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.
Explanation of Method
The same $t=1/x$, $Y=u/t$ reduction as the previous problem now yields the constant-coefficient equation $u''-c^2u=0$, producing exponential rather than trigonometric solutions.
Step 1: Same change of variable as before
This equation differs from the previous problem only by the sign in front of $c^2$, so the same substitution chain applies verbatim: set $t=1/x$, $Y(t)=y(x)$. Exactly as before,
\[ y' = -t^2Y', \qquad y'' = 2t^3Y' + t^4Y'', \]
and substituting into $x^4y''-c^2y=0$ with $x^4=1/t^4$ gives
\[ Y'' + \frac{2}{t}Y' - c^2Y = 0. \]
Step 2: Clearing the first-derivative term: $Y=u/t$
With $Y=u/t$, the identities $Y'=u'/t-u/t^2$ and $Y''=u''/t-2u'/t^2+2u/t^3$ are unchanged from before, and the same pairwise cancellation of the $u'/t^2$ and $u/t^3$ terms occurs — only the sign carried on $c^2$ survives the substitution:
\[ \frac{u''}{t} - c^2\frac{u}{t} = 0 \;\Longrightarrow\; u'' - c^2u = 0. \]
Step 3: Solving the constant-coefficient equation
Now the characteristic equation $r^2-c^2=0$ has real roots $r=\pm c$, so
\[ u(t) = C_1e^{ct} + C_2e^{-ct}. \]
This is the essential difference from Problem 191: flipping the sign of $c^2$ in $x^4y''\pm c^2y=0$ flips the character of the roots of the auxiliary equation for $u$ from purely imaginary to real, exactly as it would for an ordinary constant-coefficient equation.
Step 4: Back-substitution to $x$
Undoing $Y=u/t$ and then $t=1/x$ (so $1/t=x$, $ct=c/x$): $y(x) = \frac{u(t)}{t} = x\bigl[C_1e^{ct}+C_2e^{-ct}\bigr]$.
Step 5: Verification
Take $y_1=xe^{c/x}$. Differentiating, \[y_1' = e^{c/x} - \frac{c}{x}e^{c/x} = \left(1-\frac{c}{x}\right)e^{c/x}, \qquad y_1'' = \frac{c^2}{x^3}e^{c/x}.\]
so that $x^4y_1'' = c^2xe^{c/x} = c^2y_1$, i.e. $x^4y_1''-c^2y_1=0$ identically. Similarly, $y_2=xe^{-c/x}$ satisfies $y_2''=\dfrac{c^2}{x^3}e^{-c/x}$, giving $x^4y_2''-c^2y_2 = c^2xe^{-c/x}-c^2xe^{-c/x}=0$ as well. Both basis solutions check out exactly.
For independence, with $y_1'=(1-c/x)e^{c/x}$ and $y_2'=(1+c/x)e^{-c/x}$, the Wronskian is
\[ W = y_1y_2' - y_1'y_2 = xe^{c/x}\left(1+\frac{c}{x}\right)e^{-c/x} - \left(1-\frac{c}{x}\right)e^{c/x}\cdot xe^{-c/x} = x\left[\left(1+\frac{c}{x}\right)-\left(1-\frac{c}{x}\right)\right]. \]
\[ W = x\cdot\frac{2c}{x} = 2c \ne 0 \quad (c\ne 0). \]
A nonzero constant Wronskian confirms independence of $y_1,y_2$ for all $x\ne0$, $c\ne0$.
Result
Final answer:
\[\boxed{\,y = x\left[C_1e^{c/x} + C_2e^{-c/x}\right]\,}\]
Remark
Comparison with Problem 191. The two equations $x^4y''\pm c^2y=0$ transform, via $t=1/x$ and $Y=u/t$, into $u''\pm c^2u=0$ — the same reduction, differing only by the sign carried through from the original equation. Just as a constant-coefficient equation $r^2+k=0$ gives oscillatory solutions for $k>0$ and exponential solutions for $k<0$, here the $+c^2$ equation produces the oscillatory pair $x\cos(c/x),\,x\sin(c/x)$ while the $-c^2$ equation produces the exponential pair $xe^{c/x},\,xe^{-c/x}$: the sign of the forcing term plays exactly the role that the sign of the discriminant plays for constant coefficients.