Answer to Question #314506 in Differential Equations for Jyo

Question

Solve the initial value problem:

𝑥^2𝑦" + 𝑥𝑦′ + 9𝑦 = 0; 𝑦(1) = 2, 𝑦′(1) = 0

Answer 1

$\text{Solve } x^{2} \frac{d^{2} y(x)}{d x^{2}}+x \frac{d y(x)}{d x}+9 y(x)=0,\text{ such that } y(1)=2 \text{ and } y^{\prime}(1)=0:\\[3mm] \text{Assume a solution to this Euler-Cauchy equation will be proportional to } x^{\lambda} \text{ for some constant } \lambda.\\[2mm] \text{ Substitute } y(x)=x^{\lambda} \text{ into the differential equation:}\\ x^{2} \frac{d^{2}}{d x^{2}}\left(x^{\lambda}\right)+x \frac{d}{d x}\left(x^{\lambda}\right)+9 x^{\lambda}=0\\[2mm] \text{ Substitute }\frac{d^{2}}{d x^{2}}\left(x^{\lambda}\right)=\lambda(\lambda-1) x^{\lambda-2} \text{ and }\frac{d}{d x}\left(x^{\lambda}\right)=\lambda x^{\lambda-1} x^{2} x^{\lambda}+9 x^{\lambda}=0\\[4mm] \text{ Factor out }x^{x} :\\[2mm] \left(\lambda^{2}+9\right) x^{\lambda}=0\\[3mm] \text{ Assuming } x \neq 0, \text{ the zeros must come from the polynomial:}\\ x^{2}+9=0\\[2mm] \text{ Solve for } \lambda :\\[1.5mm] \lambda=3 i \text { or } \lambda=-3 i$

$f\text{The roots } x=\pm 3 i \text{ give } y_{1}(x)=c_{1} x^{3 i}, y_{2}(x)=c_{2} x^{-3 i} \text{ as solutions, where } c_{1} \text{ and } c_{2} \text{ are arbitrary constants.}\\[2mm] \text{ The general solution is the sum of the above solutions:}\\[2mm] y(x)=y_{1}(x)+y_{2}(x)=c_{1} x^{3 i}+c_{2} x^{-3 i}\\[4mm] \text{ Using } x^{\lambda}=e^{\lambda \log (x)}, \text{ apply Euler's identity } e^{\alpha+i \beta}=e^{\alpha} \cos (\beta)+i e^{\alpha} \sin (\beta) :\\[2mm] y(x)=c_{1}(\cos (3 \log (x))+i \sin (3 \log (x)))+c_{2}(\cos (3 \log (x))-i \sin (3 \log (x)))\\[2mm] \text{ Regroup terms:}\\[1.5mm] y(x)=\left(c_{1}+c_{2}\right) \cos (3 \log (x))+i\left(c_{1}-c_{2}\right) \sin (3 \log (x))\\[4mm] \text{ Redefine } c_{1}+c_{2} \text{ as } c_{1} \text{ and } i\left(c_{1}-c_{2}\right) \text{ as } c_{2}, \text{ since these are arbitrary constants:}\\[2mm] y(x)=c_{1} \cos (3 \log (x))+c_{2} \sin (3 \log (x))\\[2mm] \text{ Solve for the unknown constants using the initial conditions:}\\[2mm] \text{ Compute }\frac{d y(x)}{d x} :\\[2mm] \frac{d y(x)}{d x}=\frac{d}{d x}\left(c_{1} \cos (3 \log (x))+c_{2} \sin (3 \log (x))\right)\\[1.5mm] =-\frac{3 c_{1} \sin (3 \log (x))}{x}+\frac{\left.3 c_{2} \cos 3 \log (x)\right)}{x}\\[2mm] \text{ Substitute } y(1)=2 \text{ into } y(x)=\cos (3 \log (x)) c_{1}+\sin (3 \log (x)) c_{2}:\\[2mm] c_{1}=2$

$\text{ Substitute } y^{\prime}(1)=0 \text{ into } \frac{d y(x)}{d x}=-\frac{3 \sin (3 \log (x)) c_{1}}{x}+\frac{3 \cos (3 \log (x)) c_{2}}{x} : 3 c_{2}=0\\[2mm] \text{ Solve the system:}\\ \begin{aligned} &c_{1}=2 \\ &c_{2}=0 \end{aligned} \\[2mm] \text{ Substitute } c_{1}=2 \text{ and } c_{2}=0 \text{ into } y(x)=\cos (3 \log (x)) c_{1}+\sin (3 \log (x)) c_{2} :\\[4mm] \text{ Answer:}\\[2mm] y(x)=2 \cos (3 \log (x))$