How Do You Solve This First Order ODE with Given Condition?

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SUMMARY

The first order ordinary differential equation (ODE) discussed is x²(dy/dx) + y² = 0 with the initial condition y(1) = 3. The equation is rearranged to -1/y² dy = 1/x² dx, allowing for separation of variables and integration. The solution derived is y = -x/(Cx + 1). By applying the initial condition, C is determined to be -4/3, leading to the final solution y = 3x/(4x - 3), which matches the book's answer.

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The question is x^2dy/dx + y^2=0 , y(1)=3

I re-arrange the equation to get -1/y^2dy=1/x^2dx

separated them, then I integrate both sides to get 1/y=-1/x + c

Now I don't get how they got the answer y=3x/(4x-3), as when I try use the condition I get a different answer, could anyone help? I might have done something wrong in the integration?
 
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You just solve it for y: [tex]y=\frac{-x}{Cx+1}[/tex]. Then set y(1) equal to 3 (since that's the condition you're given) and solve (using good ol' algebra) for C: [tex]y(1)=\frac{-1}{C+1}=3[/tex]
So, C=-4/3 and after you plug C into y, it is what your book says it is.
 

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