# Determine which of the following differential equations are homogeneous, then

Determine which of the following differential equations are homogeneous, then solve the general solution. If initial condition is given, find the particular solution:
$$\displaystyle{x}{y}{\left.{d}{x}\right.}-{\left({x}+{2}{y}\right)}^{{{2}}}{\left.{d}{y}\right.}={0}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Jim Hunt

$$\displaystyle{x}{y}{\left.{d}{x}\right.}-{\left({x}+{2}{y}\right)}^{{{2}}}{\left.{d}{y}\right.}={0}$$
$$\displaystyle{x}{y}{\left.{d}{x}\right.}={\left({x}+{2}{y}\right)}^{{{2}}}{\left.{d}{y}\right.}$$
$$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{x}^{{{2}}}+{4}{y}^{{{2}}}+{4}{x}{y}}}{{{x}{y}}}}$$
$$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\left({\frac{{{x}}}{{{y}}}}\right)}+{4}{\left({\frac{{{y}}}{{{x}}}}\right)}+{4}$$
It is a Homogeneous differential equation.
Let $$\displaystyle{\frac{{{x}}}{{{y}}}}={v}\Rightarrow{x}={v}{y}$$
So, $$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={v}+{y}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}$$
$$\displaystyle\therefore v+{y}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}=v+{\frac{{{4}}}{{{v}}}}+{4}$$
$$\displaystyle{y}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}={4}{\left({\frac{{{v}+{1}}}{{{v}}}}\right)}$$
$$\displaystyle{\frac{{{v}{d}{v}}}{{{v}+{1}}}}={4}{\frac{{{\left.{d}{y}\right.}}}{{{y}}}}$$
It becomes variable separable equation. So, now integrate both sides.
$$\displaystyle\int{\frac{{{v}{d}{v}}}{{{v}+{1}}}}={4}\int{\frac{{{\left.{d}{y}\right.}}}{{{y}}}}$$
$$\displaystyle\int{\left({\frac{{{v}+{1}-{1}}}{{{v}+{1}}}}\right)}{d}{v}={4}\int{\frac{{{\left.{d}{y}\right.}}}{{{y}}}}$$
$$\displaystyle\int{\left({1}-{\frac{{{1}}}{{{v}+{1}}}}\right)}{d}{v}={4}{\ln}{\left|{y}\right|}+{\ln{{c}}}$$
$$\displaystyle{v}-{\ln}{\left|{v}+{1}\right|}={\ln{{\left({y}^{{{4}}}{c}\right)}}}$$
$$\displaystyle\therefore{\frac{{{x}}}{{{y}}}}+{\ln}{\left|{\frac{{{x}}}{{{y}}}}+{1}\right|}={\ln}{\left|{y}^{{{4}}}{c}\right|}$$
$$\displaystyle{\frac{{{x}}}{{{y}}}}={\ln}{\left|{y}^{{{4}}}{c}\right|}-{\ln}{\left|{\frac{{{x}+{y}}}{{{y}}}}\right|}$$
$$\displaystyle{\frac{{{x}}}{{{y}}}}={\ln}{\left|{\frac{{{\left({y}^{{{4}}}{c}\right)}}}{{{\left({\frac{{{x}+{y}}}{{{y}}}}\right)}}}}\right|}$$
$$\displaystyle{x}={y}{\ln}{\left|{\frac{{{c}{y}^{{{5}}}}}{{{x}+{y}}}}\right|}$$

###### Not exactly what you’re looking for?
Serita Dewitt
$$\displaystyle{x}{y}{\left.{d}{x}\right.}-{\left({x}+{2}{y}\right)}^{{{2}}}{\left.{d}{y}\right.}={0}$$
$$\displaystyle{x}{y}{\left.{d}{x}\right.}={\left({x}+{2}{y}\right)}^{{{2}}}{\left.{d}{y}\right.}$$
$$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{\left({x}+{2}{y}\right)}^{{{2}}}}}{{{x}{y}}}}$$
$$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{x}^{{{2}}}+{4}{y}^{{{2}}}+{4}{x}{y}}}{{{x}{y}}}}$$
$$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={\frac{{{x}}}{{{y}}}}+{4}{\frac{{{y}}}{{{x}}}}+{4}$$.
Substituting; $$\displaystyle{x}={y}{v}$$
Diff. w.r.t. y
$$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}={v}+{y}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}$$
$$\displaystyle{v}+{y}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}={v}+{4}{\frac{{{1}}}{{{v}}}}+{4}$$
$$\displaystyle\neg{\left\lbrace{v}\right\rbrace}+{y}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}=\neg{\left\lbrace{v}\right\rbrace}+{4}{\left({\frac{{{1}}}{{{v}}}}+{1}\right)}$$
$$\displaystyle{\frac{{{y}}}{{{4}}}}{\frac{{{d}{v}}}{{{\left.{d}{y}\right.}}}}={\frac{{{1}+{v}}}{{{v}}}}$$
Integrating $$\displaystyle\int{\frac{{{v}}}{{{v}+{1}}}}{d}{v}={4}\int{\frac{{{1}}}{{{y}}}}{\left.{d}{y}\right.}$$
$$\displaystyle\int{\frac{{{v}+{1}-{1}}}{{{v}+{1}}}}{d}{v}={4}\int{\frac{{{1}}}{{{y}}}}{\left.{d}{y}\right.}$$
$$\displaystyle{v}-{\ln}{\left|{v}+{1}\right|}={4}{\log{{y}}}+{c}$$
applying $$\displaystyle{v}={\frac{{{x}}}{{{y}}}}$$
$$\displaystyle{\frac{{{x}}}{{{y}}}}-{\ln}{\left|{\frac{{{x}}}{{{y}}}}+{1}\right|}={4}{\log{{y}}}+{c}$$
karton

$$\begin{array}{} xydx-(x+2y)^{2}dy=0 \\xydx=(x+2y)^{2}dy \\\frac{dx}{dy}=\frac{x^{2}+4y^{2}+4xy}{xy} \\\frac{dx}{dy}=(\frac{x}{y})+4(\frac{y}{x})+4 \\\frac{x}{y}=v \Rightarrow x=vy \\\frac{dx}{dy}=v+y \frac{dv}{dy} \\\therefore \not{v}+y \frac{dv}{dy}=\not{v}+\frac{4}{v}+4 \\y \frac{dv}{dy}=4 (\frac{v+1}{v}) \\\frac{vdv}{v+1}=4 \frac{dy}{y} \\\int \frac{vdv}{v+1}=4 \int \frac{dy}{y} \\\int (\frac{v+1-1}{v+1})dv=4 \int \frac{dy}{y} \\\int (1-\frac{1}{v+1})dv=4 \ln |y|+\ln c \\v-\ln |v+1|=\ln (y^{4}c) \\\therefore \frac{x}{y}+\ln |\frac{x}{y}+1|=\ln |y^{4}c| \\\frac{x}{y}=\ln |y^{4}c|-\ln |\frac{x+y}{y}| \\\frac{x}{y}=\ln |\frac{(y^{4}c)}{(\frac{x+y}{y})}| \\x=y \ln |\frac{cy^{5}}{x+y}| \end{array}$$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}