If y = y(x) is the solution of the differential equation 

\frac{dy}{dx}=(tanx-y)sec^{2}x

x\epsilon (\frac{-\pi}{2},\frac{\pi}{2}), such that y(0) = 0 , then 

y(\frac{-\pi}{4}) is equal to : 

  • Option 1)

    e-2

  • Option 2)

    \frac{1}{2}-e

  • Option 3)

    2+\frac{1}{e}

  • Option 4)

    \frac{1}{e}-2

 

Answers (1)
V Vakul

Given differential eqn

\frac{dy}{dx}=(tanx-y)sec^{2}x

=>\frac{dy}{dx}+y sec^{2}x+=tanxsec^{2}x

I.F. = e^{\int sec^{2}x dx }e^{tanx}

Solution of given differential eqn

y\cdot e^{tanx}=\int tanx \cdot sec^{2}x\cdot e^{tanx} dx

                =tanx\cdot -\int sec^{2}x\cdot e^{tanx} dx

               = e^{tanx} (tanx-1)

Given that y(0)=0 => C=1

y\cdot e^{tanx}=e^{tanx}(tanx-1)+1

put x=\frac{-\pi}{4}

y(\frac{-\pi}{4})=e-2

So, correct option is (1)


Option 1)

e-2

Option 2)

\frac{1}{2}-e

Option 3)

2+\frac{1}{e}

Option 4)

\frac{1}{e}-2

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