Question

solve the IVP x2y'=cos x - 2xy , y(pi)=0, x>0 .

Answer 1

Answer on Question #40490 – Math – Differential Calculus

Solve the IVP $x^{2}y' = \cos x - 2xy, y(\pi) = 0, x > 0$ .

Solution:

$x^{2}y' = \cos x - 2xy$, whence

$$y' + \frac{2y}{x} = \frac{\cos x}{x^{2}},$$

Let $y = uv, u = u(x), v = v(x), y' = u'v + v'u$.

Let substitute $u'v + v'u + \frac{2uv}{x} = \frac{cosx}{x^2}, u'v + u(v' + \frac{2v}{x}) = \frac{cosx}{x^2}$ , then

1) $v' + \frac{2v}{x} = 0$,

and, 2) $u'v = \frac{\cos x}{x^2}$.

Let solve 1):

$$\frac{dv}{dx} = -\frac{2v}{x}, \frac{dv}{v} = -\frac{2dx}{x},$$ $$lnv = \ln x^{-2}, \quad v = \frac{1}{x^{2}}.$$

Let solve 2):

$$\frac{u'}{x^{2}} = \frac{\cos x}{x^{2}}, \quad u' = \cos x, \quad du = \cos x \, dx, \quad \text{whence}$$

$u = \sin x + c$, where $c$ be some arbitrary const.

So, $y = uv = \frac{c + \sin x}{x^{2}}$,

Let solve IVP $y(\pi) = 0, x > 0$ :

$$0 = \frac{c + \sin \pi}{\pi^{2}}, \quad c = -\sin \pi = 0.$$ $$y = \frac{\sin x}{x^{2}}.$$

Answer: $y = \frac{\sin x}{x^{2}}$.