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Q.
If the differentiable equation $\frac{d y}{d x}-y=y^2(\sin x+\cos x)$ with $y(0)=1$ then $y(\pi)$ has the value equal to
Differential Equations
Solution:
$\frac{d y}{d x}-y=y^2(\sin x+\cos x)$
$\frac{1}{ y ^2} \frac{ dy }{ dx }-\frac{1}{ y }=\sin x +\cos x ; $ Let $-\frac{1}{ y }= t \Rightarrow \frac{1}{ y ^2} \frac{ dy }{ dx }=\frac{ dt }{ dx }$ $\frac{ dt }{ dx }+ t =\sin x +\cos x$
I.F. $e^x$
$\therefore t \cdot e ^{ x }=\int e ^{ x }(\sin x +\cos x ) dx ; -\frac{1}{ y } e ^{ x }= e ^{ x } \sin x + C$
if $x =0, y =1 \Rightarrow C =-1$
$-\frac{e^x}{y}=e^x \sin x-1$
if $x=\pi$ then $-\frac{e^x}{y}=-1 \Rightarrow y=e^\pi$