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Confused! kindly explain, - Differential equations - JEE Main-5

Let y=y(x) be the solution of the fifferential equation,

\frac{\mathrm{d} y}{\mathrm{d} x}+ytanx=2x+x^{2}tanx, x\epsilon (\frac{-\pi}{2},\frac{\pi}{2}),

such that y(0)=1. Then :

  • Option 1)

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

  • Option 2)

    y'(\frac{\pi}{4})+y'(\frac{-\pi}{4})=-\sqrt2

  • Option 3)

    y(\frac{\pi}{4})-y(\frac{-\pi}{4})=\sqrt2

  • Option 4)

    y'(\frac{\pi}{4})-y'(\frac{-\pi}{4})=\pi-\sqrt2

 
Answers (4)
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Option A

C

Option 1

\frac{\mathrm{d} y}{\mathrm{d} x}+ytanx=2x+x^{2}tanx,

I.F.=e^{\int tanxdx}

I.F.=e^{\int tanxdx}=e^{\ln secx}=secx

Solution of DE

y\: secx=\int (2x+x^{2}tanx)secx \: dx+C

y\: secx=\int (2xsecx\: dx)+\int x^{2}tanx\: secx \: dx+C

y\: secx=x^{2}secx+C

Given y(0)=1

=>y\: =0^{2}+C\: cos0

=> C\: =1

y\: =x^{2}+cosx

y'=2x-sinx

y'(\frac{\pi}{4})=2(\frac{\pi}{4})-sin(\frac{\pi}{4})=2(\frac{\pi}{4})-\frac{1}{\sqrt2}

y'(-\frac{\pi}{4})=2(-\frac{\pi}{4})-sin(-\frac{\pi}{4})=-2(\frac{\pi}{4})+\frac{1}{\sqrt2}


Option 1)

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

Option 2)

y'(\frac{\pi}{4})+y'(\frac{-\pi}{4})=-\sqrt2

Option 3)

y(\frac{\pi}{4})-y(\frac{-\pi}{4})=\sqrt2

Option 4)

y'(\frac{\pi}{4})-y'(\frac{-\pi}{4})=\pi-\sqrt2

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