Answer to Question #92901 in Differential Equations for Subhasis Padhy

Question

Solve the IVP.
x² dy/dx = cosx - 2xy , y(π) = 0, x>0.

Answer 1

1 STEP: solve the homogeneous equation.

$\left.x^2\cdot\displaystyle\frac{dy}{dx}=-2xy\right|\cdot\displaystyle\frac{dx}{yx^2}\longrightarrow\quad \displaystyle\frac{dy}{y}=-2\cdot\displaystyle\frac{dx}{x}\longrightarrow\quad$

$\int\displaystyle\frac{dy}{y}=-2\cdot\int\displaystyle\frac{dx}{x}\longrightarrow\quad \ln|y|=-2\cdot\ln|x|+\ln|C|\longrightarrow\quad$

$\boxed{y_{hom}(x)=\frac{C}{x^2}}$

2 STEP: solve the inhomogeneous equation.

We use the constant variation method

$y(x)=\displaystyle\frac{C(x)}{x^2}\longrightarrow\boxed{\frac{dy}{dx}=\frac{C'(x)}{x^2}-2\cdot\frac{C(x)}{x^3}}$

Substitute the found derivative into the initial equation

$x^2\cdot\displaystyle\frac{dy}{dx}=-2xy\longrightarrow x^2\cdot\left(\frac{C'(x)}{x^2}-2\cdot\frac{C(x)}{x^3}\right)=-2x\cdot\frac{C(x)}{x^2}+\cos(x)\longrightarrow$

$C'(x)-2\cdot\displaystyle\frac{C(x)}{x}=-2\cdot\displaystyle\frac{C(x)}{x}+\cos(x)\longrightarrow$

$C'(x)=\cos(x)\longrightarrow\boxed{C(x)=\sin(x)+C_1}$

Conclusion,

$y(x)=\displaystyle\frac{\sin(x)+C_1}{x^2}$

It remains to determine the integration constant, for this we use the initial condition:

$y(\pi)=0\longrightarrow0=\displaystyle\frac{C_1+\sin(\pi)}{\pi^2}\longrightarrow0=\displaystyle\frac{C_1}{\pi^2}\longrightarrow\boxed{C_1=0}$

ANSWER

$y(x)=\displaystyle\frac{\sin(x)}{x^2}$