Answer to Question #162382 in Differential Equations for Kanchan Roy

Question

x^3*dy/dx – x^2*y = – y^4*cosx

Answer 1

$x^3\frac{dy}{dx}-x^2y=-y^4\cos{x}$

Solution:

Divide the left and right sides of the equation by $x^2y^4$ :

$\frac{x}{y^4}\frac{dy}{dx}-\frac{1}{y^3}=-\frac{\cos{x}}{x^2}$

$\frac{1}{y^3}=z$ , $z'=-3\frac{y'}{y^4}$ , $\frac{y'}{y^4}=-\frac{z'}{3}$ .

$x(-\frac{z'}{3})-z=-\frac{\cos{x}}{x^2}$

$x\frac{z'}{3}+z=\frac{\cos{x}}{x^2}$

$z=uv$ , $z'=u'v+v'u$

$x\frac{u'v+v'u}{3}+uv=\frac{\cos{x}}{x^2}$

$x\frac{u'v}{3}+u(\frac x3v'+v)=\frac{\cos{x}}{x^2}$

Let's compose and solve the system:

$\begin{cases} \frac x3v'+v=0 \\ x\frac{u'v}{3}=\frac{\cos{x}}{x^2} \end{cases}$

From the first equation:

$\frac{dv}{v}=-3\frac{dx}{x}$

$\ln|v|=-3\ln|x|$

$v=\frac{1}{x^3}$

Substitute $v$ into the second equation:

$x\frac{u'}{3x^3}=\frac{\cos{x}}{x^2}$

$u'=3\cos{x}$

$u=3\sin{x}+C$

$z=uv=\frac{1}{x^3}(3\sin{x}+C)$

$\frac{1}{y^3}=\frac{1}{x^3}(3\sin{x}+C)$

$y^3=\frac{x^3}{3\sin{x}+C}$

Answer: $y^3=\frac{x^3}{3\sin{x}+C}$ .