Hi
I have attached my question
For part a) i got final answer as y=-4/5 +ce^(5/3x)
i got this using intergrating factors
However i cannot seem to do part b)
could someone please help
thank you
$\displaystyle x*y'+(1-2x)y=4x$ so $\displaystyle y'+\frac{1-2x}{x}y=4$
The integrating factor is then $\displaystyle e^{\int\frac{1-2x}{x}dx}$
So now, $\displaystyle \int\frac{1-2x}{x}dx=\int\left(\frac{1}{x}-\frac{2x}{x}\right)dx=ln|x|-2x$ and we ignore the +c...
So the integrating factor is $\displaystyle e^{ln|x|-2x}=e^{ln|x|}e^{-2x}=xe^{-2x}$
So $\displaystyle xe^{-2x}y'+e^{-2x}(1-2x)y=4xe^{-2x}$
and the left side of the equation is just $\displaystyle \frac{dy}{dx}(xe^{-2x}y)$
So $\displaystyle \frac{dy}{dx}(xe^{-2x}y)=4xe^{-2x}$
Now integrate both sides, but this time be sure to include the +c...
$\displaystyle xe^{-2x}y=-(2x+1)e^{-2x}+c$
and now you just have to use your initial condition to find c (that integration was done by parts which I trust you can do)