Simpler way to solve differential equation for initial value problem?

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Homework Help Overview

The discussion revolves around solving the differential equation \(\frac{dy}{dx}=x{\sqrt{y}}\) with an initial condition of \(f(3)=25\). The original poster expresses uncertainty about their integration process and seeks confirmation on their approach and potential simpler methods.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to separate variables and integrate, questioning their work and the possibility of simpler methods. Other participants confirm the correctness of the approach but also explore the idea of alternative methods.

Discussion Status

The discussion includes confirmations of the original poster's integration steps, with some participants suggesting that the variable separation method is the most straightforward approach. There is no explicit consensus on alternative methods, but the conversation remains open to exploration.

Contextual Notes

The original poster indicates they are new to integration, which may influence their confidence in the solution process. The initial condition provided is also a key aspect of the problem context.

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
 
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Looks right to me.

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

Thanks
 
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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