Q. 37 B5.0( 1 Vote )

# Solve each of the following initial value problems:

, y(1) = 0

Answer :

, y(1) = 0

Given and y(1) = 0

This is a first order linear differential equation of the form

Here, and

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

We have

[∵ m log a = log a^{m}]

∴ I.F = x^{–1} [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

∴ y = –log x – 1 + cx

However, when x = 1, we have y = 0

⇒ 0 = –log 1 – 1 + c(1)

⇒ 0 = –0 – 1 + c

⇒ 0 = –1 + c

∴ c = 1

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

y = –log x – 1 + (1)x

⇒ y = –log x – 1 + x

∴ y = x – 1 – log x

Thus, the solution of the given initial value problem is y = x – 1 – log x

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