Question

$$\frac{dy}{dx}+y=4x$$

Answer 1

We have,

$$\frac{dy}{dx}+y=4x.....\left(1\right)$$

Clearly, it is a linear differential equation of the form

$$\frac{dy}{dx}+Py=Q$$

$$\text{where }P=1\text{ and }Q=4x$$

$$\therefore I.F.=e^{\int Pdx}$$

$$=e^{\int dx}$$

$$=e^x$$

$$\text{Multiplying both sides of (1) by }I.F.=e^x,\text{ we get}$$

$$e^x(\frac{dy}{dx}+y)=e^{x}4x$$

$$\Rightarrow e^x\frac{dy}{dx}+e^{x}y=e^{x}4x$$

Integrating both sides with respect to x, we get

$$y{e}^x=4\int x{e}^{x}dx+C$$

$$\Rightarrow y{e}^x=4x\int e^{x}dx-4\int [\frac{d}{dx}\left(x\right)\int e^{x}dx]dx+C$$

$$\Rightarrow y{e}^x=4x{e}^x-4\int e^{x}dx+C$$

$$\Rightarrow y{e}^x=4x{e}^x-4e^x+C$$

$$\Rightarrow y{e}^x=4(x-1)e^x+C$$

$$\Rightarrow y=4(x-1)+C{e}^{-x}$$

$$\text{Hence, }y=4(x-1)+C{e}^{-x}\text{ is the required solution.}$$