Answer to Question #313850 in Differential Equations for sly

Question

y''-y=3x^2e^x

Answer 1

$\text{Solve } \frac{d^{2} y(x)}{d x^{2}}-y(x)=3 e^{x} x^{2}\\[2mm] \text{The general solution will be the sum of the complementary solution and particular solution.}\\ \text{Find the complementary solution by solving } \frac{d^{2} y(x)}{d x^{2}}-y(x)=0 :\\[2mm] \text{Assume a solution will be proportional to } e^{\lambda x} \text{ for some constant } \lambda.\\[2mm] \text{ Substitute } y(x)=e^{\lambda x} \text{ into the differential equation:}\\ \frac{d^{2}}{d x^{2}}\left(e^{\lambda x}\right)-e^{\lambda x}=0\\[2mm] \text{ Substitute } \frac{d^{2}}{d x^{2}}\left(e^{\lambda x}\right)=\lambda^{2} e^{\lambda x} :\\[2mm] x^{2} e^{x x}-e^{x x}=0\\[4mm] \text{ Factor out }e^{\lambda x} : \left(\lambda^{2}-1\right) e^{\lambda x}=0\\$

$\text{Since } e^{\lambda x} \neq 0 \text{ for any finite } \lambda, \text{ the zeros must come from the polynomial:}\\ \lambda^{2}-1=0\\[2mm] \text{ Factor:}\\[2mm] (\lambda-1)(\lambda+1)=0\\[2mm] \text{Solve for } \lambda :\\[2mm] \lambda=-1 \text { or } \lambda=1\\[4mm] \text{The root } \lambda=-1 \text{ gives } y_{1}(x)=c_{1} e^{-x} \text{ as a solution, where } c_{1} \text{ is an arbitrary constant.}\\[2mm] \text{The root } \lambda=1 gives y_{2}(x)=c_{2} e^{x} \text{ as a solution, where } c_{2} \text{ is an arbitrary constant. The general solution is the sum of the above solutions:}\\[2mm] y(x)=y_{1}(x)+y_{2}(x)=c_{1} e^{-x}+c_{2} e^{x}\\[4mm] \text{Determine the particular solution to } \frac{d^{2} y(x)}{d x^{2}}-y(x)=3 e^{x} x^{2} \text{ by the method of undetermined coefficients:}\\[2mm] \text{The particular solution to } \frac{d^{2} y(x)}{d x^{2}}-y(x)=3 e^{x} x^{2} \text{ is of the form:}\\[2mm] y_{p}(x)=x\left(a_{1} e^{x}+a_{2} e^{x} x+a_{3} e^{x} x^{2}\right), \text{ where }a_{1} e^{x}+a_{2} e^{x} x+a_{3} e^{x} x^{2}\\[2mm] \text{was multiplied by x to account for } e^{x} \text{ in the complementary solution.}$

$\text{Solve for the unknown constants } a_{1}, a_{2}, and a_{3} :\\[2mm] \text{ Compute }\frac{d^{2} y_{p}(x)}{d x^{2}} :\\ \begin{aligned} \frac{d^{2} y_{p}(x)}{d x^{2}} &=\frac{d^{2}}{d x^{2}}\left(a_{1} e^{x} x+a_{2} e^{x} x^{2}+a_{3} e^{x} x^{3}\right) \\ &=a_{1}\left(2 e^{x}+e^{x} x\right)+a_{2}\left(2 e^{x}+e^{x} x^{2}+4 e^{x} x\right)+a_{3}\left(e^{x} x^{3}+6 e^{x} x^{2}+6 e^{x} x\right) \end{aligned} \\[2mm] \text{Substitute the particular solution} y_{p}(x) \text{ into the differential equation:}\\ \begin{aligned} &\frac{d^{2} y_{p}(x)}{d x^{2}}-y_{p}(x)=3 e^{x} x^{2} \\ &a_{1}\left(2 e^{x}+e^{x} x\right)+a_{2}\left(2 e^{x}+e^{x} x^{2}+4 e^{x} x\right)+ \\ &\quad a_{3}\left(e^{x} x^{3}+6 e^{x} x^{2}+6 e^{x} x\right)-\left(a_{1} e^{x} x+a_{2} e^{x} x^{2}+a_{3} e^{x} x^{3}\right)=3 e^{x} x^{2} \end{aligned} \\[2mm] \text{ Simplify:}\\[1mm] \left(2 a_{1}+2 a_{2}\right) e^{x}+\left(4 a_{2}+6 a_{3}\right) e^{x} x+6 a_{3} e^{x} x^{2}=3 e^{x} x^{2}\\[2mm] \text{Equate the coefficients of } e^{x} \text{ on both sides of the equation:}\\[2mm] 2 a_{1}+2 a_{2}=0\\[2mm] \text{ Equate the coefficients of } e^{x} x \text{ on both sides of the equation:}\\[1mm] 4 a_{2}+6 a_{3}=0$

$\text{Equate the coefficients of } e^{x} x^{2} \text{ on both sides of the equation:}\\ 6 a_{3}=3\\ \text{ Solve the system:}\\ \begin{aligned} &a_{1}=\frac{3}{4} \\ &a_{2}=-\frac{3}{4} \\ &a_{3}=\frac{1}{2} \end{aligned} \\[2mm] \text{Substitute } a_{1}, a_{2}, \text{ and } a_{3} \text{ into } y_{p}(x)=e^{x} x a_{1}+e^{x} x^{2} a_{2}+e^{x} x^{3} a_{3} :\\ y_{p}(x)=\frac{e^{x} x^{3}}{2}-\frac{3 e^{x} x^{2}}{4}+\frac{3 e^{x} x}{4}\\[4mm] \text{ The general solution is:}\\[2mm] y(x)=y_{c}(x)+y_{p}(x)=\frac{e^{x} x^{3}}{2}-\frac{3 e^{x} x^{2}}{4}+\frac{3 e^{x} x}{4}+c_{1} e^{-x}+c_{2} e^{x}$