How to solve $vv''+v'^2-6t^2=0$

Question

Hi, I have been trying to solve this equation for a week now.

However, I keep ending up with the same result everytime, it may be because my knowledge on this chapter is not great.

I am unable to figure it out, please help.

Below is the question and the conditions.

$$ \begin{cases} vv''+v'^2-6t^2=0 \\ v'(0)=0,\\ v(0)=1,\\ v>0\quad\forall t \end{cases}$$

They are asking to use the substitution $z=vv'$ to solve the equation, and here you can find my attempt, but I am obviously unsure if it is the correct way of solving the exercise.

Thank you so much, any help is appreciated.

Answer 1

You can notice the integral of $z = v_tv$ is $u = \int z dt = \frac{v^2}{2}$. So using the substitution $u = v^2$. You then get $u'' -12t^2=0$. This is then solvable with the standard methods of solving differential equations.

$u'' = 12t^2 \implies u = t^4+c_1t+c_2 \implies v = \sqrt{t^4+c_1t+c_2}$

Now, we plug in initial conditions and we are done:

$v(0) = 1 = \sqrt{0^4+c_1(0)+c_2} \implies c_2 = 1$

$v'(0) = 0 = \frac{4(0)^3+c_1}{2\sqrt{(0^4+c_1(0)+1}} \implies c_1=0$

So the solution is:

$v = \sqrt{t^4+1}$

Answer 2

Hint: $$vv''+v'^2=6t^2$$ Rewrite it as: $$(v'v)'=6t^2$$ $$(v^2)''=12t^2$$ And integrate.

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With $z=vv'$ rewrite the DE as: $$z'=6t^2$$