In this differential equation, the degree of both polynomials $x^2y$ and $x^3+y^3$ are same. Hence, they are called homogeneous functions. The homogeneous function $x^2y$ is multiplied by a differential $dx$ and another homogeneous function $x^3+y^3$ is multiplied by another differential $dy$.

$x^2y\,dx$ $\,=\,$ $(x^3+y^3)\,dy$

It is given that the products of them are equal. Thus, a homogenous differential equation of the order one is formed in this problem. So, let’s learn how to solve this first order and first degree homogeneous differential equation.

$\implies$ $(x^3+y^3)\,dy$ $\,=\,$ $x^2y\,dx$

On both sides of the equation, the homogeneous functions $x^2y$ and $x^3+y^3$ in terms of the variables $x$ and $y$ are multiplying with differentials. The differentials must be separated from the homogenous functions for solving given homogenous differential equation.

$\implies$ $\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{x^2y}{x^3+y^3}$

The variables are separated from the differentials successfully but it is not possible to separate a variable from another variable in the both homogeneous functions. So, it is quite complicated for solving this first order differential equation. However, it can overcome by taking a variable as a product of two variables.

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

Now, eliminate the variable $y$ by substituting its equivalent value in both expressions of the equation.

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{x^2(vx)}{x^3+(vx)^3}$

$\,\,=\,$ $\dfrac{x^2 \times v \times x}{x^3+v^3 \times x^3}$

$\,\,=\,$ $\dfrac{x^2 \times x \times v}{x^3+v^3 \times x^3}$

Now, get the product of the functions in terms of $x$ by the product rule of the exponents in the numerator.

$\,\,=\,$ $\dfrac{x^3 \times v}{x^3+v^3 \times x^3}$

$\,\,=\,$ $\dfrac{x^3 \times v}{x^3 \times 1+v^3 \times x^3}$

In the denominator, each term contains the cube of $x$ as a factor. The common factor can be taken out from them.

$\,\,=\,$ $\dfrac{x^3 \times v}{x^3 \times (1+v^3)}$

$\,\,=\,$ $\dfrac{\cancel{x^3} \times v}{\cancel{x^3} \times (1+v^3)}$

$\implies$ $\dfrac{d(vx)}{dx}$ $\,=\,$ $\dfrac{v}{1+v^3}$

Use the product rule of derivatives to find the differentiation of the product of two variables on the left-hand side of the equation.

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

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

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

It is time to use the separation of the variables to start the procedure of solving the differential equation.

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

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

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

$\implies$ $x\dfrac{dv}{dx}$ $\,=\,$ $\dfrac{v-v-v^4}{1+v^3}$

$\implies$ $x\dfrac{dv}{dx}$ $\,=\,$ $\dfrac{\cancel{v}-\cancel{v}-v^4}{1+v^3}$

$\implies$ $x\dfrac{dv}{dx}$ $\,=\,$ $\dfrac{-v^4}{1+v^3}$

It is time to separate the variables by shifting the functions in terms of $x$ on one side and remaining functions are left on other side of the equation.

$\implies$ $\dfrac{1+v^3}{-v^4}\,dv$ $\,=\,$ $\dfrac{1}{x}\,dx$

$\implies$ $-\dfrac{1+v^3}{v^4}\,dv$ $\,=\,$ $\dfrac{1}{x}\,dx$

Now, let’s find the indefinite integration on both sides of the equation.

$\implies$ $\displaystyle \int{\bigg(-\dfrac{1+v^3}{v^4}\bigg)}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$

$\implies$ $\displaystyle \int{\bigg((-1) \times \dfrac{1+v^3}{v^4}\bigg)}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$

The constant factors can be separated from the integration as per the constant multiple rule of the integrals.

$\implies$ $(-1) \times \displaystyle \int{\bigg(\dfrac{1+v^3}{v^4}\bigg)}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$

There is nothing to simplify the expression on the right-hand side of the equation. So, let us simplify the expression on the left hand side of the equation.

$\implies$ $-\displaystyle \int{\bigg(\dfrac{1}{v^4}+\dfrac{v^3}{v^4}\bigg)}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$

$\implies$ $-\displaystyle \int{\bigg(\dfrac{1}{v^4}+\dfrac{\cancel{v^3}}{\cancel{v^4}}\bigg)}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$

$\implies$ $-\displaystyle \int{\bigg(\dfrac{1}{v^4}+\dfrac{1}{v}\bigg)}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$

$\implies$ $-\displaystyle \int{\bigg(v^{-4}+\dfrac{1}{v}\bigg)}\,dv$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$

The integral of sum of two functions can be evaluated by the sum of their integrals as per the sum rule of the integrals.

