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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)

best_answer

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

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

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

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)

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

Posted by

Aadil

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JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Option 1

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subhadeep

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C

Posted by

Kaveri rodage

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Option A

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Pooja borole

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