Answer to Question #317306 in Differential Equations for Sharlene

Question

Solve

(D^2 + 1) y = x Sin x

Answer 1

$\text{Solve }\frac{d^{2} y(x)}{d x^{2}}+y(x)=x \sin (x)\\[2mm] \text{The general solution will be the sum of the complementary solution and particular solution.}\\ \text{Find the complementary solution by solving } \frac{d^{2} y(x)}{d x^{2}}+y(x)=0 :\\[2mm] \text{Assume a solution will be proportional to } e^{\lambda x} \text{ for some constant } \lambda.\\[3mm] \text{Substitute } y(x)=e^{\lambda x} \text{ into the differential equation:}\\ \frac{d^{2}}{d x^{2}}\left(e^{x x}\right)+e^{x x}=0\\[2.5mm] \text{Substitute } \frac{d^{2}}{d x^{2}}\left(e^{\lambda x}\right)=\lambda^{2} e^{\lambda x}\\[2mm] \lambda^{2} e^{\lambda x}+e^{\lambda x}=0\\[3mm] \text{Factor out }e^{\lambda x} :\\ \left(\lambda^{2}+1\right) e^{\lambda x}=0\\[2mm] \text{Since } e^{\lambda x} \neq 0 \text{ for any finite } \lambda, \text{ the zeros must come from the polynomial: } x^{2}+1=0\\$

$\text{ Solve for } \lambda :\\[1.5mm] \lambda=i \text { or } \lambda=-i\\[2mm] \text{The roots } \lambda=\pm i \text{ give } y_{1}(x)=c_{1} e^{i x}, y_{2}(x)=c_{2} e^{-i x} \text{ as solutions, where } c_{1} \text{ and } c_{2} \text{ are arbitrary constants.}\\[2mm] \text{The general solution is the sum of the above solutions:}\\[2mm] y(x)=y_{1}(x)+y_{2}(x)=c_{1} e^{i x}+c_{2} e^{-i x}\\[4mm] \text{Apply Euler's identity } e^{\alpha+i \beta}=e^{\alpha} \cos (\beta)+i e^{\alpha} \sin (\beta) :\\[2mm] y(x)=c_{1}(\cos (x)+i \sin (x))+c_{2}(\cos (x)-i \sin (x))\\[2mm] \text{Regroup terms:}\\[2mm] y(x)=\left(c_{1}+c_{2}\right) \cos (x)+i\left(c_{1}-c_{2}\right) \sin (x)\\[2mm] \text{Redefine } c_{1}+c_{2} \text{ as } c_{1} \text{ and } i\left(c_{1}-c_{2}\right) \text{ as } c_{2}, \text{ since these are arbitrary constants: } y(x)=c_{1} \cos (x)+c_{2} \sin (x)\\ \text{Determine the particular solution to } \frac{d^{2} y(x)}{d x^{2}}+y(x)=x \sin (x) \text{ by the method of undetermined coefficients:}\\[3mm] \text{The particular solution to } \frac{d^{2} y(x)}{d x^{2}}+y(x)=x \sin (x) \text{ is of the form:}\\[2mm] y_{p}(x)=x\left(a_{1} \cos (x)+a_{2} x \cos (x)+a_{3} \sin (x)+a_{4} x \sin (x)\right),\\ \text{ where } a_{1} \cos (x)+a_{2} x \cos (x)+a_{3} \sin (x) + a_{4} x \sin (x)\\ \text{ was multiplied by } x \text{ to account for } \sin (x) \text{ in the complementary solution. }$

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

$\text{Equate the coefficients of } x \sin (x) \text{ on both sides of the equation:}\\ -4 a_{2}=1[4mm] \text{Solve the system:}\\ \begin{aligned} &a_{1}=0 \\ &a_{2}=-\frac{1}{4} \\ &a_{3}=\frac{1}{4} \\ &a_{4}=0 \end{aligned} \text{Substitute } a_{1}, a_{2}, a_{3}, \text{ and }a_{4} \text{ into } y_{p}(x)=x \cos (x) a_{1}+x^{2} \cos (x) a_{2}+x \sin (x) a_{3}+ x^{2} \sin (x) a_{4}\\[2mm] y_{p}(x)=-\frac{1}{4} x^{2} \cos (x)+\frac{1}{4} x \sin (x)\\[4mm] \text{The general solution is:}\\[2mm] y(x)=y_{\mathrm{c}}(x)+y_{p}(x)=-\frac{1}{4} x^{2} \cos (x)+\frac{1}{4} x \sin (x)+c_{1} \cos (x)+c_{2} \sin (x)$