1. ## differential equaiton

i am asked to form a differential equation using dy/dx = 1 + y + (x^2 ) + y(x^2) , but i gt stucked here , hw to proceed?

2. ## Re: differential equaiton

\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 + y + x^2 + x^2\,y \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 + x^2 + \left( 1 + x^2 \right) \,y \\ \frac{\mathrm{d}y}{\mathrm{d}x} - \left( 1 + x^2 \right) \, y &= 1 + x^2 \end{align*}

Now since this is first order linear you can solve with an integrating factor...

3. ## Re: differential equaiton

Originally Posted by Prove It
\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 + y + x^2 + x^2\,y \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 + x^2 + \left( 1 + x^2 \right) \,y \\ \frac{\mathrm{d}y}{\mathrm{d}x} - \left( 1 + x^2 \right) \, y &= 1 + x^2 \end{align*}

Now since this is first order linear you can solve with an integrating factor...
how if i don't wanna solve it with integrating factor ? but just using the substuition of y = Vx , is it feasible ?

4. ## Re: differential equaiton

Is the original equation homogeneous? If so, you can use that substitution. You performed the test in another thread, so you know how to do it.

5. ## Re: differential equaiton

Originally Posted by Archie
Is the original equation homogeneous? If so, you can use that substitution. You performed the test in another thread, so you know how to do it.
what do you mean by the original equation ? i am only given dy/dx = 1 + y + (x^2 ) + y(x^2) ,
btw , i found that solving by using integrating factor , it is very troublesome . Is there any simple way ?

6. ## Re: differential equaiton

Yes, that was the original equation. And that is dy/dx= 1+ y+ x^2+ yx^2= (1+ y)+ x^2(1+ y)= (1+ y)(1+ x^2) so that
dy/(1+ y)= (1+ x^2)dx