Question

Use the substitution of x=e^{t} to transform the given Cauchy-Euler differential equation to a differential equation with constant coefficients then solve that by any method. (finding the characterristic (auxiliary) equation or variation of parameters). Do not forget to re-subtitute for x after you solve the equation.

x^{2}y^{''}+9xy^{'}-20y=0

Answer 1

By the chain rule,

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

which follows from $x=e^t\implies t=\ln x\implies\dfrac{\mathrm dt}{\mathrm dx}=\dfrac1x$.

$\dfrac{\mathrm dy}{\mathrm dt}$ is then a function of $x$; denote this function by $f(x)$. Then by the product rule,

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dt}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac1x\dfrac{\mathrm df}{\mathrm dx}$

and by the chain rule,

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

so that

$\dfrac{\mathrm d^2y}{\mathrm dt^2}-\dfrac{\mathrm dy}{\mathrm dt}=x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}$

Then the ODE in terms of $t$ is

$\dfrac{\mathrm d^2y}{\mathrm dt^2}+8\dfrac{\mathrm dy}{\mathrm dt}-20y=0$

The characteristic equation

$r^2+8r-20=(r+10)(r-2)=0$

has two roots at $r=-10$ and $r=2$, so the characteristic solution is

$y_c(t)=C_1e^{-10t}+C_2e^{2t}$

Solving in terms of $x$ gives

$y_c(x)=C_1e^{-10\ln x}+C_2e^{2\ln x}\implies\boxed{y_c(x)=C_1x^{-10}+C_2x^2}$