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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b $y=e^x(a\;\cos x+b\;\sin x)$

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Toolbox:
  • When a function occurs in product we can apply the product rule which states $uv = uv' + vu'$
Step 1:
Given : $y = e^x(a\cos x + b\sin x)$
Differentiating on both sides by applying the product rule we get,
$y' = e^x(a\cos x + b\sin x) + e^x(-a\sin x + b\cos x)$
$y - y' = e^x(b\cos x - a\sin x)$------(1)
Step 2:
Again differentiating y' on both sides we get
$y'' = e^x(a\cos x +b\sin x) + e^x(-\sin x + b\cos x) + e^x (-a\sin x + b\cos x) - e^x(a\cos x + b\sin x)$
On simplifying we get
$y'' = 2e^x(b\cos x - a\sin x)$
Substituting equ(1) we get y'' = 2(y' - y)
$y'' -2y' +2y = 0$
answered Aug 16, 2013 by sreemathi.v
 

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