- 665
- 68
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
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