Answer to Question #222707 in Differential Equations for kim

Question

x²y''-4xy'+4y=4x²-6x³

Answer 1

Solution. Denote by D the linear differential operator $x^2\frac{d^2}{dx^2}-4x\frac{d}{dx}+4$.

For all $k\in\mathbb{Z}$ $D(x^k)=k(k-1)x^k-4kx^k+4x^k=(k^2-5k+4)x^k=(k-1)(k-4)x^k$

One can conclude, that the functions $x^k$ are eigenfunctions for this operator and that the linear independent functions $x$ and $x^4$ belong to the kernel (null space) of $D$.

Since $D$ is an ordinary differential operator of the 2nd order, its kernel is 2-dimensional. Therefore, $\ker D={\rm span}\{x, x^4\}=\{Ax+Bx^4| A,B\in\mathbb{R}\}$.

It remains to find any partial solution to form the general solution of the given ODE.

One can notice, that the right-hand function in this ODE lies in the span of eigenfunctions $x^2$ and $x^3$:

$D(c_1x^2+c_2x^3)=3c_1x^2-2c_2x^3=4x^2-6x^3$. Therefore, $c_1=4/3$, $c_2=3$, the function $\frac{4}{3}x^2+3x^3$ is a partial solution of ODE, and hence, the general solution is

$y(x)=\frac{4}{3}x^2+3x^3+Ax+Bx^4$

Answer. $y(x)=\frac{4}{3}x^2+3x^3+Ax+Bx^4$