Question

2y'y"=1 , y(0)=2 , y'(0)=1​

Answer 1

The left side is the derivative of $(y')^2$:

$\left((y')^2\right)'=2y'y''$

So we can integrate both sides of

$\left((y')^2\right)'=1\implies (y')^2=x+C\implies y'=\pm\sqrt{x+C}$

Then integrate again to solve for $y$:

$y=\pm\dfrac23(x+C_1)^{3/2}+C_2$

With the given initial conditions, we find

$y(0)=2\implies 2=\pm\dfrac23{C_1}^{3/2}+C_2$

$y'(0)=1\implies 1=\pm\sqrt{C_1}$

The second equation says $C_1$ is either 1 or -1, but in the latter case, we would get $(-1)^{3/2}=\sqrt{-1}$ in the first equation, which is undefined over the real numbers, so $C_1=1$.

So there are two candidate solutions,

$y_1=\dfrac23(x+1)^{3/2}+\dfrac43$

$y_2=-\dfrac23(x+1)^{3/2}+\dfrac83$

However, the second equation doesn't satisfy the initial value of the derivative, since ${y_2}'(0)=-1\neq1$. So the solution is

$\boxed{y(x)=\dfrac23(x+1)^{3/2}+\dfrac43}$