Question

x²y dx -(x³ + y³) dy = 0

Answer 1

solution

x²ydx-(x³+y³)dy=0

using homogeneous method

then we let y=vx

⇒dy/dx=(x³+y³)/x²y

dy/dx=(x³+ (vx)³)/x²(vx)

⇒ dy/dx=(x³+v³x³)/x³v

dy/dx=x³(1+v³)/x³v

dy/dx=(1+v³)/v

there for dy/dx=1+vdv/dx

⇒ 1+vdv/dx=(1+v³)/v

⇒vdv/dx=(1+v³)/v(-1)

dv/dx=(1/v+v²)-1

dv/dx=(1/v²+v-1/v)

⇒dv/(v²+v∧-1-v)=dx

⇒ (v³/3)+(inv)-(v²/2)=x

but v=y/x

⇒ (y/x)³/3 + (in(y/x)) - (y/x)²/2=x+c

Answer 2

Answer:

The solution of the given equation is

 $ln|y|- \dfrac{x^{3} }{ 3y^3}=C$ .

Step-by-step explanation:

Given the homogeneous differential equation,

$x^2y dx -(x^3 + y^3) dy = 0$

Rewriting the equation,

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

Taking $x^{3}$ common from numerator and denominator of RHS

$\dfrac{dy}{dx}=\dfrac{x^3\bigg(\dfrac{x^2y}{x^3} \bigg)}{x^3 \bigg(1+ \dfrac{y^3}{x^3} \bigg)}$

$\implies \dfrac{dy}{dx}=\dfrac{\bigg(\dfrac{y}{x} \bigg)}{ 1+ \bigg(\dfrac{y}{x} \bigg)^3}$                         ...(1)

Let

$\dfrac{y}{x} =v$                                               ...(2)

$\implies y=vx$        

Differentiating with respect to $x$, we get

$\dfrac{dy}{dx} =v+x\dfrac{dv}{dx}$                                   ...(3)

Using equations (2) and (3) in equation (1),

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

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

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

Separating the variables,

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

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

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

Integrating on both sides,

${\displaystyle \int \dfrac{1}{ v^4}dv+ {\displaystyle \int \dfrac{ 1}{ v}dv=- {\displaystyle \int \dfrac{1}{ x}{dx}$

$- \dfrac{1}{ 3v^3}+ln|v|=-ln|x|+C$

$ln|x|- \dfrac{1}{ 3v^3}+ln|v|=C$

$ln|x|- \dfrac{1}{ 3\Big(\dfrac{y}{x}\Big)^3}+ln\Big|\dfrac{y}{x}\Big|=C$

$\implies ln|y|- \dfrac{x^{3} }{ 3y^3}=C$

Therefore, the required solution of the given equation is

 $ln|y|- \dfrac{x^{3} }{ 3y^3}=C$

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