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Login Now**Definition of Initial Value Problem:**

A differential equation together with initial condition(s) is called the initial value problem. For example,

\(\frac{d^2y}{d^2x} = 2\frac{dy}{dx} + y = 0, y(0) = -1, y^{‘}(0) = 1\)

Here \(y(0) = -1, y^{‘}(0) = 1\) is an initial conditions

**Problem Part:**

Solution:

Given that

y^{‘} + 5y = 1 …… (i)

y(0) = 2 …….(ii)

Since (i) is linear differential equation of first order.

Comparing (i) with y’ + Py = Q then

P = 5, Q = 1

Then

\(\int P dx = 5 \int dx = 5x\)

Here, the integrating factor (I.F) of (i) is

\(I.F. = e^{\int Pdx} = e^{5x}\)

Now, Multiplying (i) by I.F. and then integrating we get,

\(y \times I.F. = \int(Q \times I.F.)dx + C\)

\(i.e \enspace ye^{5x} = \int (1) (e^{5x}) dx + C = \frac{e^{5x}}{5} + C\) …..(iii)

By (ii), we have y(0) = 2, then (iii) becomes,

\(ye^{5x} = \frac{e^{5x}}{5} + \frac{9}{5}\)

\(\Rightarrow y = \frac{1}{5} (1 + 9e^{-5x})\)

This is the solution of (i) that stratifies (ii).

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