2 b \,{\mathrm e}^{a x}}{2 a}} \\ \end{align*}
Verification of solutions
\[ y = \left (c_{1} {\mathrm e}^{\frac {\sqrt {a^{2}}\, x}{2}}+c_{2} {\mathrm e}^{-\frac {\sqrt {a^{2}}\, x}{2}}\right ) {\mathrm e}^{\frac {a^{2} x +2 b \,{\mathrm e}^{a x}}{2 a}} \] Verified OK.
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- 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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 39
dsolve(diff(y(x),x$2)-(a+2*b*exp(a*x))*diff(y(x),x)+b^2*exp(2*a*x)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{\frac {a^{2} x +2 \,{\mathrm e}^{a x} b}{2 a}} \left (c_{1} \sinh \left (\frac {a x}{2}\right )+c_{2} \cosh \left (\frac {a x}{2}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.079 (sec). Leaf size: 35
DSolve[y''[x]-(a+2*b*Exp[a*x])*y'[x]+b^2*Exp[2*a*x]*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {e^{\frac {b e^{a x}}{a}} \left (b c_2 e^{a x}+a c_1\right )}{a} \]