Answer to Question #222278 in Differential Equations for mandy

Question

use the transformation x=e^t to transform the equation x²d²y/dx²-2xdy/dx+2y=x³ to a linear equation with constant coefficients hence solve the equation

Answer 1

Let us solve the differential equation $x^2\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+2y=x^3.$ Let us use the transformation $x=e^t.$ Then $y'_x=y'_t\cdot t'_x=y'_t\cdot \frac{1}{x'_t}=y'_t\cdot e^{-t},$ $y''_{x^2}=(y''_{t^2}\cdot e^{-t}-y'_t\cdot e^{-t})e^{-t}=(y''_{t^2}-y'_t)e^{-2t}.$

It follows that we have the following equation $e^{2t}(y''_{t^2}-y'_t)e^{-2t}-2e^ty'_te^{-t}+2y=e^{3t},$ which is equivalent to the equation $y''_{t^2}-3y'_t+2y=e^{3t}.$ The characteristic equation $k^2-3k+2=0$ is equivalent to $(k-1)(k-2)=0,$ and hence has the roots $k_1=1$ and $k_2=2.$ The general solution of the differential equation $y''_{t^2}-3y'_t+2y=e^{3t}$ is $y(t)=C_1e^t+C_2e^{2t}+y_p(t),$ where $y_p(t)=ae^{3t}.$ It follows that $y_p'(t)=3ae^{3t},y_p''(t)=9ae^{3t}.$ We have that $9ae^{3t}-9ae^{3t}+2ae^{3t}=e^{3t}.$ Then $2a=1,$ and thus $a=\frac{1}{2}.$ Therefore, $y(t)=C_1e^t+C_2e^{2t}+\frac{1}{2}e^{3t}.$ We conclude that the general solutions of the differential equation $x^2\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+2y=x^3$ is $y(x)=C_1x+C_2x^2+\frac{1}{2}x^3.$