# Is this general solution for ODE correct?

Find the general solution of the following ODE:

dx/dt = 3x^(2) cos t

Make x the subject of the solution.

Heres my solution, is this correct?

dx/dt = 3x^(2) cos t

dx/3x^(2) = cos t dt

Integrating both sides gives:

ln (3x^(2)) = sin t + C

3x^(2) = e^(sin t + C)

3x^(2) = Ae^(sin t)

x^(2) = (Ae^(sin t))/3)

x = SQRT(Ae^(sin t))/3)

## The Attempt at a Solution

rock.freak667
Homework Helper
Heres my solution, is this correct?

dx/dt = 3x^(2) cos t

dx/3x^(2) = cos t dt

Integrating both sides gives:

ln (3x^(2)) = sin t + C

here is where you went wrong,

1/3x2 can be written as x-2/3.

You know that ∫xn dx = xn+1/(n+1) + C for n≠-1

So from what you have said:

∫dx/3x^(2) = -1/3x + C or (-x^(-1)/3) + C

Giving a solution of:

-1/3x + C = sin t

rock.freak667
Homework Helper
That should be correct.

vela
Staff Emeritus