Question

Solve the following initial value problem:-

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

Answer 1

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