Answer to Question #96321 in Differential Equations for Olajide Olaitan

Question

Obtain the solution for the initial value problem y'+(cotx)y=xcscx, y left frac{pi}{2} right=1

Answer 1

We find the general solution of the equation

$\frac{dy}{dx} +(cotx)y=x csc(x)$

This equation is a linear equation of the form

$\frac{dy}{dx} + P(x)y = Q(x)$

where

$P(x) = cot (x),𝑄(𝑥) = 𝑥 csc(x)$

Integrating factor :

$IF(x) = e^ {\int 𝑃(𝑥)𝑑x }=e^ {\int cot (x) 𝑑x }=\\ =e^ {\int \frac{𝑑(sin𝑥) }{sin x} }= e^ {ln|sin𝑥| + 𝐶}$

We are looking for a particular integrating factor, so we take C=0.

Then we get 

$IF (x)= sin(𝑥)$

Multiply both sides of the equation by $IF (x)= sin(𝑥)$

$sin(𝑥) \frac{dy}{dx} +(sin(𝑥) ⋅cot(x))y=x csc(x)⋅ sin(𝑥)$

or

$sin(𝑥) \frac{dy}{dx} + (cos(𝑥))𝑦 = 𝑥$

i.e.

$\frac{d}{dx}(𝑦 ⋅ sin𝑥) = 𝑥.$

Integrate both sides of this equation we get the general solution

$\int \frac{d}{dx}(𝑦 ⋅ sin𝑥)dx = \int 𝑥 dx \Rightarrow 𝑦 ⋅ sin𝑥 = \frac{𝑥^2}{2} +C$ ,

where C - constant defined by the initial conditions. 

Now we find C and solve the initial value problem.

Substitute the initial condition

$y(\frac{\pi}{2}) =1$

into the general solution:

$y(\frac{\pi}{2})⋅ sin(\frac{\pi}{2})= \frac{(\frac{\pi}{2})^2}{2} +C \Rightarrow \\1⋅ sin(\frac{\pi}{2})= \frac{1}{2} (\frac{\pi}{2})^2+C,$

or

$1= \frac{\pi^2}{8}+C \Rightarrow C= 1- \frac{\pi^2}{8}.$

Then we obtain the solution for the initial value problem

$𝑦 ⋅ sin𝑥 = \frac{𝑥^2}{2} +1- \frac{\pi^2}{8}$

or

$𝑦 =( \frac{𝑥^2}{2} +1- \frac{\pi^2}{8})csc(x)$

The solution for the initial value problem is

$𝑦 =( \frac{𝑥^2}{2} +1- \frac{\pi^2}{8})csc(x).$