Question

find solution for diff eqn: y'=1+x+y+xy

Answer 1

Try letting \(y=vx\), then \(y'=xv'+v\):
\[\begin{align*}xv'+v&=1+x+vx+vx^2\\
xv'&=1+x+vx-v+vx^2\\
xv'&=1+x+v(x^2+x-1)\\
v'&=\frac{1}{x}+1+\frac{v(x^2+x-1)}{x}\\
v'-\frac{x^2+x-1}{x}v&=\frac{1}{x}+1
\end{align*}\]
Now you have a linear equation.

Answer 2

I'm confused

Answer 3

By what? The substitution? or how to solve a linear differential equation?

Answer 4

When you say "I'm confused," it doesn't help at all to not address what's causing the confusion. I'm willing to clarify any points in my first post, but it's on you to tell me what to clarify.