Question

Solve the following differential equation:

x^2 d^2y/dx^2 - x dy/dx + 4y = cos(logx) + x sin(logx)

Answer 1

$x^2 y''-xy'+4y=cos(logx)+xsin(logx)\\ x=e^t, y'=e^{-t}y'_t , y''=e^{-2t}(y''_t-y'_t)\\ e^{2t}e^{-2t}(y''_t-y'_t)-e^te^{-t}y'_t +4y=cost+e^tsint\\ y''_t-2y'_t+4y=cost+e^tsint$

$1) y''_t-2y'_t+4y=0\\ \lambda^2 -2\lambda+4=0\\ D=4-16=-12=12i^2\\ \lambda_1=\frac{2-2\sqrt3 i}{2}=1-\sqrt3 i\\ \lambda_2=\frac{2+2\sqrt3 i}{2}=1+\sqrt3 i\\ y_1=C_1e^tcos\sqrt3 t+C_2e^tsin\sqrt3 t$

$2) a) f_1(t)=cost\\ y_2=acost+bsint\\ y'_2=-asint+bcost\\ y''_2=-acost-bsint\\ -acost-bsint+2asint-2bcost+4acost+4bsint=cost\\ cost:-a-2b+4a=1\\ sint:-b+2a+4b=0\\ 3a-2b=1\\ 2a+3b=0\\ a=\frac{3}{13}, b=-\frac{2}{13}\\ y_2=\frac{3}{13}cost-\frac{2}{13}sint$

$b) f_2(t)=e^tsint\\ y_3=ae^tcost+be^tsint=e^t(acost+bsint)\\ y'_3=e^t(acost+bsint)+e^t(-asint+bcost)=\\ =e^t((a+b)cost+(b-a)sint)\\ y''_3=e^t((a+b)cost+(b-a)sint)+\\ +e^t(-(a+b)sint+(b-a)cost)=\\ =e^t(2bcost-2asint)$

$e^t(2bcost-2asint)-2e^t((a+b)cost+(b-a)sint)+\\ +4e^t(acost+bsint)=e^tsint\\ cost:2b-2(a+b)+4a=0\\ sint:-2a-2(b-a)+4b=1\\ a=0\\ b=\frac{1}{2}\\ y_3=\frac{1}{2}e^tsint\\$

$y=y_1+y_2+y_3=\\ =C_1e^tcos\sqrt3 t+C_2e^tsin\sqrt3 t+\\ +\frac{3}{13}cost-\frac{2}{13}sint+\frac{1}{2}e^tsint$

Return to $x$

$y=C_1xcos(\sqrt3 log|x|)+C_2xsin(\sqrt3 log|x|)+\\ +\frac{3}{13}cos(log|x|)-\frac{2}{13}sin(log|x|)+\frac{1}{2}xsin(log|x|)$