Thread: Solve the following ODE .. #5

1. Solve the following ODE .. #5

Problem:
Solve the following equation:
$\dfrac{dy}{dx}+2 \, \dfrac{sin(x)}{x} \, y = x y^2$

Solution:
Its Bernoulli's Equation in y

Let $t=y^{-1} \implies \dfrac{dy}{dx}=\dfrac{-1}{t^2} \, \dfrac{dt}{dx}$

Multiply the equation by $y^{-2}$ :

$y^{-2} \, \dfrac{dy}{dx}+2 \, \dfrac{sin(x)}{x} \, y^{-1} = x$

Which is:

$-t^2 \, \dfrac{1}{t^2} \, \dfrac{dt}{dx} + 2 \, \dfrac{sin(x)}{x} \, t = x$

Or:

$\dfrac{dt}{dx} - 2 \dfrac{sin(x)}{x} \, t = -x$

Which is a Linear Equation in t, the problem here is I can't find the Integrating Factor

Since I can't find $\int \dfrac{sin(x)}{x} \, dx$

Maybe I did something wrong?

2. What you have found so far is fine.

I suggest writing $\displaystyle{\frac{\sin{x}}{x}}$ as a Taylor series, then you can integrate it term by term...

3. Thanks.
I said there is something wrong.
I will check my TA, since he said in this course we will solve ,through the ODEs, only elementary integrals.