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Solving the Homogeneous differential equations

A method of solving a differential equation that consists of homogeneous functions in two variables is called the method of solving homogeneous differential equations.

Method

There are four simple steps that we need to follow for solving any homogenous differential equation.

  1. Separate the differentials from the homogeneous functions.
  2. Eliminate the dependent variable by its equivalent value from the homogenous functions.
  3. Evaluate the derivative of product of the functions by the product rule of differentiation.
  4. Solve the differential equation by the variable separable method.

Let’s understand the above steps from an example homogeneous differential equation problem.

$(x^2-y^2)dx-xy dy \,=\, 0$

Separate differentials and homogeneous functions

In this example problem, the differentials $dx$ and $dy$ are mixed with the homogenous functions $x^2-y^2$ and $xy$. The differentials must be separated from the homogeneous functions for solving it.

$\implies$ $(x^2-y^2)dx$ $\,=\,$ $xy dy$

$\implies$ $xy dy$ $\,=\,$ $(x^2-y^2)dx$

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{x^2-y^2}{xy}$

It is given that the separated homogeneous functions and differentials in some problems. In such cases, skip to second step. In such cases, skip the first step.

Eliminate a variable by its equivalent value

The mathematical formation of a homogenous function creates the problem for the integration. In order to overcome it, a variable should be eliminated from the function and it is possible by expressing it in terms of another variable. In mathematics, the variables $x$ and $y$ are used to represent an independent variable and dependent variable respectively. In other words, the value of $y$ always depends on the value of the variable $x$.

The mathematical relation between the variables $x$ and $y$ is unknown. Hence, let us take $y \,=\, vx$ where $v$ is also a variable. Now, the variable $y$ can be eliminated by substituting its equivalent value.

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x^2-(vx)^2}{x(vx)}$

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x^2-v^2x^2}{x \times v \times x}$

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x^2-v^2x^2}{v \times x \times x}$

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x^2-v^2x^2}{v \times x^2}$

Now, simplify the expression on the right-hand side of the equation.

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x^2 \times 1-v^2 \times x^2}{v \times x^2}$

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x^2 \times (1-v^2)}{v \times x^2}$

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{\cancel{x^2} \times (1-v^2)}{v \times \cancel{x^2}}$

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{1-v^2}{v}$

Find the differentiation of the product

The expression on the right-hand side of the equation is simplified and it is time to evaluate the left-hand side expression. It is a product of two functions and its differentiation can be calculated by the product rule.

$\implies$ $v \times \dfrac{dx}{dx}$ $+$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1-v^2}{v}$

$\implies$ $v \times 1$ $+$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1-v^2}{v}$

$\implies$ $v$ $+$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1-v^2}{v}$

Solve the equation by the variables separable

Use the separation of variables method for expressing the function in one variable with its corresponding differential on one side of the equation and the function in another variable with its respective differential on other side of the equation.

$\implies$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1-v^2}{v}-v$

$\implies$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1-v^2-v^2}{v}$

$\implies$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1-2v^2}{v}$

$\implies$ $\dfrac{1}{\dfrac{1-2v^2}{v}} \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1}{x}$

$\implies$ $\dfrac{1}{\dfrac{1-2v^2}{v}} \times dv$ $\,=\,$ $\dfrac{1}{x} \times dx$

$\implies$ $\dfrac{v}{1-2v^2} \times dv$ $\,=\,$ $\dfrac{1}{x} \times dx$

$\implies$ $\dfrac{v}{1-2v^2}\,dv$ $\,=\,$ $\dfrac{1}{x}\,dx$

$\implies$ $\displaystyle \int{\dfrac{v}{1-2v^2}\,}dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

Take $z \,=\, 1-2v^2$ and differentiate both sides of the equation with respect to $v$.

