Answer to Question #96319 in Differential Equations for Olajide Olaitan

Question

The general solution of the equation is y'+(cotx)y=xcscx

Answer 1

$y'+(cotx)y=xcscx \newline \frac{dy}{dx}+(cotx)y=(cscx)x$

This equation is a linear equation of the form

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

To solve this equation, we multiply both sides by the integrating factor I(x):

$I(x)=e^{\int P(x)dx}=e^{\int cotxdx} \newline \int cotxdx = \int \frac{cosx}{sinx}dx = \int \frac{d(sinx)}{sinx} = ln|sinx|+c$

We take c=0. Then we get:

$I(x)=e^{ln|sinx|}=|sinx|$

Let's multiply both sides by sinx (absolute value doesn't really change anything):

$sin(x)\frac{dy}{dx}+sinx\cdot cotx \cdot y=x(cscx \cdot sinx) \newline sinx\frac{dy}{dx}+cosx \cdot y=x \newline \frac{d}{dx}(y\cdot sinx)=x$

Integrating both sides we get the answer:

$y\cdot sinx=\frac{x^2}{2}+c \newline y=\frac{\frac{x^2}{2}+c}{sinx}$