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

Mathematics

1 answer:

seraphim [82]2 years ago

3 0

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 $t$ :

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