$\implies$ $-\bigg(\displaystyle \int{v^{-4}}\,dv$ $+$ $\displaystyle \int{\dfrac{1}{v}}\,dv\bigg)$ $\,=\,$ $\displaystyle \int{\dfrac{1}{x}}\,dx$

The integration of negative exponent function can be evaluated by using the power rule of integrals. Similarly, the integral of reciprocal of a variable can be calculated as per the reciprocal rule of the integration.

$\implies$ $-\bigg(\dfrac{v^{\displaystyle -4+1}}{-4+1}+c_1$ $+$ $\log_{e}{v}+c_2\bigg)$ $\,=\,$ $\log_{e}{x}$ $+$ $c_3$

The given first order and first degree homogeneous differential equation is successfully solved and it is time for simplification.

$\implies$ $-\bigg(\dfrac{v^{\displaystyle -3}}{-3}+c_1$ $+$ $\log_{e}{v}+c_2\bigg)$ $\,=\,$ $\log_{e}{x}$ $+$ $c_3$

$\implies$ $-\bigg(-\dfrac{v^{\displaystyle -3}}{3}+c_1$ $+$ $\log_{e}{v}+c_2\bigg)$ $\,=\,$ $\log_{e}{x}$ $+$ $c_3$

$\implies$ $\dfrac{v^{\displaystyle -3}}{3}$ $-$ $c_1$ $-$ $\log_{e}{v}$ $-$ $c_2$ $\,=\,$ $\log_{e}{x}$ $+$ $c_3$

The first term on the left-hand side of the equation can be written as follows as per negative exponent rule.

$\implies$ $\dfrac{1}{3v^3}$ $-$ $c_1$ $-$ $\log_{e}{v}$ $-$ $c_2$ $\,=\,$ $\log_{e}{x}$ $+$ $c_3$

$\implies$ $\dfrac{1}{3v^3}$ $\,=\,$ $\log_{e}{x}$ $+$ $c_3$ $+$ $c_1$ $+$ $\log_{e}{v}$ $+$ $c_2$

$\implies$ $\dfrac{1}{3v^3}$ $\,=\,$ $\log_{e}{x}$ $+$ $\log_{e}{v}$ $+$ $c_1$ $+$ $c_2$ $+$ $c_3$

The sum of the constants is also a constant and it is simply denoted by another constant $c_4$.

$\implies$ $\dfrac{1}{3v^3}$ $\,=\,$ $\log_{e}{x}$ $+$ $\log_{e}{v}$ $+$ $c_4$

On right-hand side of the equation, the two logarithmic terms are connected a plus sign. So, the sum of them can be evaluated as per the product rule of logarithms. For simplifying the expression further, the constant term c4 can be denoted by the logarithm of a constant $c$.

$\implies$ $\dfrac{1}{3v^3}$ $\,=\,$ $\log_{e}{x}$ $+$ $\log_{e}{v}$ $+$ $\log_{e}{c}$

$\implies$ $\dfrac{1}{3v^3}$ $\,=\,$ $\log_{e}{(x \times v \times c)}$

In this problem, the homogeneous differential equation is given in terms of $x$ and $y$ but the solution is in terms of $x$ and $v$. So, replace the value of $v$ in terms of $x$ and $y$.

$\implies$ $\dfrac{1}{3\Big(\dfrac{y}{x}\Big)^3}$ $\,=\,$ $\log_{e}{\Big(x \times \dfrac{y}{x} \times c\Big)}$

Now, use the multiplication of the fractions to get the product of the factors inside the logarithm.

$\implies$ $\dfrac{1}{3\Big(\dfrac{y}{x}\Big)^3}$ $\,=\,$ $\log_{e}{\Big(\dfrac{x \times y \times c}{x}\Big)}$

$\implies$ $\dfrac{1}{3\Big(\dfrac{y}{x}\Big)^3}$ $\,=\,$ $\log_{e}{\Big(\dfrac{\cancel{x} \times y \times c}{\cancel{x}}\Big)}$

$\implies$ $\dfrac{1}{3\Big(\dfrac{y}{x}\Big)^3}$ $\,=\,$ $\log_{e}{(y \times c)}$

$\implies$ $\dfrac{1}{3\Big(\dfrac{y}{x}\Big)^3}$ $\,=\,$ $\log_{e}{(c \times y)}$

$\implies$ $\dfrac{1}{3\Big(\dfrac{y}{x}\Big)^3}$ $\,=\,$ $\log_{e}{(cy)}$

$\implies$ $\dfrac{1}{3 \times \Big(\dfrac{y}{x}\Big)^3}$ $\,=\,$ $\log_{e}{(cy)}$

$\implies$ $\dfrac{1}{\Big(\dfrac{y}{x}\Big)^3}$ $\,=\,$ $3 \times \log_{e}{(cy)}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\Big(\dfrac{x}{y}\Big)^3$ $\,=\,$ $3\log_{e}{(cy)}$

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