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

In this differential equation, $xy$ and $x^2+2y^2$ are two homogeneous functions. The homogeneous function $xy$ is multiplied by the differential $dx$ and the homogeneous function $x^2+2y^2$ is multiplied by another differential $dy$. It is given that the difference of the products is equal to zero.

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

This differential equation is an example for a homogenous differential equation. So, let’s learn how to solve the homogenous equation mathematically from the step by step procedure.

Separate the differentials from homogenous functions

The expressions in both variables $x$ and $y$ are multiplied by the differentials. So, it is not mathematically acceptable to integrate a function in one variable with respect to another variable. Hence, it is recommendable to separate the differentials from homogenous functions. It can be done by shifting the differentials to one side of the equation and moving the homogenous functions to another side of the equation.

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

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

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

The differentials $dx$ and $dy$ are successfully separated from the homogeneous functions $xy$ and $x^2+2y^2$ by expressing them in ratio form.

Eliminate the dependent variable by its equivalent value

$\implies$ $\dfrac{d}{dx}{\,(y)}$ $\,=\,$ $\dfrac{xy}{x^2+2y^2}$

In this equation, the variable $x$ is an independent variable, but the variable $y$ is a dependable on $x$. Hence, the variable $y$ can be eliminated by taking an equivalent value.

Take $y \,=\, vx$ where $v$ is also a variable.

Now, eliminate the variable $y$ from the equation by substituting its equivalent value.

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

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

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

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

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

Look at the expression on the right-hand side of the equation. In this expression, $x^2$ is a factor in the numerator and it is also a factor in each term of the denominator. So, take the factor $x^2$ common from both terms in the denominator for cancelling it by the factor in the numerator.

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

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

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

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

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

Find the derivative of the product

The expression on the right-hand side of the equation is simplified in the previous step. It is time to simplify the expression on the left-hand side of the equation.

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

The derivative of product of two functions can be evaluated by the product rule of differentiation.

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

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

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

Solve the differential equation by variable separable

Firstly, let us simplify the differential equation.

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

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

$\implies$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{v-v \times 1- v \times2v^2}{1+2v^2}$

$\implies$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{v-v-2v^3}{1+2v^2}$

$\implies$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{\cancel{v}-\cancel{v}-2v^3}{1+2v^2}$

$\implies$ $x \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{-2v^3}{1+2v^2}$

The expressions on the both sides of the equation are simplified successfully. So, it is time to solve the differential equation and it can be done by the separation of variables method. In fact, the function in one variable with respective differential should be separated from the function in another variable with respective differential.

$\implies$ $\dfrac{1}{\dfrac{-2v^3}{1+2v^2}} \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1}{x}$

$\implies$ $\dfrac{1+2v^2}{-2v^3} \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1}{x}$

$\implies$ $-\dfrac{1+2v^2}{2v^3} \times \dfrac{dv}{dx}$ $\,=\,$ $\dfrac{1}{x}$

$\implies$ $-\dfrac{1+2v^2}{2v^3} \times dv$ $\,=\,$ $\dfrac{1}{x} \times dx$

Now, we can solve the differential equation by integrating both sides of the equation.

$\implies$ $\displaystyle \int{\Bigg(-\dfrac{1+2v^2}{2v^3}\Bigg)\,}dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $-\displaystyle \int{\Bigg(\dfrac{1+2v^2}{2v^3}\Bigg)\,}dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $-\displaystyle \int{\bigg(\dfrac{1}{2v^3}+\dfrac{2v^2}{2v^3}\bigg)\,}dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $-\displaystyle \int{\bigg(\dfrac{1}{2v^3}+\dfrac{\cancel{2v^2}}{\cancel{2v^3}}\bigg)\,}dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $-\displaystyle \int{\bigg(\dfrac{1}{2v^3}+\dfrac{1}{v}\bigg)\,}dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $-\displaystyle \int{\bigg(\dfrac{1 \times 1}{2 \times v^3}+\dfrac{1}{v}\bigg)\,}dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $-\displaystyle \int{\bigg(\dfrac{1}{2} \times \dfrac{1}{v^3} +\dfrac{1}{v}\bigg)\,}dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $-\displaystyle \int{\bigg(\dfrac{1}{2} \times v^{-3}+\dfrac{1}{v}\bigg)\,}dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $-\bigg(\displaystyle \int{\dfrac{1}{2} \times v^{-3}\,}dv$ $+$ $\displaystyle \int{\dfrac{1}{v}\,}dv\bigg)$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $-\displaystyle \int{\dfrac{1}{2} \times v^{-3}\,}dv$ $-$ $\displaystyle \int{\dfrac{1}{v}\,}dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

$\implies$ $-\dfrac{1}{2} \times \displaystyle \int{v^{-3}\,}dv$ $-$ $\displaystyle \int{\dfrac{1}{v}\,}dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}\,}dx$

It is time to find the indefinite integral of each function. The integral of the function in the first term of the left-hand side of the equation can be evaluated by the power rule of integration. The integral of the reciprocal of variable can be calculated by the reciprocal rule of integration.

$\implies$ $-\dfrac{1}{2} \times \Bigg(\dfrac{v^{-3+1}}{-3+1}+c_1\Bigg)$ $-$ $\bigg(\log_e{(v)}+c_2\bigg)$ $\,=\,$ $\log_e{(x)}$ $+$ $c_3$

Now, concentrate on simplifying this mathematical equation to get the required solution for the given homogeneous differential equation.

