2.2.26 Problem 26

Internal problem ID [9149]
Book : Second order enumerated odes
Section : section 2
Problem number : 26
Date solved : Sunday, March 30, 2025 at 02:23:57 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Solved as second order ode using change of variable on x method 2

Time used: 1.367 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+(1-\frac{1}{x})y^{\prime }+4x^{2}y{\mathrm{e}}^{-2x}& =4(x^3+x^2){\mathrm{e}}^{-3x}\end{array}$$

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 $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=0$, and $y_p$ is a particular solution to the non-homogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=f(x)$. $y_h$ is the solution to

$$y^{\prime \prime }+\frac{(x-1)y^{\prime }}{x}+4x^{2}y{\mathrm{e}}^{-2x}=0$$

In normal form the ode

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+\frac{(x-1)y^{\prime }}{x}+4x^{2}y{\mathrm{e}}^{-2x}& =0\end{array}$$

Becomes

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(x\right)y^{\prime }+q\left(x\right)y& =0\end{array}$$

Where

$$\begin{array}{rl}p\left(x\right)& =\frac{x-1}{x}\\ q\left(x\right)& =4x^2{\mathrm{e}}^{-2x}\end{array}$$

Applying change of variables $\tau =g\left(x\right)$ to (2) gives

$$\begin{array}{}\text{(3)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+p_1\left(\frac{d}{d\tau }y\left(\tau \right)\right)+q_{1}y\left(\tau \right)& =0\end{array}$$

Where $\tau$ is the new independent variable, and

$$\begin{array}{}\text{(4)}& p_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\text{(5)}& q_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\end{array}$$

Let $p_1=0$. Eq (4) simplifies to

$$\begin{array}{rl}{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)& =0\end{array}$$

This ode is solved resulting in

$$\begin{array}{rl}\tau & =\int {\mathrm{e}}^{-\int p\left(x\right)dx}dx\\ & =\int {\mathrm{e}}^{-\int \frac{x-1}{x}dx}dx\\ & =\int e^{-x+\ln \left(x\right)}dx\\ & =\int x{\mathrm{e}}^{-x}dx\\ \text{(6)}& & =-(x+1){\mathrm{e}}^{-x}\end{array}$$

Using (6) to evaluate $q_1$ from (5) gives

$$\begin{array}{rl}{q}_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\\ & =\frac{4x^2{\mathrm{e}}^{-2x}}{x^2{\mathrm{e}}^{-2x}}\\ \text{(7)}& & =4\end{array}$$

Substituting the above in (3) and noting that now $p_1=0$ results in

$$\begin{array}{rl}\frac{d^2}{d{\tau }^2}y\left(\tau \right)+q_{1}y\left(\tau \right)& =0\\ \frac{d^2}{d{\tau }^2}y\left(\tau \right)+4y\left(\tau \right)& =0\end{array}$$

The above ode is now solved for $y\left(\tau \right)$.This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{y}^{″}(\tau )+B{y}^{\prime }(\tau )+Cy(\tau )=0$$

Where in the above $A=1,B=0,C=4$. Let the solution be $y=e^{\lambda \tau }$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{\tau \lambda }+4{\mathrm{e}}^{\tau \lambda }=0\end{array}$$

Since exponential function is never zero, then dividing Eq(2) throughout by $e^{\lambda \tau }$ gives

$$\begin{array}{}\text{(2)}& {\lambda }^2+4=0\end{array}$$

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=0,C=4$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{0}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{0^2-(4)\left(1\right)\left(4\right)}\\ & =\pm 2i\end{array}$$

Hence

$$\begin{array}{rl}{\lambda }_1& =+2i\\ {\lambda }_2& =-2i\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =2i\\ {\lambda }_2& =-2i\end{array}$$

Since roots are complex conjugate of each others, then let the roots be

$${\lambda }_{1,2}=\alpha \pm i\beta$$

Where $\alpha =0$ and $\beta =2$. Therefore the final solution, when using Euler relation, can be written as

$$y=e^{\alpha \tau }(c_1\cos (\beta \tau )+c_2\sin (\beta \tau ))$$

Which becomes

$$y=e^0(c_1\cos \left(2\tau \right)+c_2\sin \left(2\tau \right))$$

Or

$$y=c_1\cos \left(2\tau \right)+c_2\sin \left(2\tau \right)$$

Will add steps showing solving for IC soon.

