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Verify that the given function(explicit or implicit)is a solution of the corresponding differential equation $y=x\sin x\qquad:xy'=y+x\sqrt{x^2-y^2} (x\neq0\;and\;x>y\;or\;x<-y)$

1 Answer

  • Whenever the function is a product of two functions $u$ and $v$ , we apply the product rule which states differentiation of $uv = uv' + vu'. $
Step 1:
Here it is given $y = x\sin x$
We have two functions $x$ and $\sin x$
Applying product rule:
Let $u = x$. Hence $u' = 1$ and $v = \sin x$ and $v' = \cos x$
Product rule is $uv' + vu'$
Applying this we get
$x\cos x + \sin x.1$
$y' = x\cos x + \sin x$
Step 2:
Multiply by $x$ on both sides we get
$xy' = x^2\cos x + x\sin x$
But $x\sin x = y$ and $x^2\cos^2x$, but $\cos^2x = 1- sin^2x$
Hence $x^2 - y^2 = x^2 - x^2sin^2x$
Substituting this we get
$xy' = y + x \sqrt{x^2 - y^2}$.
Hence the answer is the solution for the given function.
answered Aug 19, 2013 by sreemathi.v

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