Q. 37 C5.0( 1 Vote )

# Solve each of the following initial value problems:, y(0) = 0

, 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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