The above solution is now transformed back to $y\left(x\right)$ using (6) which results in

$$y\left(x\right)=c_1\cos \left(2(x+1){\mathrm{e}}^{-x}\right)-c_2\sin \left(2(x+1){\mathrm{e}}^{-x}\right)$$

Therefore the homogeneous solution $y_h$ is

$$y_h=c_1\cos \left(2(x+1){\mathrm{e}}^{-x}\right)-c_2\sin \left(2(x+1){\mathrm{e}}^{-x}\right)$$

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

$$\begin{array}{}\text{(1)}& y_p(x)=u_1{y}_1+u_2{y}_2\end{array}$$

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

$$\begin{array}{rl}{y}_1& =-4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\cos {\left({\mathrm{e}}^{-x}\right)}^2+2\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)+2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)\\ y_2& =4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\cos {\left({\mathrm{e}}^{-x}\right)}^2-2\cos {\left(x{\mathrm{e}}^{-x}\right)}^2-2\cos {\left({\mathrm{e}}^{-x}\right)}^2+1-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)\end{array}$$

In the Variation of parameters $u_1,u_2$ are found using

$$\begin{array}{}\text{(2)}& u_1& =-\int \frac{y_{2}f(x)}{aW(x)}\text{(3)}& u_2& =\int \frac{y_{1}f(x)}{aW(x)}\end{array}$$

Where $W(x)$ is the Wronskian and $a$ is the coefficient in front of $y^{″}$ in the given ODE. The Wronskian is given by $W=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right|$. Hence

$$W=\left|\begin{array}{cc}-4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\cos {\left({\mathrm{e}}^{-x}\right)}^2+2\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)+2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)& 4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\cos {\left({\mathrm{e}}^{-x}\right)}^2-2\cos {\left(x{\mathrm{e}}^{-x}\right)}^2-2\cos {\left({\mathrm{e}}^{-x}\right)}^2+1-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)\\ \frac{d}{dx}(-4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\cos {\left({\mathrm{e}}^{-x}\right)}^2+2\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)+2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right))& \frac{d}{dx}(4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\cos {\left({\mathrm{e}}^{-x}\right)}^2-2\cos {\left(x{\mathrm{e}}^{-x}\right)}^2-2\cos {\left({\mathrm{e}}^{-x}\right)}^2+1-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right))\end{array}\right|$$

Which gives

$$W=\left|\begin{array}{cc}-4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\cos {\left({\mathrm{e}}^{-x}\right)}^2+2\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)+2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)& 4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\cos {\left({\mathrm{e}}^{-x}\right)}^2-2\cos {\left(x{\mathrm{e}}^{-x}\right)}^2-2\cos {\left({\mathrm{e}}^{-x}\right)}^2+1-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)\\ 8\cos \left(x{\mathrm{e}}^{-x}\right)\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\sin \left(x{\mathrm{e}}^{-x}\right)+4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2{\mathrm{e}}^{-x}\cos {\left({\mathrm{e}}^{-x}\right)}^2-4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\sin {\left({\mathrm{e}}^{-x}\right)}^2{\mathrm{e}}^{-x}-4(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\cos {\left({\mathrm{e}}^{-x}\right)}^2+4\sin {\left(x{\mathrm{e}}^{-x}\right)}^2(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\cos {\left({\mathrm{e}}^{-x}\right)}^2-8\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right){\mathrm{e}}^{-x}\sin \left({\mathrm{e}}^{-x}\right)+2(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\cos {\left(x{\mathrm{e}}^{-x}\right)}^2-2\sin {\left(x{\mathrm{e}}^{-x}\right)}^2(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})-2{\mathrm{e}}^{-x}\cos {\left({\mathrm{e}}^{-x}\right)}^2+2\sin {\left({\mathrm{e}}^{-x}\right)}^2{\mathrm{e}}^{-x}& -8\cos \left(x{\mathrm{e}}^{-x}\right)\cos {\left({\mathrm{e}}^{-x}\right)}^2(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\sin \left(x{\mathrm{e}}^{-x}\right)+8\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\cos \left({\mathrm{e}}^{-x}\right){\mathrm{e}}^{-x}\sin \left({\mathrm{e}}^{-x}\right)+4\cos \left(x{\mathrm{e}}^{-x}\right)(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\sin \left(x{\mathrm{e}}^{-x}\right)-4\cos \left({\mathrm{e}}^{-x}\right){\mathrm{e}}^{-x}\sin \left({\mathrm{e}}^{-x}\right)-4(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)+4\sin {\left(x{\mathrm{e}}^{-x}\right)}^2(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)+4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right){\mathrm{e}}^{-x}\cos {\left({\mathrm{e}}^{-x}\right)}^2-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\sin {\left({\mathrm{e}}^{-x}\right)}^2{\mathrm{e}}^{-x}\end{array}\right|$$

