# Initial value problem

Gold Member
Rewrite the differential equation $$\frac{dy}{dx}=x{\sqrt{y}}$$ in the form y=f(x) given the initial condition f(3)=25.

I am new to integration so I am unsure about my work on this problem.

$$\frac{dy}{dx}=x{\sqrt{y}}$$

$$dy=(dx)(x)(\sqrt{y})$$

$$\frac{dy}{\sqrt{y}}=(dx)(x)$$

$$\int{\frac{dy}{\sqrt{y}}}=\int{(x)(dx)}$$

$$2y^{\frac{1}{2}}=\frac{1}{2}x^2+ C$$

$$10=\frac{9}{2}+C$$

$$C=\frac{11}{2}$$

$$2y^{\frac{1}{2}}=\frac{1}{2}x^2+\frac{11}{2}$$

$$y=(\frac{1}{4}x^2+\frac{11}{4})^2$$

If I did it correctly, is there an easier way to do it? If I messed up, where?

Thanks

Looks right to me.

It doesn't seem like that much work to me.

Gold Member
No, it's not a lot of work, I just thought that there might be others ways to arrive at the same answer.

Thanks

dextercioby