Find the general solution of x^4y'' + 2x^3y' - 4y = (1)/(x)

Find the general solution of $x^4y'' + 2x^3y' - 4y = \frac{1}{x}$.

Step 1: Choosing the substitution $t=1/x$

The coefficients $x^4$, $2x^3$, and $-4$ attached to $y''$, $y'$, and $y$ drop in power by exactly one each time — the signature of an equation that becomes constant-coefficient under an inversion of the independent variable. Set

\[ t = \frac{1}{x}, \qquad \frac{dt}{dx} = -\frac{1}{x^2} = -t^2, \]

and let $Y(t)=y(x)$ denote the unknown function of the new variable $t$.

Step 2: Transforming the derivatives

By the chain rule, $y' = \frac{dY}{dt}\frac{dt}{dx} = -t^2Y'(t)$.

Differentiating again, and applying the chain rule a second time to $dt/dx=-t^2$: \[y'' = \frac{d}{dx}\bigl(-t^2Y'\bigr) = \frac{d}{dt}\bigl(-t^2Y'\bigr)\cdot\frac{dt}{dx} = \bigl(-2tY'-t^2Y''\bigr)\cdot(-t^2) = 2t^3Y' + t^4Y''.\]

Step 3: Substituting into the equation

With $x=1/t$, so that $x^4=1/t^4$ and $x^3=1/t^3$, the left-hand side becomes

\[ x^4y'' + 2x^3y' - 4y = \frac{1}{t^4}\bigl(2t^3Y'+t^4Y''\bigr) + \frac{2}{t^3}\bigl(-t^2Y'\bigr) - 4Y = \frac{2Y'}{t} + Y'' - \frac{2Y'}{t} - 4Y. \]

The two $2Y'/t$ terms cancel exactly, leaving $Y''-4Y$. On the right, $1/x=t$, so the equation collapses to a constant-coefficient equation in $t$: $Y'' - 4Y = t$.

Step 4: Homogeneous solution

The characteristic equation $r^2-4=0$ has roots $r=\pm2$, so

\[ Y_h = C_1e^{2t} + C_2e^{-2t}. \]

Step 5: Particular solution

Since the forcing term $t$ is a degree-1 polynomial and $r=0$ is not a root, try $Y_p=At+B$. Then $Y_p''=0$, and substituting into $Y''-4Y=t$: \[-4(At+B) = t \;\Longrightarrow\; -4A=1,\quad -4B=0 \;\Longrightarrow\; A=-\frac14,\ B=0.\]

So $Y_p=-t/4$, and the general solution in $t$ is

\[ Y(t) = C_1e^{2t} + C_2e^{-2t} - \frac{t}{4}. \]

Step 6: Back-substituting $t=1/x$

The solution is recorded in the result below.

Step 7: Verification

Particular solution. For $y_p=-\dfrac{1}{4x}$, differentiating twice gives $y_p'=\dfrac{1}{4x^2}$ and $y_p''=-\dfrac{1}{2x^3}$. Substituting: \[x^4y_p'' + 2x^3y_p' - 4y_p = x^4\left(-\frac{1}{2x^3}\right) + 2x^3\left(\frac{1}{4x^2}\right) - 4\left(-\frac{1}{4x}\right) = -\frac{x}{2} + \frac{x}{2} + \frac{1}{x} = \frac1x.\ \checkmark\]

Homogeneous solutions. For $y=e^{2/x}$: $y'=-\dfrac{2}{x^2}e^{2/x}$, and differentiating again, $y'' = \frac{4}{x^3}e^{2/x} + \frac{4}{x^4}e^{2/x}$.

Substituting into the homogeneous operator, \[x^4y''+2x^3y'-4y = (4x+4)e^{2/x} - 4xe^{2/x} - 4e^{2/x} = \bigl(4x+4-4x-4\bigr)e^{2/x} = 0.\ \checkmark\]

The same cancellation occurs for $e^{-2/x}$ with $t\to -t$. By linearity, the boxed $y$ satisfies $x^4y''+2x^3y'-4y=1/x$ identically in $C_1,C_2$.