Therefore

$$W=(-4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\cos {\left({\mathrm{e}}^{-x}\right)}^2+2\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)+2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right))(-8\cos \left(x{\mathrm{e}}^{-x}\right)\cos {\left({\mathrm{e}}^{-x}\right)}^2(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\sin \left(x{\mathrm{e}}^{-x}\right)+8\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\cos \left({\mathrm{e}}^{-x}\right){\mathrm{e}}^{-x}\sin \left({\mathrm{e}}^{-x}\right)+4\cos \left(x{\mathrm{e}}^{-x}\right)(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\sin \left(x{\mathrm{e}}^{-x}\right)-4\cos \left({\mathrm{e}}^{-x}\right){\mathrm{e}}^{-x}\sin \left({\mathrm{e}}^{-x}\right)-4(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)+4\sin {\left(x{\mathrm{e}}^{-x}\right)}^2(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)+4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right){\mathrm{e}}^{-x}\cos {\left({\mathrm{e}}^{-x}\right)}^2-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\sin {\left({\mathrm{e}}^{-x}\right)}^2{\mathrm{e}}^{-x})-(4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\cos {\left({\mathrm{e}}^{-x}\right)}^2-2\cos {\left(x{\mathrm{e}}^{-x}\right)}^2-2\cos {\left({\mathrm{e}}^{-x}\right)}^2+1-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right))(8\cos \left(x{\mathrm{e}}^{-x}\right)\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\sin \left(x{\mathrm{e}}^{-x}\right)+4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2{\mathrm{e}}^{-x}\cos {\left({\mathrm{e}}^{-x}\right)}^2-4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\sin {\left({\mathrm{e}}^{-x}\right)}^2{\mathrm{e}}^{-x}-4(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\cos {\left({\mathrm{e}}^{-x}\right)}^2+4\sin {\left(x{\mathrm{e}}^{-x}\right)}^2(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\cos {\left({\mathrm{e}}^{-x}\right)}^2-8\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right){\mathrm{e}}^{-x}\sin \left({\mathrm{e}}^{-x}\right)+2(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})\cos {\left(x{\mathrm{e}}^{-x}\right)}^2-2\sin {\left(x{\mathrm{e}}^{-x}\right)}^2(-x{\mathrm{e}}^{-x}+{\mathrm{e}}^{-x})-2{\mathrm{e}}^{-x}\cos {\left({\mathrm{e}}^{-x}\right)}^2+2\sin {\left({\mathrm{e}}^{-x}\right)}^2{\mathrm{e}}^{-x})$$

Which simplifies to

$$W=\text{Expression too large to display}$$

Which simplifies to

$$W=-2x{\mathrm{e}}^{-x}$$

Therefore Eq. (2) becomes

$$u_1=-\int \frac{4(4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\cos {\left({\mathrm{e}}^{-x}\right)}^2-2\cos {\left(x{\mathrm{e}}^{-x}\right)}^2-2\cos {\left({\mathrm{e}}^{-x}\right)}^2+1-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)){\mathrm{e}}^{-3x}{x}^2(x+1)}{-2x{\mathrm{e}}^{-x}}dx$$

