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

The correct option is B y(π4)y(π4)=π2
I.F=etanxdx=elnsecx=secx
y.secx=(2x+x2tanx)secxdx=2xsecxdx+x2(secx.tanxdx
ysecx=x2secx+λy=x2+λcosx
y(0)=0+λ=1λ=1
y=x2+cosx
y(π4)=π216+12;y(π4)=π216+12
y(x)=2xsinx
y(π4)=π212;y(π4)=π212
y(π4)y(π4)=π2

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