# Show that

Question:

$y=e^{x}(a \cos x+b \sin x)$

Solution:

$y=e^{x}(a \cos x+b \sin x)$             ...(1)

Differentiating both sides with respect to x, we get:

$y^{\prime}=e^{x}(a \cos x+b \sin x)+e^{x}(-a \sin x+b \cos x)$

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

Again, differentiating with respect to x, we get:

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

$y^{\prime \prime}=e^{x}[2 b \cos x-2 a \sin x]$

$y^{\prime \prime}=2 e^{x}(b \cos x-a \sin x)$

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

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

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

$\Rightarrow y+\frac{y^{\prime \prime}}{2}=y^{\prime}$

$\Rightarrow 2 y+y^{\prime \prime}=2 y^{\prime}$

$\Rightarrow y^{\prime \prime}-2 y^{\prime}+2 y=0$

This is the required differential equation of the given curve.

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