How Do You Solve a differential equation of the form: $y'=yx+x$ How do you solve a differential equation of the form: $y'=yx+x$
In this case you cannot separate the equation indeed: $y'=yx+x \iff \dfrac{dy}{dx}=yx+x$ And I can't separate it? How does one solve that problem.
The equation I am struggling with is $\dfrac{dy}{dx}= \dfrac{y}{1-x^2}+x \iff \dfrac{dy}{dx} = y(1-\dfrac{1}{x^2})+x$ 
I can't seperate it.
I understood how to solve $y'=yx+x$ But I can't solve the equation above.
What do I do?
 A: Let's give it a try assuming $y\neq -1$
$$y'=x(1+y)\iff {dy\over 1+y}=xdx$$
And we have separated. The solution in this case is $y(x)=ke^{x^2/ 2}-1$ for $k\neq 0$. One can see that $y=-1$ is also a solution so the general solution is
$$y(x)=ke^{{x^2\over 2}}-1, \, k\in \Bbb{R}$$
A: Hint 
Considering $$y'=yx+x=x(y+1)$$ define $z=y+1$ which makes $$\frac{dz}{dx}=x z$$ which is separable.
I am sure that you can take it from here.
A: $$\frac{dy}{dx}+\left(\frac{1}{x^2-1}\right)y=x$$
We can use a trick call integrating factor.
$$\exp\left(\int \frac{1}{x^2-1} dx\right)=\sqrt{\left|\frac{x-1}{x+1}\right|}$$
Note: we can ignore the constant in the previous integration, we are just interested in one such function.
Multiplying the integrating factor to the original equation, we have
$$\sqrt{\left|\frac{x-1}{x+1}\right|}\frac{dy}{dx}+\sqrt{\left|\frac{x-1}{x+1}\right|}\left(\frac{1}{x^2-1}\right)y=\sqrt{\left|\frac{x-1}{x+1}\right|}x$$
Notice that this is just 
$$\frac{d}{dx}\left(y\sqrt{\left|\frac{x-1}{x+1}\right|}\right)=\sqrt{\left|\frac{x-1}{x+1}\right|}x$$
Now you can integrate both sides.
$$y\sqrt{\left|\frac{x-1}{x+1}\right|}=\int \sqrt{\left|\frac{x-1}{x+1}\right|}x dx$$
$$y=\sqrt{\left|\frac{x+1}{x-1}\right|}\int \sqrt{\left|\frac{x-1}{x+1}\right|}x dx$$
