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Question

Let y=y(x) be the solution of the differential equation cosx(3sinx+cosx+3)dy=(1+ysinx(3sinx+cosx+3))dx,0xπ2,y(0)=0. Then, yπ3 is equal to:


A

2loge23+1011

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B

2loge3+72

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C

2loge33-84

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D

2loge23+96

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Solution

The correct option is A

2loge23+1011


Explanation for the correct option:

Step-1: convert in linear differential equation form and find I.F:

Given,

cosx(3sinx+cosx+3)dy=(1+ysinx(3sinx+cosx+3))dx,y(0)=0.

On solving, we get:

cosx(3sinx+cosx+3)dy=(1+ysinx(3sinx+cosx+3))dxdydx-y(tanx)=1cosx(3sinx+cosx+3)

I.F.=e-tanxdx=elncosx=cosx=cosx[0xπ2]

Step-2: Solution of linear differential equation:

Solving the differential equation:

ycosx=cosx.1cosx(3sinx+cosx+3)dxy(cosx)=1(3sinx+cosx+3)dxy(cosx)=dx32tanx21+tan2x2+1-tan2x21+tan2x2+3y(cosx)=sec2x22tan2x2+6tanx2+4dx

Let,

Step-3 : Integration and final solution of differential equation:

I1=sec2x22tan2x2+6tanx2+4dxSubstitutingtanx2=t,12sec2x2dx=dt

dtt3+3t+2=dt(t+2)(t+1)=1t+1-1t+2dt=lnt+1t+2=lntanx2+1tanx2+2+C

Finding the solution of the differential equation:

y(cosx)=lntanx2+1tanx2+2+Cy(cosx)=lntanx2+1tanx2+2+Cfor0x<π2

Now, it is given that y(0)=0.

Therefore,

0=ln12+CC=ln2y(cosx)=ln1+tanx22+tanx2+ln2Forx=π3y12=ln1+132+13+ln2[cosπ3=12]y=2ln23+1011[Rationlizethedenominator]

Hence, the correct answer is option (A).


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