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Question

Find the particular solution of differential equation : 

dydx=x+y cos x1+sin x, given that y=1 when x=0


Solution

Given:

     dydx=x1+sin xy cos x1+sin x

dydx+cos x1+sin x y=x1+sin x  ....................(1)

Comparing equation (1) with linear differential equation
 
dydx+P y=Q 
 
we will get, 
 
P=cos x1+sin x;Q=x1+sin x

Intergrating factor (I.F.),  I.F =ePdx

I.F=ecos x1+sin xdx=elog(1+sin x)=1+sin x

For general solution of differential equation, 

(1+sin x)y=x dx+C

[y(I.F.)=Q(I.F.)dx+C]

y(1+sin x)=x22+C .............(2)

Now, we get y=1, when x=0

1(1+sin 0)=02+CC=1

Putting C=1 in equation (2), we get 

y(1+sin x)=x22+1

Hence the particular solution is,

2y(1+sin x)+x22=0

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