IVP ODE checking specifics of solution

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    Ivp Ode
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SUMMARY

The discussion focuses on solving the initial value problem (IVP) represented by the equation 3xy + y² + (x² + xy)y' = 0 with the condition y(1) = 0. The solution derived is x²y(x + ½y) = 0, leading to potential solutions x = 0, y = 0, or y = -2x. However, only y = 0 satisfies the initial condition y(1) = 0, confirming that y = 0 is the valid solution for all x.

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cambo86
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Homework Statement


I've got an IVP where,
3xy+y2+(x2+xy)y'=0, y(1)=0

The Attempt at a Solution


I've solved to get,
x2y(x+[itex]\frac{1}{2}[/itex]y)=0

Is it correct to say,
x=0 or y=0 or y=-2x,
Since y= 0 is the only solution that fits y(1)=0, then
y=0 [itex]\forall[/itex]x
 
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I didn't check your solution. But in any case, x=0 wouldn't be a solution. Otherwise, assuming your work is OK, yes.
 

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