Question

The differential equation of the family of curves y = e^x (A cos x + B sin x). Where A and B are arbitary constants is ______.

पर्याय

• y'' = 2y' + 2y = 0

• y'' + 2y' - 2y = 0

• y'' + y'² + y = 0

• y'' + 2y' - y = 0

Answer 1

The differential equation of the family of curves y = e^x (A cos x + B sin x). Where A and B are arbitary constants is y'' = 2y' + 2y = 0.

Explanation:

y = e^x(A cos x + B sin x)

⇒ y' = e^x(A cos x + B sin x) + y = e^x(B cos x - A sin x)

⇒ y' = y + e^x(B cos x - A sin x)     ...(i)

∴ y'' = y' + e^x(B cos x - A sin x) - e^x(A cos x - B sin x)

⇒ y'' = y' + (y' - y) - y    ....[From (i)]

⇒ y'' - 2y' + 2y = 0