$\implies$ $\dfrac{d}{dv}{(z)} \,=\, \dfrac{d}{dv}{(1-2v^2)}$

$\implies$ $\dfrac{dz}{dv} \,=\, \dfrac{d}{dv}{(1)}-\dfrac{d}{dv}{(2v^2)}$

$\implies$ $\dfrac{dz}{dv} \,=\, 0-2\dfrac{d}{dv}{(v^2)}$

$\implies$ $\dfrac{dz}{dv} \,=\, (-2) \times 2 \times v^{2-1}$

$\implies$ $\dfrac{dz}{dv} \,=\, -4 \times v^{1}$

$\implies$ $\dfrac{dz}{dv} \,=\, -4 \times v$

$\implies$ $\dfrac{dz}{-4} \,=\, v \times dv$

$\implies$ $v \times dv \,=\, \dfrac{dz}{-4}$

Now, express the left-hand side integral expression in terms of $z$.

$\implies$ $\displaystyle \int{\dfrac{\dfrac{dz}{-4}}{z}\,}$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $\displaystyle \int{\dfrac{\dfrac{1 \times dz}{-4}}{z}\,}$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $\displaystyle \int{\dfrac{\dfrac{1}{-4} \times dz}{z}\,}$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $\displaystyle \int{\Bigg(\dfrac{1}{-4} \times \dfrac{dz}{z}\Bigg)\,}$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $\displaystyle \int{\Bigg(\dfrac{-1}{4} \times \dfrac{dz}{z}\Bigg)\,}$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $\dfrac{-1}{4} \times \displaystyle \int{\dfrac{dz}{z}\,}$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $\dfrac{-1}{4} \times \displaystyle \int{\dfrac{1}{z}\times dz\,}$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $\dfrac{-1}{4} \times \Big(\log_{e}{(z)}+c_1\Big)$ $\,=\,$ $\log_{e}{(x)}+c_2$

$\implies$ $(-1) \times \Big(\log_{e}{(z)}+c_1\Big)$ $\,=\,$ $4\Big(\log_{e}{(x)}+c_2\Big)$

$\implies$ $(-1) \times \log_{e}{(z)}$ $+$ $(-1) \times c_1$ $\,=\,$ $4 \times \log_{e}{(x)}$ $+$ $4 \times c_2$

$\implies$ $-\log_{e}{(z)}$ $-$ $c_1$ $\,=\,$ $4 \times \log_{e}{(x)}$ $+$ $4c_2$

$\implies$ $-\log_{e}{(z)}$ $-$ $c_1$ $\,=\,$ $\log_{e}{(x^4)}$ $+$ $4c_2$

$\implies$ $-\log_{e}{(1-2v^2)}$ $-$ $c_1$ $\,=\,$ $\log_{e}{(x^4)}$ $+$ $4c_2$

$\implies$ $-$ $c_1$ $-$ $4c_2$ $\,=\,$ $\log_{e}{(x^4)}$ $+$ $\log_{e}{(1-2v^2)}$

$\implies$ $\log_{e}{(x^4)}$ $+$ $\log_{e}{(1-2v^2)}$ $\,=\,$ $-$ $c_1$ $-$ $4c_2$

$\implies$ $\log_{e}{(x^4 \times (1-2v^2))}$ $\,=\,$ $-$ $c_1$ $-$ $4c_2$

$\implies$ $\log_{e}{(x^4 \times (1-2v^2))}$ $\,=\,$ $-(c_1+4c_2)$

$\implies$ $x^4 \times (1-2v^2)$ $\,=\,$ $e^{\displaystyle -(c_1+4c_2)}$

$\implies$ $x^4 \times (1-2v^2)$ $\,=\,$ $c$

$\implies$ $x^4 \times \Bigg(1-2\bigg(\dfrac{y}{x}\bigg)^2\Bigg)$ $\,=\,$ $c$

$\implies$ $x^4 \times \Bigg(1-2\bigg(\dfrac{y^2}{x^2}\bigg)\Bigg)$ $\,=\,$ $c$

$\implies$ $x^4 \times \Bigg(\dfrac{x^2-2y^2}{x^2}\Bigg)$ $\,=\,$ $c$

$\implies$ $\dfrac{x^4 \times (x^2-2y^2)}{x^2}$ $\,=\,$ $c$

$\implies$ $\dfrac{\cancel{x^4} \times (x^2-2y^2)}{\cancel{x^2}}$ $\,=\,$ $c$

$\implies$ $x^2 \times (x^2-2y^2)$ $\,=\,$ $c$

$\,\,\,\therefore\,\,\,\,\,\,$ $x^2(x^2-2y^2)$ $\,=\,$ $c$

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