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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Differential Equations

$\large\frac{dy}{dx}$$=(4x+y+1)^2$ if $dy(0)=1$

$(a)\;4x+y+1= 2\tan (2x+\frac{\pi}{4}) \\ (b)\;4x+y+1= 2\tan (x+\frac{\pi}{4}) \\ (c)\;4x+y+1= 2 \cot (x+\frac{\pi}{4}) \\ (d)\;4x+y+1= 2 \cot (2x+\frac{\pi}{4}) $

1 Answer

$4x+y+1=t$
$4 + \large\frac{dy}{dx}= \frac{dt}{dx}$
$\large\frac{dt}{dx}$$=4+t^2$
$\int \large\frac{dt}{4+t^2}$$=\int dx$
$\large\frac{1}{2}$$ \tan ^{-1}(\large\frac{t}{2})$$=x+c$
$\tan ^{-1} \bigg(\large\frac{4x+y+1}{2}\bigg) $$=2x+2c$
$\large\frac{4x+y+1}{2} $$=\tan (2x+c)$
$y(0)=1$
$\large\frac{1+1}{2}$$=\tan c$
$c= \large\frac{\pi}{4}$
$4x+y+1= 2\tan (2x+\frac{\pi}{4})$
Hence a is the correct answer.
answered Feb 4, 2014 by meena.p
 

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