Question

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b:$y=e^x(a\cos x+b\sin x)$

Answer 1

$y=e^x(a\cos x+b\sin x)$...(1)
Differentiating both sides with respect to x,we get:
$y'=e^x(a\cos x+b\sin x)+e^x (-a \sin x+b\cos x)$

$\Rightarrow y'=e^x\bigg[(a+b)\cos x-(a-b)\sin x\bigg]$...(2)
Again, differentiating with respect to x, we get:

$y''=e^x\bigg[(a+b)\cos x-(a-b)\sin x\bigg]+e^x\bigg[-(a+b)\sin x-(a-b)\cos x\bigg]$

$y''=e^x\bigg[2b\cos x-2a \sin x\bigg]$

$y''=2e^x[b\cos x-a\sin x]$

$\Rightarrow \frac{y''}{2}=e^x(b\cos x- a\sin x)$...(3)

Adding equations(1)and(3),we get:

$y+\frac{y''}{2}=e^x[(a+b)\cos x-(a-b)\sin x]$

$\Rightarrow$ $y+\frac{y''}{2}=y'$

$\Rightarrow$ 2y+y''=2y'

$\Rightarrow$ y''-2y'+2y=0

This is the required differential equation of the given curve.