dragonblood
Sep20-09, 09:32 AM
I have tried to solve the differential equation
y'=x\sqrt{y}
like this:
y^{-\frac{1}{2}}y'=x
\int{y^{-\frac{1}{2}}}dy=\int{xdx}
y^{\frac{1}{2}}=\frac{x^2 +C}{4}
y=\left(\frac{x^2+C}{4}\right)^2
Is this the right way to solve it? Because the answer in my textbook says that the answer is
y=\pm\sqrt{x^2+C}
But I really can't see where I've gone wrong.
y'=x\sqrt{y}
like this:
y^{-\frac{1}{2}}y'=x
\int{y^{-\frac{1}{2}}}dy=\int{xdx}
y^{\frac{1}{2}}=\frac{x^2 +C}{4}
y=\left(\frac{x^2+C}{4}\right)^2
Is this the right way to solve it? Because the answer in my textbook says that the answer is
y=\pm\sqrt{x^2+C}
But I really can't see where I've gone wrong.