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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation

$x+y = \tan^{-1}y$ $\:$:$\:$ $y^2y'+y^2+1=0$

$\begin{array}{1 1} (A)\;y'y^2 + y^2 + 1 = 0 \\ (B)\;y'y - y^2 + 1 = 0 \\(C)\;y'y^2 - y^2 - 1 = 0 \\ (D)\;y'y^2 + y^2 - 1 = 0\end{array}$

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1 Answer

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Toolbox:
  • Differentiation of $\tan^{-1} y$ is $[\large\frac{1}{1+y^2}]$$ . y'$
Step 1:
Given: $x + y = \tan^{-1}y$
Differentiating on both sides we get,
$1 + y' =\large\frac{ 1}{1 + y^2}$$ . y'$
$y'\big[\large\frac{ 1}{1+y^2} \big]$$- 1 = 1$
Step 2:
On simplifying we get
$y'\large\frac{[1 -1 -y^2]}{1 + y^2}$$ = 1$
$y' \big[ \large\frac{-y^2}{1+y^2}\big] $$= 1$
On cross multiplication we get,
$y'y^2 + y^2 + 1 = 0$
Hence the solution is verified.
answered Aug 16, 2013 by sreemathi.v
 

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