Question

Solve the equation x^3 y^'" + 5x^2 y" + 7xy' + 8y = 0

Answer 1

Make the substitution $t=\ln x$, then compute the derivatives of $y$ with respect to $t$ via the chain rule.

• First derivative

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dt}$

• Second derivative

Let $f(t)=\frac{\mathrm dy}{\mathrm dt}$.

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac fx\right]=\dfrac{x\frac{\mathrm df}{\mathrm dx}-f}{x^2}$

$\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\mathrm df}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dt^2}$

$\implies\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dt^2}-\dfrac{\mathrm dy}{\mathrm dt}\right)$

• Third derivative

Let $g(t)=\frac{\mathrm df}{\mathrm dt}=\frac{\mathrm d^2y}{\mathrm dt^2}$.

$\dfrac{\mathrm d^3y}{\mathrm dx^3}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac{g-f}{x^2}\right]=\dfrac{x^2\left(\frac{\mathrm dg}{\mathrm dx}-\frac{\mathrm df}{\mathrm dx}\right)-2x(g-f)}{x^4}$

$\dfrac{\mathrm dg}{\mathrm dx}=\dfrac{\mathrm dg}{\mathrm dt}\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x\dfrac{\mathrm d^3y}{\mathrm dt^3}$

$\implies\dfrac{\mathrm d^3y}{\mathrm dx^3}=\dfrac{x^2\left(\frac1x\frac{\mathrm dg}{\mathrm dt}-\frac1x\frac{\mathrm df}{\mathrm dt}\right)-2x(g-f)}{x^4}=\dfrac1{x^3}\left(\dfrac{\mathrm d^3y}{\mathrm dt^3}-3\dfrac{\mathrm d^2y}{\mathrm dt^2}+2\dfrac{\mathrm dy}{\mathrm dt}\right)$

Substituting $y(t)$ and its derivatives into the ODE gives a new one that is linear in $t$:

$\left(\dfrac{\mathrm d^3y}{\mathrm dt^3}-3\dfrac{\mathrm d^2y}{\mathrm dt^2}+2\dfrac{\mathrm dy}{\mathrm dt}\right)+5\left(\dfrac{\mathrm d^2y}{\mathrm dt^2}-\dfrac{\mathrm dy}{\mathrm dt}\right)+7\dfrac{\mathrm dy}{\mathrm dt}+8y=0$

$\dfrac{\mathrm d^3y}{\mathrm dt^3}+2\dfrac{\mathrm d^2y}{\mathmr dt^2}+4\dfrac{\mathrm dy}{\mathrm dt}+8y=0$

$y'''+2y''+4y'+8y=0$

which has characteristic equation

$r^3+2r^2+4r+8=(r+2)(r^2+4)=0$

with roots $r=-2$ and $r=\pm2i$, so that the characteristic solution is

$y_c(t)=C_1e^{-2t}+C_2\cos2t+C_3\sin2t$

Replace $t=\ln x$ to solve for $y(x)$:

$y_c(x)=C_1e^{-2\ln x}+C_2\cos(2\ln x)+C_3\sin(2\ln x)$

$\boxed{y(x)=\dfrac{C_1}{x^2}+C_2\cos(2\ln x)+C_3\sin(2\ln x)}$