# Solve the Following Initial Value Problem:- Y ′ + Y = E X , Y ( 0 ) = 1 2 - Mathematics and Statistics

Sum

Solve the following initial value problem:-

$y' + y = e^x , y\left( 0 \right) = \frac{1}{2}$

#### Solution

We have,
$y' + y = e^x$
$\Rightarrow \frac{dy}{dx} + y = e^x . . . . . \left( 1 \right)$
Clearly, it is a linear differential equation of the form
$\frac{dy}{dx} + Py = Q$
$\text{ where }P = 1\text{ and }Q = e^x$
$\therefore I . F . = e^{\int P\ dx}$
$= e^{\int1 dx}$
$= e^x$
$\text{ Multiplying both sides of }\left( 1 \right)\text{ by }I . F . = e^x ,\text{ we get }$
$e^x \left( \frac{dy}{dx} + y \right) = e^x e^x$
$\Rightarrow e^x \frac{dy}{dx} + e^x y = e^{2x}$
Integrating both sides with respect to x, we get
$y e^x = \int e^{2x} dx + C$
$\Rightarrow y e^x = \frac{e^{2x}}{2} + C . . . . . \left( 2 \right)$
Now,
$y\left( 0 \right) = \frac{1}{2}$
$\therefore \frac{1}{2} e^0 = \frac{e^0}{2} + C$
$\Rightarrow C = 0$
$\text{ Putting the value of C in }\left( 2 \right),\text{ we get }$
$y e^x = \frac{e^{2x}}{2}$
$\Rightarrow e^x = \frac{e^x}{2}$
$\text{ Hence, }y = \frac{e^x}{2}\text{ is the required solution.}$

Concept: Basic Concepts of Differential Equation
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Exercise 22.1 | Q 37.01 | Page 107