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

Beginner2021-12-05Added 9 answers

Step 1

Given,

$\frac{dy}{dx}=\frac{x+1}{xy}$ , where x=1, y=4

Step 2

Now,

$\therefore \frac{dy}{dx}=\frac{x+1}{xy}$

$\Rightarrow ydy=\frac{x+1}{x}dx$

Integrating both sides, we have

$\Rightarrow \int ydy=\int \frac{x+1}{x}dx$

$\Rightarrow \frac{{y}^{2}}{2}=\int (x+\frac{1}{x})dx$

$\Rightarrow \frac{{y}^{2}}{2}=\frac{{x}^{2}}{2}+\mathrm{ln}\left|x\right|+C$

Put x=1, y=4, then

$\Rightarrow \frac{{\left(4\right)}^{2}}{2}=\frac{{\left(1\right)}^{2}}{2}+\mathrm{ln}\left|1\right|+C$

$\Rightarrow 8=\frac{1}{2}+C$

$\Rightarrow 8-\frac{1}{2}=C$

$\Rightarrow C=\frac{15}{2}$

The solution becomes :

$\Rightarrow \frac{{y}^{2}}{2}=\frac{{x}^{2}}{2}+\mathrm{ln}\left|x\right|+\frac{15}{2}$

$\Rightarrow {y}^{2}={x}^{2}+2\mathrm{ln}\left|x\right|+15$

$\Rightarrow y=\sqrt{{x}^{2}+2\mathrm{ln}\left|x\right|+15}$

$\therefore y=\sqrt{{x}^{2}+2\mathrm{ln}\left|x\right|+15}$

Given,

Step 2

Now,

Integrating both sides, we have

Put x=1, y=4, then

The solution becomes :

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