Is this general solution for ODE correct?

[andrey21]

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)

 

[rock.freak667]

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

 

[andrey21]

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]

That should be correct.

 

[vela]

You can easily check your answer by plugging it back into the original differential equation and seeing if it works.