Expert Answer to "Find the particular solution to the differential equation...."

vetrila10

Answered question

2021-12-04

Find the particular solution to the differential equation. (Remember to use absolute values where appropriate.)

\frac{d y}{d x} = \frac{x + 1}{x y}

when x=1 , y=4

Upout1940

Beginner2021-12-05Added 9 answers

Step 1
Given,

\frac{d y}{d x} = \frac{x + 1}{x y}

, where x=1, y=4
Step 2
Now,

∴ \frac{d y}{d x} = \frac{x + 1}{x y}

\Rightarrow y d y = \frac{x + 1}{x} d x

Integrating both sides, we have

\Rightarrow \int y d y = \int \frac{x + 1}{x} d x

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

\Rightarrow \frac{y^{2}}{2} = \frac{x^{2}}{2} + \ln | x | + C

Put x=1, y=4, then

\Rightarrow \frac{{(4)}^{2}}{2} = \frac{{(1)}^{2}}{2} + \ln | 1 | + 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} + \ln | x | + \frac{15}{2}

\Rightarrow y^{2} = x^{2} + 2 \ln | x | + 15

\Rightarrow y = \sqrt{x^{2} + 2 \ln | x | + 15}

∴ y = \sqrt{x^{2} + 2 \ln | x | + 15}

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