Question

Solve the following initial value problem:-

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

Answer 1

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, 
\[y\left( 0 \right) = \frac{1}{2}\]
\[ \therefore \frac{1}{2} e^0 = \frac{e^0}{2} + C\]
\[ \Rightarrow C = 0\]
\[\text{ Putting the value of C in }\left( 2 \right),\text{ we get }\]
\[y e^x = \frac{e^{2x}}{2}\]
\[ \Rightarrow e^x = \frac{e^x}{2}\]
\[\text{ Hence, }y = \frac{e^x}{2}\text{ is the required solution.}\]