Which simplifies to

$$u_1=-\int 2x(x+1){\mathrm{e}}^{-2x}(\sin \left(2x{\mathrm{e}}^{-x}\right)\sin \left(2{\mathrm{e}}^{-x}\right)-\cos \left(2{\mathrm{e}}^{-x}\right)\cos \left(2x{\mathrm{e}}^{-x}\right))dx$$

Hence

$$u_1=i\sinh (2i{\mathrm{e}}^{-x}+2ix{\mathrm{e}}^{-x}){\mathrm{e}}^{-x}x-\frac{\cosh (2i{\mathrm{e}}^{-x}+2ix{\mathrm{e}}^{-x})}{2}+i\sinh (2i{\mathrm{e}}^{-x}+2ix{\mathrm{e}}^{-x}){\mathrm{e}}^{-x}$$

And Eq. (3) becomes

$$u_2=\int \frac{4(-4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\cos {\left({\mathrm{e}}^{-x}\right)}^2+2\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)+2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)){\mathrm{e}}^{-3x}{x}^2(x+1)}{-2x{\mathrm{e}}^{-x}}dx$$

Which simplifies to

$$u_2=\int 2x(x+1){\mathrm{e}}^{-2x}(\sin \left(2{\mathrm{e}}^{-x}\right)\cos \left(2x{\mathrm{e}}^{-x}\right)+\cos \left(2{\mathrm{e}}^{-x}\right)\sin \left(2x{\mathrm{e}}^{-x}\right))dx$$

Hence

$$u_2=2(\frac{x{\mathrm{e}}^{-x}}{2}+\frac{{\mathrm{e}}^{-x}}{2})\cosh (2i{\mathrm{e}}^{-x}+2ix{\mathrm{e}}^{-x})+\frac{i\sinh (2i{\mathrm{e}}^{-x}+2ix{\mathrm{e}}^{-x})}{2}$$

Therefore the particular solution, from equation (1) is

$$y_p(x)=(i\sinh (2i{\mathrm{e}}^{-x}+2ix{\mathrm{e}}^{-x}){\mathrm{e}}^{-x}x-\frac{\cosh (2i{\mathrm{e}}^{-x}+2ix{\mathrm{e}}^{-x})}{2}+i\sinh (2i{\mathrm{e}}^{-x}+2ix{\mathrm{e}}^{-x}){\mathrm{e}}^{-x})(-4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right)-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\cos {\left({\mathrm{e}}^{-x}\right)}^2+2\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)+2\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right))+(4\cos {\left(x{\mathrm{e}}^{-x}\right)}^2\cos {\left({\mathrm{e}}^{-x}\right)}^2-2\cos {\left(x{\mathrm{e}}^{-x}\right)}^2-2\cos {\left({\mathrm{e}}^{-x}\right)}^2+1-4\sin \left(x{\mathrm{e}}^{-x}\right)\cos \left(x{\mathrm{e}}^{-x}\right)\sin \left({\mathrm{e}}^{-x}\right)\cos \left({\mathrm{e}}^{-x}\right))(2(\frac{x{\mathrm{e}}^{-x}}{2}+\frac{{\mathrm{e}}^{-x}}{2})\cosh (2i{\mathrm{e}}^{-x}+2ix{\mathrm{e}}^{-x})+\frac{i\sinh (2i{\mathrm{e}}^{-x}+2ix{\mathrm{e}}^{-x})}{2})$$

Which simplifies to

$$y_p(x)=(x+1){\mathrm{e}}^{-x}$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =(c_1\cos \left(2(x+1){\mathrm{e}}^{-x}\right)-c_2\sin \left(2(x+1){\mathrm{e}}^{-x}\right))+\left((x+1){\mathrm{e}}^{-x}\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y\left(x\right)& =(x+1){\mathrm{e}}^{-x}+c_1\cos \left(2(x+1){\mathrm{e}}^{-x}\right)-c_2\sin \left(2(x+1){\mathrm{e}}^{-x}\right)\end{array}$$

