Find the general solution of x^4y'' + c^2y = 0 , with c ≠ 0…

Find the general solution of $x^4y'' + c^2y = 0$, with $c \ne 0$ a nonzero constant.

Step 1: Change of independent variable

The coefficient $x^4$ suggests the substitution $t=1/x$, which turns the singular point at $x=0$ into the ordinary point $t=0$... at infinity. Writing $Y(t)=y(x)$, the chain rule with $dt/dx=-t^2$ gives

\[ y' = \frac{dY}{dt}\frac{dt}{dx} = -t^2Y', \qquad y'' = \frac{d}{dx}\bigl(-t^2Y'\bigr) = \bigl(-2tY'-t^2Y''\bigr)(-t^2) = 2t^3Y' + t^4Y''. \]

Step 2: Transforming the equation

Since $x=1/t$, we have $x^4=1/t^4$. Substituting into $x^4y''+c^2y=0$: \[\frac{1}{t^4}\bigl(2t^3Y' + t^4Y''\bigr) + c^2Y = 0 \;\Longrightarrow\; \frac{2Y'}{t} + Y'' + c^2Y = 0,\]

i.e. a Bessel-like equation in $t$ with variable coefficients: $Y'' + \frac{2}{t}Y' + c^2Y = 0$.

Step 3: Removing the first-derivative term: $Y=u/t$

The combination $2/t$ is exactly what the substitution $Y=u/t$ is designed to clear. With

\[ Y = \frac{u}{t}, \qquad Y' = \frac{u'}{t} - \frac{u}{t^2}, \qquad Y'' = \frac{u''}{t} - \frac{2u'}{t^2} + \frac{2u}{t^3}, \]

substitution into $Y''+\frac{2}{t}Y'+c^2Y=0$ gives

\[ \left(\frac{u''}{t} - \frac{2u'}{t^2} + \frac{2u}{t^3}\right) + \frac{2}{t}\left(\frac{u'}{t} - \frac{u}{t^2}\right) + c^2\frac{u}{t} = 0. \]

The $u'/t^2$ terms cancel ($-2u'/t^2+2u'/t^2=0$) and the $u/t^3$ terms cancel ($2u/t^3-2u/t^3=0$), leaving simply

\[ \frac{u''}{t} + c^2\frac{u}{t} = 0 \;\Longrightarrow\; u'' + c^2u = 0. \]

Step 4: Solving the constant-coefficient equation

The characteristic equation $r^2+c^2=0$ has purely imaginary roots $r=\pm ci$, so

\[ u(t) = C_1\cos(ct) + C_2\sin(ct). \]

Step 5: Back-substitution to $x$

Undo the two substitutions in turn: $Y=u/t$, then $t=1/x$ so that $1/t=x$ and $ct=c/x$: \[y(x) = Y(t) = \frac{u(t)}{t} = x\bigl[C_1\cos(ct)+C_2\sin(ct)\bigr].\]

Step 6: Verification

Take $y_1 = x\cos(c/x)$. Differentiating, \[y_1' = \cos\frac{c}{x} + \frac{c}{x}\sin\frac{c}{x}, \qquad y_1'' = -\frac{c^2}{x^3}\cos\frac{c}{x}.\]

so that $x^4y_1'' = -c^2x\cos(c/x) = -c^2y_1$, i.e. $x^4y_1''+c^2y_1=0$ identically. The companion function $y_2=x\sin(c/x)$ satisfies $y_2''=-\dfrac{c^2}{x^3}\sin(c/x)$, giving $x^4y_2''+c^2y_2 = -c^2x\sin(c/x)+c^2x\sin(c/x)=0$ as well. Both solutions, and hence the whole family, check out exactly.

For independence, compute the Wronskian of $y_1=x\cos(c/x)$, $y_2=x\sin(c/x)$: \[W = y_1y_2' - y_1'y_2 = x\cos\frac{c}{x}\left(\sin\frac{c}{x}-\frac{c}{x}\cos\frac{c}{x}\right) - x\left(\cos\frac{c}{x}+\frac{c}{x}\sin\frac{c}{x}\right)\sin\frac{c}{x}.\]

Expanding and using $\sin^2(c/x)+\cos^2(c/x)=1$, every cross term cancels and the $c/x$ pieces combine to give

\[ W = -c\cos^2\frac{c}{x} - c\sin^2\frac{c}{x} = -c \ne 0 \quad (c\ne 0). \]

A nonvanishing, constant Wronskian confirms $y_1,y_2$ are independent for every $x\ne0$ and every $c\ne0$, so the boxed formula is indeed the general solution.