Solve the differential equation :

\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [ \frac{x+y\cos x }{1+\sin x} \right ]

 

 

 

 
 
 
 
 

Answers (1)

\frac{\mathrm{d} y}{\mathrm{d} x}=-\left [ \frac{x+y\cos x }{1+\sin x} \right ]

\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=-\frac{x}{1+\sin x}-\frac{y\cos x}{1+\sin x }\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+\frac{y\cos x}{1+\sin x}=\frac{-x}{1+\sin x}

Here, P=\frac{\cos x }{1+\sin x}\: \: \: \: and\: \: \: \: Q=\frac{-x}{1+\sin x }

Then, 

1F =e^{ \int Pdx}

=e^{\int \frac{\cos x}{1+\sin x}dx}=e^{\log \left | 1+\sin x \right |}

=1+\sin x

Then, 

y\times IF=\int Q\times IF\: \: dx+c

y(1+\sin x)=\int \frac{-x}{1+\sin x}\times (1+\sin x)dx

\Rightarrow \int -xdx

y(1+\sin x )=\frac{-x^2}{2}+c

 

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