Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The solution of x \frac{d y}{d x}+y=e^{x} is

(a) y=\frac{e^{x}}{x}+\frac{k}{x}
(b) y=x e^{x}+c x
(c) y=x e^{x}+k
(d) x=\frac{e^{y}}{y}+\frac{k}{y}

Answers (1)

The answer is the option (a) y = \frac{e^{x}}{x} + \frac{k}{x}

Explanation: -

\Rightarrow \quad \frac{d y}{d x}+\frac{y}{x}=\frac{e^{x}}{x}
This is a linear differential equation. On comparing it with \frac{d y}{d x}+P y=Q, we get
P=\frac{1}{x} \text { and } Q=\frac{e^{x}}{x}
\therefore \quad \mathrm{IF}_{0}=e^{\int \frac{1}{x} d x}=e^{(\log x)}=x
So, the general solution is:
\\y \cdot x=\int \frac{e^{x}}{x} x d x \\\Rightarrow \quad y \cdot x=\int e^{x} d x \\\Rightarrow \quad \quad y \cdot x=e^{x}+k \\\Rightarrow y=\frac{e^{x}}{x}+\frac{k}{x}

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question