vetrila10

2021-12-04

Find the particular solution to the differential equation. (Remember to use absolute values where appropriate.)
$\frac{dy}{dx}=\frac{x+1}{xy}$ when x=1 , y=4

Upout1940

Step 1
Given,
$\frac{dy}{dx}=\frac{x+1}{xy}$, where x=1, y=4
Step 2
Now,
$\therefore \frac{dy}{dx}=\frac{x+1}{xy}$
$⇒ydy=\frac{x+1}{x}dx$
Integrating both sides, we have
$⇒\int ydy=\int \frac{x+1}{x}dx$
$⇒\frac{{y}^{2}}{2}=\int \left(x+\frac{1}{x}\right)dx$
$⇒\frac{{y}^{2}}{2}=\frac{{x}^{2}}{2}+\mathrm{ln}|x|+C$
Put x=1, y=4, then
$⇒\frac{{\left(4\right)}^{2}}{2}=\frac{{\left(1\right)}^{2}}{2}+\mathrm{ln}|1|+C$
$⇒8=\frac{1}{2}+C$
$⇒8-\frac{1}{2}=C$
$⇒C=\frac{15}{2}$
The solution becomes :
$⇒\frac{{y}^{2}}{2}=\frac{{x}^{2}}{2}+\mathrm{ln}|x|+\frac{15}{2}$
$⇒{y}^{2}={x}^{2}+2\mathrm{ln}|x|+15$
$⇒y=\sqrt{{x}^{2}+2\mathrm{ln}|x|+15}$
$\therefore y=\sqrt{{x}^{2}+2\mathrm{ln}|x|+15}$

Do you have a similar question?