Solve this equation :
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The 'standard solution' is... $\displaystyle y(x)= e^{- \int \sin x\cdot dx} \{ \int \sin x \cdot e^{\int \sin x\cdot dx} + c\}= e^{\sin x + \cos x} + c\cdot e^{\cos x}$ Kind regards $\displaystyle \chi$ $\displaystyle \sigma$
Originally Posted by dhiab Solve this equation : This is of variables seperable type: $\displaystyle y'=\sin(x)(1-y)$ so: $\displaystyle \frac{y'}{1-y}=\sin(x)$ CB
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