[" (i) "y=e^(x)(a cos x+b sin x)],[" (N.C.E.R.T."" Manipur B.2019; Assam B."]

x=a cos t, y=b sin t

x = e^(-t) (a cos t + b sin t )

If y=e^(-x)(A cos x + B sin x) then y is a situation of

If y = e^(2x)(A cos 3x - B sin 3x) , then y satisfies

If y=e^(-x)(A cos x + B sin x) then y is a solution of

Derivative e^(x) cos x w.r.t. e^(x) sin x , " at" x=0 is

The differential equation of the family of curves y=e^(x)(A cos x+B sin x), where A and B are arbitrary constants is

Find the differential equation from y = e^x (A cos x + B sin x) , where A and B are arbitrary constants.

In y = e^(x) ( a cos x + b sin x) form the differential equation by eliminating arbitrary constants a and b .