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.