Painevg

## Answered question

2021-12-31

Solve for G.S. / P.S. for the following differential equations using the method of solution for homogeneous equations.
$-\left(3x+25y\right)dx+\left(25x+3y\right)dy=0$

### Answer & Explanation

servidopolisxv

Beginner2022-01-01Added 27 answers

Step 1
Given differential equation is $-\left(3x+25y\right)dx+\left(25x+3y\right)dy=0$
Given differential equation can be written as:
$\frac{dy}{dx}=\frac{3x+25y}{25x+3y}$
This homogeneous differential equation.
Put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
Step 2
Substituting the value, we get
$v+x\frac{dv}{dx}=\frac{3x+25vx}{25x+3vx}$
$v+x\frac{dv}{dx}=\frac{3+25v}{25+3v}$
$x\frac{dv}{dx}=\frac{3+25v}{25+3v}-v$
$x\frac{dv}{dx}=\frac{3+25v-v\left(25+3v\right)}{25+3v}$
$x\frac{dv}{dx}=\frac{3-3{v}^{2}}{25+3v}$
$\left(\frac{25+3v}{3-3{v}^{2}}\right)dv=\frac{1}{x}dx$
Step 3
Integrating both sides, we get
$\int \left(\frac{25+3v}{3-3{v}^{2}}\right)dv=\int \frac{1}{x}dx$
$\frac{1}{3}\left[\int \frac{25}{\left(1-{v}^{2}\right)}dv+\int \frac{3v}{\left(1-{v}^{2}\right)}dv\right]=\mathrm{ln}x+c$
$\frac{1}{3}\left[\frac{25}{2}\mathrm{ln}\left(\frac{1+v}{1-v}\right)-\frac{3}{2}\mathrm{ln}\left(1-{v}^{2}\right)\right]=\mathrm{ln}x+c$

Jim Hunt

Beginner2022-01-02Added 45 answers

$\frac{dy}{dx}=\frac{\left(3x+25y\right)}{\left(25x+3y\right)}$
Let $y=vx⇒\frac{dy}{dx}=v+x\frac{dv}{dx}$
$⇒v+x\frac{dv}{dx}=\frac{\left(3x+25vx\right)}{\left(25x+3vx\right)}=\frac{x\left(3+25v\right)}{x\left(25+3v\right)}$
$=\frac{\left(3+25v\right)}{\left(25+3v\right)}$
$⇒x\frac{dv}{dx}=\frac{\left(3+25v\right)}{\left(25+3v\right)}-v=\frac{3+25v-25v-3{v}^{2}}{\left(25+3v\right)}$
$⇒x\frac{dv}{dx}=\frac{3-3{v}^{2}}{25+3v}$
$⇒\frac{25+3v}{3-3{v}^{2}}dv=\frac{dx}{x}$
Take, integration both sides
$⇒\int PSK\frac{25+v}{3-3{v}^{2}}dv=\int \frac{dx}{x}$ (1)
Take, $\int \frac{25+3v}{3-3{v}^{2}}dv=\int \left(\frac{-v}{{v}^{2}-1}-\frac{25}{3\left({v}^{2}-1\right)}\right)dv$
$=-\int \frac{v}{{v}^{2}-1}dv-\frac{25}{3}\int \frac{1}{{v}^{2}-1}dv$
Take, $\int \frac{v}{{v}^{2}-1}dv=\frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}\mathrm{ln}\left(|t|\right)+{c}_{1}$.

Vasquez

Expert2022-01-09Added 669 answers

$\begin{array}{}-\left(3x+25y\right)dx+\left(25x+3y\right)dy=0\\ \frac{dy}{dx}=\frac{3x+25y}{25x+3y}\\ y=vx\\ \frac{dy}{dx}=v+x\frac{dv}{dx}\\ v+x\frac{dv}{dx}=\frac{3x+25vx}{25x+3vx}\\ v+x\frac{dv}{dx}=\frac{3+25v}{25+3v}\\ x\frac{dv}{dx}=\frac{3+25v}{25+3v}-v\\ x\frac{dv}{dx}=\frac{3+25v-v\left(25+3v\right)}{25+3v}\\ x\frac{dv}{dx}=\frac{3-3{v}^{2}}{25+3v}\\ \left(\frac{25+3v}{3-3{v}^{2}}\right)dv=\frac{1}{x}dx\\ \int \left(\frac{25+3v}{3-3{v}^{2}}\right)dv=\int \frac{1}{x}dx\\ \frac{1}{3}\left[\int \frac{25}{\left(1-{v}^{2}\right)}dv+\int \frac{3v}{\left(1-{v}^{2}\right)}dv\right]=\mathrm{ln}x+c\\ \frac{1}{3}\left[\frac{25}{2}\mathrm{ln}\left(\frac{1+v}{1-v}\right)-\frac{3}{2}\mathrm{ln}\left(1-{v}^{2}\right)\right]=\mathrm{ln}x+c\\ 25\mathrm{ln}\left(\frac{1+v}{1-v}\right)-3\mathrm{ln}\left(1-{v}^{2}\right)=6\mathrm{ln}x+C\\ v=\frac{y}{x}\\ 25\mathrm{ln}\left(\frac{1+\frac{y}{x}}{1-\frac{y}{x}}\right)-3\mathrm{ln}\left(1-\frac{{y}^{2}}{{x}^{2}}\right)=6\mathrm{ln}x+C\\ 25\mathrm{ln}\left(\frac{x+y}{x-y}\right)-3\mathrm{ln}\left(\frac{{x}^{2}-{y}^{2}}{{x}^{2}}\right)=6\mathrm{ln}x+C\end{array}$

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