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Could someone help me solve the following equation?

$$x^2y'' - 5xy' - 7y = x^4$$

I know this is an Euler-Cauchy Equation and I have found the solution of the homogenous part to be $$y(x) = Ax^7 + Bx^{-1}$$

I just don't know how to deal with the inhomogenous part of the equation. (I know how to use both the method of undetermined coefficients and the method of variation of paramters if that helps)

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Below is an approach using Variation of Parameters:


You found the complementary solution correctly. Now, let's list the basis solutions: $\frac{1}{x}$ and $x^{7}$. Hence, the Wronskian is: $$W(x)=\begin{vmatrix} \frac{1}{x} & x^7 \\ -\frac{1}{x^2} & 7x^6 \end{vmatrix}=8x^5$$

Important: To deal with the inhomogeneous part, we should put the ODE in the form: $$y''+p(x)y'+q(x)=r(x)$$

Hence, we must divide the ODE by $x^2$: $$y''-\frac{5}{x}y'-\frac{7}{x^2}\cdot y=x^2$$ Hence $r(x)=x^2$, not $x^4$.


Thus, we have: $$y_p(x)=v_1(x)\cdot \frac{1}{x}+v_2(x)\cdot x^7$$ Where: $$v_1(x)=-\int \frac{r(x)\cdot x^7}{W(x)}~dx=-\int \frac{x^4}{8}~dx$$ $$v_2(x)=\int \frac{r(x)\cdot \frac{1}{x}}{W(x)}~dx=\int \frac{1}{8x^4}~dx$$ From here, it should be easy to solve.

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You can set $x=e^u$ if you find it easier to work with equations with constant coefficients. But let's try it without.

The interesting thing about operators of the form $x^n (d/dx)^n$ is when you feed them a power of $x$, you get the same power of $x$ out, multiplied by a constant (they are eigenfunctions, just as $e^{kx}$ is an eigenfunction of $(d/dx)^n$). Therefore we should try $y=ax^4$ as a particular solution, and find out the value of $a$. We find $$ x^2 y'' -5xy' -7y = x^2 (4 \cdot 3ax^2) - 5x (4ax^3) - 7ax^4 = (12-20-7)ax^4 = -15ax^4. $$ Therefore $a=-1/15$ is the right number to give the right-hand side.

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