Solved as second order ode using change of variable on x method 1

Time used: 0.728 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }+(1-\frac{1}{x})y^{\prime }+4x^{2}y{\mathrm{e}}^{-2x}& =4(x^3+x^2){\mathrm{e}}^{-3x}\end{array}$$

This is second order non-homogeneous ODE. In standard form the ODE is

$$A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=f(x)$$

Where $A=1,B=\frac{x-1}{x},C=4x^2{\mathrm{e}}^{-2x},f(x)=4{\mathrm{e}}^{-3x}{x}^2(x+1)$. Let the solution be

$$y=y_h+y_p$$

Where $y_h$ is the solution to the homogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=0$, and $y_p$ is a particular solution to the non-homogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=f(x)$. Solving for $y_h$ from

$$y^{\prime \prime }+\frac{(x-1)y^{\prime }}{x}+4x^{2}y{\mathrm{e}}^{-2x}=0$$

In normal form the ode

$$\begin{array}{}\text{(1)}& y^{\prime \prime }+\frac{(x-1)y^{\prime }}{x}+4x^{2}y{\mathrm{e}}^{-2x}& =0\end{array}$$

Becomes

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(x\right)y^{\prime }+q\left(x\right)y& =0\end{array}$$

Where

$$\begin{array}{rl}p\left(x\right)& =\frac{x-1}{x}\\ q\left(x\right)& =4x^2{\mathrm{e}}^{-2x}\end{array}$$

Applying change of variables $\tau =g\left(x\right)$ to (2) results

$$\begin{array}{}\text{(3)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+p_1\left(\frac{d}{d\tau }y\left(\tau \right)\right)+q_{1}y\left(\tau \right)& =0\end{array}$$

Where $\tau$ is the new independent variable, and

$$\begin{array}{}\text{(4)}& p_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\text{(5)}& q_1\left(\tau \right)& =\frac{q\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\end{array}$$

Let $q_1=c^2$ where $c$ is some constant. Therefore from (5)

$$\begin{array}{rl}{\tau }^{\prime }& =\frac{1}{c}\sqrt{q}\\ \text{(6)}& & =\frac{2\sqrt{x^2{\mathrm{e}}^{-2x}}}{c}{\tau }^{″}& =\frac{2x{\mathrm{e}}^{-2x}-2x^2{\mathrm{e}}^{-2x}}{c\sqrt{x^2{\mathrm{e}}^{-2x}}}\end{array}$$

Substituting the above into (4) results in

$$\begin{array}{rl}{p}_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(x\right)+p\left(x\right){\tau }^{\prime }\left(x\right)}{{{\tau }^{\prime }\left(x\right)}^2}\\ & =\frac{\frac{2x{\mathrm{e}}^{-2x}-2x^2{\mathrm{e}}^{-2x}}{c\sqrt{x^2{\mathrm{e}}^{-2x}}}+\frac{x-1}{x}\frac{2\sqrt{x^2{\mathrm{e}}^{-2x}}}{c}}{{\left(\frac{2\sqrt{x^2{\mathrm{e}}^{-2x}}}{c}\right)}^2}\\ & =0\end{array}$$

Therefore ode (3) now becomes

$$\begin{array}{rl}y{\left(\tau \right)}^{″}+p_{1}y{\left(\tau \right)}^{\prime }+q_{1}y\left(\tau \right)& =0\\ \text{(7)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+c^{2}y\left(\tau \right)& =0\end{array}$$

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

$$\begin{array}{rl}y\left(\tau \right)& =c_1\cos \left(c\tau \right)+c_2\sin \left(c\tau \right)\end{array}$$

Now from (6)

$$\begin{array}{rl}\tau & =\int \frac{1}{c}\sqrt{q}dx\\ & =\frac{\int 2\sqrt{x^2{\mathrm{e}}^{-2x}}dx}{c}\\ & =-\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{cx}\end{array}$$

