Q. 37 C5.0( 1 Vote )

Solve each of the following initial value problems:

, y(0) = 0

Answer :

, y(0) = 0


Given and y(0) = 0



This is a first order linear differential equation of the form



Here, P = 2 and Q = e–2xsin x


The integrating factor (I.F) of this differential equation is,





We have


I.F = e2x


Hence, the solution of the differential equation is,





Recall


ye2x = –cos x + c


ye2x × e–2x = (–cos x + c) × e–2x


y = (–cos x + c)e–2x


However, when x = 0, we have y = 0


0 = (–cos 0 + c)e0


0 = (–1 + c) × 1


0 = –1 + c


c = 1


By substituting the value of c in the equation for y, we get


y = (–cos x + 1)e–2x


y = (1 – cos x)e–2x


Thus, the solution of the given initial value problem is y = (1 – cos x)e–2x


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