# Thread: Seperation of variables/Linear Differential

1. ## Seperation of variables/Linear Differential

I am right with my assumptions regarding these equations?

x3(dy/dx) + 2x^2y = 5 -----> use linear differential to solve equation?

dy/dt = 1 +y^2/y(1+t^2) ------> use the separating the variables method? also in this one can i get rid of the brackets on the bottom to leave me with y + t^2?

cheers

2. For the separating variables equation I have:

g(t) = 1 + y^2
h(t) = y + t^2

So

1/y+t^2 (dy/dt) = 1/y^2

(Integration) 1/y+t^2 (dy/dt) dt = (Integration) 1 +y^2 dt

(Integration) 1/y+t^2 dt = (Integration) 1 +y^2 dt

Arctan t = ????

What is the integral for 1 + y^2?

and am I along the right lines?

3. Originally Posted by poundedintodust
x3(dy/dx) + 2x^2y = 5 -----> use linear differential to solve equation?
This is a homogeneous equation. Simply do it by substitution. Take $y=Vx$. Then separate the variables.

4. sorry to be ignorant but could you give a bit more of an explanation? just this is all new to me!!!!

5. I am sorry. You can't do it by that method. I didn't see that 5. It isn't a homogeneous equation.

6. I have yet to do non-homogeneous equations. I am a noobest noob in ODEs.

7. Originally Posted by poundedintodust
I am right with my assumptions regarding these equations?

x3(dy/dx) + 2x^2y = 5 -----> use linear differential to solve equation?
Written backwards as $y' + \frac{{2y}}
{x} = \frac{5}
{{x^3 }},$
and this is a linear ODE, you can tackle it by using the integrating factor method.

As for your 2nd. question, the equation is separable. Take it to the form $f(y)\,dy=f(t)\,dt.$