differential equation

given Equation is:

x\left(y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx} \right)^2 \right)=3y\frac{dy}{dx}

Now take z=y\frac{dy}{dx}

So,
\frac{dz}{dx}=\left(y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx} \right)^2 \right)

Equation now becomes:
x\frac{dz}{dx}=3z

Solving this we get z=kx^3 , and now we substitute back y

We have:
y\frac{dy}{dx}=kx^3

Given y(1)=1 , and y'(1)=1

so we get k=1

Integrating, we obtain:
\frac{y^2}{2}=\frac{x^4}{4}+c

From given conditions c=\frac{1}{4}

y^2(2)=\frac{17}{2}