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Questions  >>  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\[y=\sqrt{1+x^2}\qquad:\;y'\;=\frac{xy}{1+x^2}\]

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Toolbox:
  • Differentiation of $\sqrt {1+ x^2}$ is $\large\frac{2x}{2\sqrt{ (1+x^2)} } =\large\frac{ x}{1+x^2}$
Step 1:
Given $y = \sqrt{1 + x^2}$
Differentiating on both sides we get
$y' =\large\frac{ 1}{2}$$ (1 + x^2)^{\Large\frac{-1}{2}}$$ \times 2x$
$y' = \large\frac{x}{\sqrt{1 + x^2}}$
Step 2:
Multiply the numerator and denominator by $\sqrt{1+x^2}$ we get
$y' =\large\frac{x\sqrt{ (1+x^2)}}{(1+x^2)} = \frac{xy}{1+x^2}$
Hence the answer is a solution to the given function.
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