Math, asked by guddy16, 1 year ago

x²y dx -(x³ + y³) dy = 0

Answers

Answered by mustaphaismail304
8

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



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Answered by talasilavijaya
0

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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