y''(x) = t^2 + 1, y'(x) = t^3 + 3t

Question

Find x(t) given that x(1) = 4

Answer 1

Your statements contradict each other; your y' does not differentiate to give your stated y''.  Please comment or message me with the correct question for further help.

Answer 2

It is a question given in a past prelim paper, so is written correct. I had a brain wave, tell me if you think this would work?

y'(x) = t^3 + 3t      ,       y''(x) = 3t^3 + 3t

 

Now differentiate y'(x) to get,

y''(x) = (3t^2 + 3)(dt/dx) = 3(t^2 +1)(dt/dx)

 

Therefore, dt/dx = 1/3

So integrating both sides you get,

t = (1/3)x + C

So for x(1) = 4, 1 = 4/3 + C

Therefore, C = - 1/3

 

So, x(t) = 3t + 1