Substituting the above into the solution obtained gives

$$y=c_1\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)-c_2\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)$$

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

$$\begin{array}{}\text{(1)}& y_p(x)=u_1{y}_1+u_2{y}_2\end{array}$$

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

$$\begin{array}{rl}{y}_1& =\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)\\ y_2& =-\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)\end{array}$$

In the Variation of parameters $u_1,u_2$ are found using

$$\begin{array}{}\text{(2)}& u_1& =-\int \frac{y_{2}f(x)}{aW(x)}\text{(3)}& u_2& =\int \frac{y_{1}f(x)}{aW(x)}\end{array}$$

Where $W(x)$ is the Wronskian and $a$ is the coefficient in front of $y^{″}$ in the given ODE. The Wronskian is given by $W=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right|$. Hence

$$W=\left|\begin{array}{cc}\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)& -\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)\\ \frac{d}{dx}(\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right))& \frac{d}{dx}(-\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right))\end{array}\right|$$

Which gives

$$W=\left|\begin{array}{cc}\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)& -\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)\\ -(\frac{2\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}+\frac{(x+1)(2x{\mathrm{e}}^{-2x}-2x^2{\mathrm{e}}^{-2x})}{\sqrt{x^2{\mathrm{e}}^{-2x}}x}-\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x^2})\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)& -(\frac{2\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}+\frac{(x+1)(2x{\mathrm{e}}^{-2x}-2x^2{\mathrm{e}}^{-2x})}{\sqrt{x^2{\mathrm{e}}^{-2x}}x}-\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x^2})\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)\end{array}\right|$$

Therefore

$$W=(\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right))(-(\frac{2\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}+\frac{(x+1)(2x{\mathrm{e}}^{-2x}-2x^2{\mathrm{e}}^{-2x})}{\sqrt{x^2{\mathrm{e}}^{-2x}}x}-\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x^2})\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right))-(-\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right))(-(\frac{2\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}+\frac{(x+1)(2x{\mathrm{e}}^{-2x}-2x^2{\mathrm{e}}^{-2x})}{\sqrt{x^2{\mathrm{e}}^{-2x}}x}-\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x^2})\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right))$$

Which simplifies to

$$W=\frac{2x^2{\mathrm{e}}^{-2x}({\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)}^2+{\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)}^2)}{\sqrt{x^2{\mathrm{e}}^{-2x}}}$$

Which simplifies to

$$W=\frac{2x^2{\mathrm{e}}^{-2x}}{\sqrt{x^2{\mathrm{e}}^{-2x}}}$$

Therefore Eq. (2) becomes

$$u_1=-\int \frac{-4\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right){\mathrm{e}}^{-3x}{x}^2(x+1)}{\frac{2x^2{\mathrm{e}}^{-2x}}{\sqrt{x^2{\mathrm{e}}^{-2x}}}}dx$$

Which simplifies to

$$u_1=-\int -2(x+1){\mathrm{e}}^{-x}\sqrt{x^2{\mathrm{e}}^{-2x}}\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)dx$$

Hence

$$u_1=-{\int }_0^x-2(\alpha +1){\mathrm{e}}^{-\alpha }\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}\sin \left(\frac{2(\alpha +1)\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}}{\alpha }\right)d\alpha$$

And Eq. (3) becomes

$$u_2=\int \frac{4\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right){\mathrm{e}}^{-3x}{x}^2(x+1)}{\frac{2x^2{\mathrm{e}}^{-2x}}{\sqrt{x^2{\mathrm{e}}^{-2x}}}}dx$$

Which simplifies to

$$u_2=\int 2(x+1){\mathrm{e}}^{-x}\sqrt{x^2{\mathrm{e}}^{-2x}}\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)dx$$

Hence

$$u_2={\int }_0^{x}2(\alpha +1){\mathrm{e}}^{-\alpha }\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}\cos \left(\frac{2(\alpha +1)\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}}{\alpha }\right)d\alpha$$

