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Question

Let y=y(x) be the solution of the differential equation, dydx+ytanx=2x+x2tanx, x(π2,π2), such that y(0)=1. Then :

A
y(π4)y(π4)=2
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B
y(π4)+y(π4)=π22+2
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C
y(π4)y(π4)=π2
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D
y(π4)+y(π4)=2
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Solution

The correct option is C y(π4)y(π4)=π2
Given equation is linear differential of the form dydx+Py=Q
Here, P=tanx and Q=2x+x2tanx

I.F.=etanxdx=elnsecx=secx

ysecx=(2x+x2tanx)secxdx
ysecx=x2secx+λ
y=x2+λcosx
y(0)=0+λ=1
λ=1
y=x2+cosx
y=2xsinx
y(π4)=π212
y(π4)=π2+12

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