$\implies$ $-\dfrac{1}{2} \times \Bigg(\dfrac{v^{-2}}{-2}+c_1\Bigg)$ $-$ $\log_e{(v)}$ $-$ $c_2$ $\,=\,$ $\log_e{(x)}$ $+$ $c_3$

$\implies$ $\bigg(-\dfrac{1}{2}\bigg) \times \dfrac{v^{-2}}{-2}$ $+$ $\bigg(-\dfrac{1}{2}\bigg) \times c_1$ $-$ $\log_e{(v)}$ $-$ $c_2$ $\,=\,$ $\log_e{(x)}$ $+$ $c_3$

$\implies$ $\bigg(\dfrac{-1}{2}\bigg) \times \dfrac{v^{-2}}{-2}$ $-$ $\dfrac{1}{2} \times c_1$ $-$ $\log_e{(v)}$ $-$ $c_2$ $\,=\,$ $\log_e{(x)}$ $+$ $c_3$

$\implies$ $\bigg(\dfrac{-1}{2 \times (-2)}\bigg) \times v^{-2}$ $-$ $\dfrac{1 \times c_1}{2}$ $-$ $\log_e{(v)}$ $-$ $c_2$ $\,=\,$ $\log_e{(x)}$ $+$ $c_3$

$\implies$ $\bigg(\dfrac{-1}{-4}\bigg) \times v^{-2}$ $-$ $\dfrac{c_1}{2}$ $-$ $\log_e{(v)}$ $-$ $c_2$ $\,=\,$ $\log_e{(x)}$ $+$ $c_3$

$\implies$ $\bigg(\dfrac{1 \times (-1)}{4 \times (-1)}\bigg) \times v^{-2}$ $-$ $\dfrac{c_1}{2}$ $-$ $\log_e{(v)}$ $-$ $c_2$ $\,=\,$ $\log_e{(x)}$ $+$ $c_3$

$\implies$ $\bigg(\dfrac{1 \times \cancel{(-1)}}{4 \times \cancel{(-1)}}\bigg) \times v^{-2}$ $-$ $\dfrac{c_1}{2}$ $-$ $\log_e{(v)}$ $-$ $c_2$ $\,=\,$ $\log_e{(x)}$ $+$ $c_3$

$\implies$ $\bigg(\dfrac{1 \times 1}{4 \times 1}\bigg) \times v^{-2}$ $-$ $\dfrac{c_1}{2}$ $-$ $\log_e{(v)}$ $-$ $c_2$ $\,=\,$ $\log_e{(x)}$ $+$ $c_3$

$\implies$ $\dfrac{1}{4} \times v^{-2}$ $-$ $\dfrac{c_1}{2}$ $-$ $\log_e{(v)}$ $-$ $c_2$ $\,=\,$ $\log_e{(x)}$ $+$ $c_3$

$\implies$ $\dfrac{1}{4} \times v^{-2}$ $-$ $\dfrac{c_1}{2}$ $-$ $\log_e{(v)}$ $-$ $c_2$ $-$ $c_3$ $\,=\,$ $\log_e{(x)}$

$\implies$ $\dfrac{1}{4} \times \dfrac{1}{v^{2}}$ $-$ $\dfrac{c_1}{2}$ $-$ $\log_e{(v)}$ $-$ $c_2$ $-$ $c_3$ $\,=\,$ $\log_e{(x)}$

There is a logarithmic term on the right-hand side of the equation. So, shift the log term on the left-hand side of the equation to right-hand side of the equation.

$\implies$ $\dfrac{1}{4} \times \dfrac{1}{v^{2}}$ $-$ $\dfrac{c_1}{2}$ $-$ $c_2$ $-$ $c_3$ $\,=\,$ $\log_e{(v)}$ $+$ $\log_e{(x)}$

The sum of the logarithmic terms can be combined by the product rule of logarithms.

$\implies$ $\dfrac{1}{4} \times \dfrac{1}{v^{2}}$ $-$ $\dfrac{c_1}{2}$ $-$ $c_2$ $-$ $c_3$ $\,=\,$ $\log_e{(v \times x)}$

$\implies$ $\dfrac{1}{4} \times \dfrac{1}{v^{2}}$ $-$ $\dfrac{c_1}{2}$ $-$ $c_2$ $-$ $c_3$ $\,=\,$ $\log_e{(vx)}$

$\implies$ $\dfrac{1}{4} \times \dfrac{1}{v^{2}}$ $+$ $\Bigg(-\dfrac{c_1}{2}-c_2-c_3\Bigg)$ $\,=\,$ $\log_e{(vx)}$

The second term on the left-hand side of the equation is a constant. Hence, the constant expression in the second term on the left-hand side of the equation is simply denoted by a constant c.

$\implies$ $\dfrac{1}{4} \times \dfrac{1}{v^{2}}$ $+$ $c$ $\,=\,$ $\log_e{(vx)}$

We have taken $y = vx$ earlier for eliminating the variable $y$ from the equation. If $y = vx$, then $v \,=\, \dfrac{y}{x}$. So, it is time to eliminate the variable $v$ by expressing it in terms of the variables $x$ and $y$.

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

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

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

$\implies$ $\dfrac{1}{4} \bigg(\dfrac{x}{y}\bigg)^{2}$ $+$ $c$ $\,=\,$ $\log_e{(y)}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\log_e{(y)}$ $\,=\,$ $\dfrac{1}{4} \bigg(\dfrac{x}{y}\bigg)^{2}$ $+$ $c$

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