Therefore the particular solution, from equation (1) is

$$y_p(x)=-{\int }_0^x-2(\alpha +1){\mathrm{e}}^{-\alpha }\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}\sin \left(\frac{2(\alpha +1)\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}}{\alpha }\right)d\alpha \cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)-\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right){\int }_0^{x}2(\alpha +1){\mathrm{e}}^{-\alpha }\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}\cos \left(\frac{2(\alpha +1)\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}}{\alpha }\right)d\alpha$$

Which simplifies to

$$y_p(x)=2{\int }_0^x(\alpha +1){\mathrm{e}}^{-\alpha }\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}\sin \left(\frac{2(\alpha +1)\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}}{\alpha }\right)d\alpha \cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)-2\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right){\int }_0^x(\alpha +1){\mathrm{e}}^{-\alpha }\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}\cos \left(\frac{2(\alpha +1)\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}}{\alpha }\right)d\alpha$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =(c_1\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)-c_2\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right))+(2{\int }_0^x(\alpha +1){\mathrm{e}}^{-\alpha }\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}\sin \left(\frac{2(\alpha +1)\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}}{\alpha }\right)d\alpha \cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)-2\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right){\int }_0^x(\alpha +1){\mathrm{e}}^{-\alpha }\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}\cos \left(\frac{2(\alpha +1)\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}}{\alpha }\right)d\alpha )\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =2{\int }_0^x(\alpha +1){\mathrm{e}}^{-\alpha }\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}\sin \left(\frac{2(\alpha +1)\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}}{\alpha }\right)d\alpha \cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)-2\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right){\int }_0^x(\alpha +1){\mathrm{e}}^{-\alpha }\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}\cos \left(\frac{2(\alpha +1)\sqrt{{\alpha }^2{\mathrm{e}}^{-2\alpha }}}{\alpha }\right)d\alpha +c_1\cos \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)-c_2\sin \left(\frac{2(x+1)\sqrt{x^2{\mathrm{e}}^{-2x}}}{x}\right)\end{array}$$

✓ Maple. Time used: 0.009 (sec). Leaf size: 37

ode:=diff(diff(y(x),x),x)+(1-1/x)*diff(y(x),x)+4*x^2*y(x)*exp(-2*x) = 4*(x^3+x^2)*exp(-3*x); 
dsolve(ode,y(x), singsol=all);
 

$$y=\sin \left(2(x+1){\mathrm{e}}^{-x}\right)c_2+\cos \left(2(x+1){\mathrm{e}}^{-x}\right)c_1+(x+1){\mathrm{e}}^{-x}$$

Maple trace

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)] 
   <- linear_1 successful 
<- solving first the homogeneous part of the ODE successful
 

✓ Mathematica. Time used: 4.276 (sec). Leaf size: 142

ode=D[y[x],{x,2}]+(1-1/x)*D[y[x],x]+4*x^2*y[x]*Exp[-2*x]==4*(x^2+x^3)*Exp[-3*x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \cos \left(2e^{-x}(\log \left(e^x\right)+1)\right){\int }_1^{e^x}\frac{2\log (K[1])(\log (K[1])+1)\sin \left(\frac{2(\log (K[1])+1)}{K[1]}\right)}{K[1{]}^3}dK[1]-\sin \left(2e^{-x}(\log \left(e^x\right)+1)\right){\int }_1^{e^x}\frac{2\cos \left(\frac{2(\log (K[2])+1)}{K[2]}\right)\log (K[2])(\log (K[2])+1)}{K[2{]}^3}dK[2]+c_1\cos \left(2e^{-x}(\log \left(e^x\right)+1)\right)-c_2\sin \left(2e^{-x}(\log \left(e^x\right)+1)\right)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(4*x**2*y(x)*exp(-2*x) + (1 - 1/x)*Derivative(y(x), x) - (4*x**3 + 4*x**2)*exp(-3*x) + Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE -x*(4*x**3 - 4*x**2*y(x)*exp(x) + 4*x**2 - exp(3*x)*Derivative(y(x), (x, 2)))*exp(-3*x)/(x - 1) + Derivative(y(x), x) cannot be solved by the factorable group method