Solve the following homogeneous differential equations \frac{dy}{dx}=\frac{x-y}{x+y}

Solve the following homogeneous differential equations
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}-{y}}}{{{x}+{y}}}}$$

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$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}-{y}}}{{{x}+{y}}}}$$ (1)
We have to solve the equation
Rearrange the equation
$$\displaystyle{\left({x}+{y}\right)}{\left.{d}{y}\right.}={\left({x}-{y}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle{x}{\left.{d}{y}\right.}+{y}{\left.{d}{y}\right.}={x}{\left.{d}{x}\right.}-{y}{\left.{d}{x}\right.}$$
$$\displaystyle{x}{\left.{d}{y}\right.}+{y}{\left.{d}{x}\right.}+{y}{\left.{d}{y}\right.}-{y}{\left.{d}{x}\right.}={0}$$
$$\displaystyle{d}{\left({x}{y}\right)}+{y}{\left.{d}{y}\right.}-{x}{\left.{d}{x}\right.}={0}$$
Integrating we get
$$\displaystyle{x}{y}+{\frac{{{y}^{{{2}}}}}{{{2}}}}-{\frac{{{x}^{{{2}}}+{c}}}{{{2}}}}={0}$$
$$\displaystyle{x}{y}+{\frac{{{y}^{{{2}}}}}{{{2}}}}-{\frac{{{x}^{{{2}}}}}{{{2}}}}+{c}={0}$$
It is solution of differential equation.
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Hector Roberts
Put, $$\displaystyle{y}={v}{x}$$
$$\displaystyle\Rightarrow{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={v}+{x}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}$$
$$\displaystyle\Rightarrow{v}+{x}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}={\frac{{{1}-{v}}}{{{1}+{v}}}}$$
$$\displaystyle\Rightarrow{x}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}={\frac{{{1}-{2}{v}-{v}^{{{2}}}}}{{{1}+{v}}}}$$
$$\displaystyle\Rightarrow\int{\frac{{{v}+{1}}}{{{\left({v}+{1}\right)}^{{{2}}}-{2}}}}{\left.{d}{x}\right.}=-\int{\frac{{{1}}}{{{x}}}}{\left.{d}{x}\right.}$$
$$\displaystyle\Rightarrow{\frac{{{1}}}{{{2}}}}{\ln}{\left|{\left({v}+{1}\right)}^{{{2}}}-{2}\right|}=-{\ln}{\left|{x}\right|}+{\ln}{\left|{c}_{{{1}}}\right|}$$
$$\displaystyle\Rightarrow{\ln}{\left|{\left[{\left({v}+{1}\right)}^{{{2}}}-{2}\right]}\right|}={2}{\ln}{\left|{\frac{{{c}_{{{1}}}}}{{{x}}}}\right|}$$
$$\displaystyle\Rightarrow{\left({v}+{1}\right)}^{{{2}}}-{2}={\frac{{{{c}_{{{1}}}^{{{2}}}}}}{{{x}^{{{2}}}}}}$$
$$\displaystyle\Rightarrow{x}^{{{2}}}{\left({v}^{{{2}}}+{2}{v}-{1}\right)}={C}$$
Where $$\displaystyle{C}={{c}_{{{1}}}^{{{2}}}}$$
Since, $$\displaystyle{v}={\frac{{{y}}}{{{x}}}}$$
$$\displaystyle\Rightarrow{y}^{{{2}}}+{2}{x}{y}-{x}^{{{2}}}={C}$$
karton

$$\begin{array}{}\text{Solution:} \\\frac{dy}{dx}=\frac{x-y}{x+y}=\frac{1-y/x}{1+y/x} (i) \\\text{Since, it is a homogeneous differential equation,} \\\text{Put}\ y=vx \\\Rightarrow \frac{dy}{dx}=v+x \frac{dv}{dx} \\\text{Hence, eq. (i) becomes} \\v+x \frac{dv}{dx}=\frac{1-v}{1+v} \\\Rightarrow x \frac{dv}{dx}=\frac{1-v}{1+v}-v \\\Rightarrow x \frac{dv}{dx}=\frac{1-2v-v^{2}}{1+v} \\\Rightarrow \frac{dx}{x}=\frac{-1}{2} (\frac{-2-2v}{1-2v-v^{2}})dv \\\text{On integration, we get} \\\int \frac{1}{x}dx=\frac{-1}{2} \int \frac{-2-2v}{1-2v-v^{2}}dv \\\Rightarrow \log c_{1}-2 \log x=\log (1-2v-v^{2}) \\\Rightarrow \log \frac{c_{1}}{x^{2}}=\log (\frac{x^{2}-2yx-y^{2}}{x^{2}}) \\\Rightarrow x^{2}-2yx-y^{2}=c_{1} \\\Rightarrow x^{2}-2yx-y^{2}+c=0 (c=-c_{1}) \end{array}$$