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Lelu [443]
3 years ago
12

Use the substitution of x=e^{t} to transform the given Cauchy-Euler differential equation to a differential equation with consta

nt 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
Mathematics
1 answer:
kherson [118]3 years ago
4 0

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}

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