2y'y"=1 , y(0)=2 , y'(0)=1

The left side is the derivative of :

So we can integrate both sides of

Then integrate again to solve for :

With the given initial conditions, we find

The second equation says is either 1 or -1, but in the latter case, we would get in the first equation, which is undefined over the real numbers, so .

So there are two candidate solutions,

However, the second equation doesn't satisfy the initial value of the derivative, since . So the solution is

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step-by-step explanation:

step-by-step explanation:

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