1. Solve y'+y^2sinx=0.

solve the differential equation:
y'+y^2sinx=0

I used the method of separation:

y'=-y^2sinx
dy/dx=-y^2sinx
dy/-y^2=sinxdx
integral of -y^-2dy=integral of sinxdx
1/y=cosx+C
y=1/(cosx+C)

Did I do this right?

2. Originally Posted by Taurus3
solve the differential equation:
y'+y^2sinx=0

I used the method of separation:

y'=-y^2sinx
dy/dx=-y^2sinx
dy/-y^2=sinxdx
integral of -y^-2dy=integral of sinxdx
1/y=cosx+C
y=1/(cosx+C)

Did I do this right?
$\displaystyle\frac{y'}{y^2}=-\sin(x)\Rightarrow \frac{-1}{y}=\cos(x)+C\Rightarrow y=\frac{-1}{\cos(x)+C}$

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y=akardari 2sinx

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