Non-linear Differential Equation.

Question

I am trying to solve this equation
$$y'' = 8yy', y(0) = 0, y'(0) = 1$$ $$v = y'$$ $$v' = 8y v$$ $$ \frac 1v dv = 8y dy $$ $$ ln(v) = 4y^2 + c $$ $$ v(0) = 1 => ln(1) = 0 + c <=> c = 0$$ $$ ln(v) = 4y^2$$ $$ v = e^{4y^2} $$ $$ v = y' => y'= e^{4y^2}$$ Now Wolfram Alpha tells me the solution is $y = \frac 12 tan(2x)$ and so I am stuck as I am getting a different answer when I solve this. Could anyone guide me? Thankyou.

Answer 1

Just integrate once in the original equation to find $$ y'=4y^2+C $$ which then can be solved via separation for all the signs of $C$.

---

For your approach in seeking solutions in the form $y'=v(y)$ you need to correctly apply the chain rule in $y''=v'(y)y'=v'(y)v(y)$ to find that $$ v'v = 8yv. $$