Initial value problem

  • Thread starter kreil
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  • #1
kreil
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Rewrite the differential equation [tex]\frac{dy}{dx}=x{\sqrt{y}}[/tex] 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.

[tex]\frac{dy}{dx}=x{\sqrt{y}}[/tex]

[tex]dy=(dx)(x)(\sqrt{y})[/tex]

[tex]\frac{dy}{\sqrt{y}}=(dx)(x)[/tex]

[tex]\int{\frac{dy}{\sqrt{y}}}=\int{(x)(dx)}[/tex]

[tex]2y^{\frac{1}{2}}=\frac{1}{2}x^2+ C[/tex]

[tex]10=\frac{9}{2}+C[/tex]

[tex]C=\frac{11}{2}[/tex]

[tex]2y^{\frac{1}{2}}=\frac{1}{2}x^2+\frac{11}{2}[/tex]

[tex]y=(\frac{1}{4}x^2+\frac{11}{4})^2[/tex]

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

Thanks
 

Answers and Replies

  • #2
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Looks right to me.

It doesn't seem like that much work to me.
 
  • #3
kreil
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No, it's not a lot of work, I just thought that there might be others ways to arrive at the same answer.

Thanks
 
  • #4
dextercioby
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kreil said:
No, it's not a lot of work, I just thought that there might be others ways to arrive at the same answer.
Thanks

Nope,there's no simpler way to integrate that diff.eq. than the variable separation method.
Nice work!!

Daniel.
 

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