Question

Solve (x³+3xy²)dx+(y³+3x²y)dy=0

Answer 1

Answer:

= (x³+3xy²)dx+ ( y³ + 3x²y)dy = 0

= dx * x³ + dx * 3xy² + dy * y³ + dy * 3x²y = 0

= dx⁴ + 3dx²y² + dy⁴ + 3dx²y² = 0

= dx² + 6dx²y² + dy⁴ = 0

Answer 2

Concept

A differential equation of the form f(x, y)$dy$ = g(x, y)$dx$ is said to be homogeneous differential equation if the degree of f(x, y) and g(x, y) is same.

Given

The differential equation is $(x^{3} +3y^{2} )dx+(y^{3} +3x^{2} y)dy=0$.

Find

We have to find the solution of the differential equation, $(x^{3} +3y^{2} )dx+(y^{3} +3x^{2} y)dy=0$.

Solution

The given differential equation is written as-

$\frac{dy}{dx}=-\frac{\left( {{x}^{3}}+3x{{y}^{2}} \right)}{\left( {{y}^{3}}+3{{x}^{2}}y \right)}$

$\frac{dy}{dx}=-\frac{3x{{y}^{2}}\left( \frac{{{x}^{3}}}{3x{{y}^{2}}}+1 \right)}{3{{x}^{2}}y\left( \frac{{{y}^{3}}}{3{{x}^{2}}y}+1 \right)}$

$\frac{dy}{dx}=-\frac{y\left( \frac{{{x}^{2}}}{3{{y}^{2}}}+1 \right)}{x\left( \frac{{{y}^{2}}}{3{{x}^{2}}}+1 \right)}$

$\Rightarrow \frac{d y}{d x}=f\left(\frac{y}{x}\right)$

So, the given differential equation is a homogeneous equation.

Now, solving for the given differential equation-

Consider $y=v x$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}$

$\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=-\frac{\text{vx}\left( \frac{{{\text{x}}^{2}}}{3{{(\text{vx})}^{2}}}+1 \right)}{\text{x}\left( \frac{{{(\text{vx})}^{2}}}{3{{\text{x}}^{2}}} \right)+1}$

$\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=-\frac{\left( \frac{1}{3{{(\text{v})}^{2}}}+1 \right)}{\left( \frac{{{(\text{v})}^{2}}}{3}+1 \right)}$

$\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=-\frac{1+3{{(\text{v})}^{2}}}{3+{{(\text{v})}^{2}}}\times \frac{1}{\text{v}}$

$\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}=-\frac{1+3{{(\text{v})}^{2}}}{3\text{v}+{{(\text{v})}^{3}}}$

$\Rightarrow \text{x}\frac{\text{dv}}{\text{dx}}=-\frac{1+3{{(\text{v})}^{2}}}{3\text{v}+{{(\text{v})}^{3}}}-\text{v}$

$\text{x}\frac{\text{dv}}{\text{dx}}=-\frac{1+3{{(\text{v})}^{2}}+3{{(\text{v})}^{2}}+{{(\text{v})}^{4}}}{3\text{v}+{{(\text{v})}^{3}}}$

$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+6{{(\text{v})}^{2}}+{{(\text{v})}^{4}}}{3\text{v}+{{(\text{v})}^{3}}}$

$\Rightarrow \frac{3 \mathrm{v}+(\mathrm{v})^{3}}{1+6(\mathrm{v})^{2}+(\mathrm{v})^{4}} \mathrm{dv}=-\frac{\mathrm{dx}}{\mathrm{x}}$

Now, integrating both the sides, we get-

$\Rightarrow \int \frac{3 \mathrm{v}+(\mathrm{v})^{3}}{1+6(\mathrm{v})^{2}+(\mathrm{v})^{4}} \mathrm{dv}=-\int \frac{\mathrm{dx}}{\mathrm{x}}+\mathrm{c}$

Now, we have $\frac{d}{d v}\left(1+6(v)^{2}+(v)^{4}\right)=12 v+4 v^{3}=4\left(3 v+v^{3}\right)$, therefore,

$\Rightarrow \frac{\ln \mid 1+6(\mathrm{v})^{2}+(\mathrm{v})^{4}}{4}+\ln |\mathrm{x}|=\ln |\mathrm{c}|$

Now, substituting the value of $y=v x$, we get-

$\Rightarrow \frac{\ln \mid 1+6\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}+\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{4}}{4}+\ln |\mathrm{x}|=\ln |\mathrm{c}|$

$\Rightarrow \mathrm{y}^{4}+6 \mathrm{x}^{2} \mathrm{y}^{2}+\mathrm{x}^{3} 4=\mathrm{C}$

where, $C$ is the constant.

Hence, $\mathrm{y}^{4}+6 \mathrm{x}^{2} \mathrm{y}^{2}+\mathrm{x}^{3} 4=\mathrm{C}$ is the required solution of the differential equation, $(x^{3} +3y^{2} )dx+(y^{3} +3x^{2} y)dy=0$.