Question

specify the type of ODE based on its characteristics and find the solution.

(2y^2 x-2y^3 )dx+(4y^3-6y^2 x+2yx^2 )dy=0​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Differential equation is}$

$\mathsf{(2xy^2-2y^3)dx+(4y^3-6xy^2+2x^2y)dy=0}$

$\underline{\textbf{To find:}}$

$\textsf{Solution of the given differential equation}$

$\underline{\textbf{Solution:}}$

$\textsf{The equation is Homogeneous differential equation}$

$\mathsf{(2xy^2-2y^3)dx+(4y^3-6xy^2+2x^2y)dy=0}$

$\textsf{This can be written as}$

$\mathsf{(4y^3-6xy^2+2x^2y)dy=-(2xy^2-2y^3)dx}$

$\mathsf{\dfrac{dy}{dx}=-\dfrac{2xy^2-2y^3}{4y^3-6xy^2+2x^2y}}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{-2xy^2+2y^3}{4y^3-6xy^2+2x^2y}}$

$\mathsf{Put\;y=vx}$

$\implies\mathsf{\dfrac{dy}{dx}=v.1+x\dfrac{dv}{dx}}$

$\mathsf{v+x\dfrac{dv}{dx}=\dfrac{-2x(v^2x^2)+2v^3x^3}{4v^3x^3-6x(v^2x^2)+2x^2(vx)}}$

$\mathsf{v+x\dfrac{dv}{dx}=\dfrac{-2v^2x^3+2v^3x^3}{4v^3x^3-6v^2x^3+2vx^3}}$

$\mathsf{v+x\dfrac{dv}{dx}=\dfrac{x^3(-2v^2+2v^3)}{x^3(4v^3-6v^2+2v)}}$

$\mathsf{x\dfrac{dv}{dx}=\dfrac{-2v^2+2v^3}{4v^3-6v^2+2v}-v}$

$\mathsf{x\dfrac{dv}{dx}=\dfrac{-2v^2+2v^3-4v^4+6v^3-2v^2}{4v^3-6v^2+2v}}$

$\mathsf{x\dfrac{dv}{dx}=\dfrac{-4v^4+8v^3-4v^2}{4v^3-6v^2+2v}}$

$\mathsf{x\dfrac{dv}{dx}=\dfrac{-4(v^4-2v^3+v^2)}{4v^3-6v^2+2v}}$

$\textsf{Separating the variable, we get}$

$\mathsf{\dfrac{4v^3-6v^2+2v}{v^4-2v^3+v^2}\;dv=-4\;\dfrac{dx}{x}}$

$\textsf{Integrating on bothsides, we get}$

$\mathsf{\displaystyle\int\dfrac{4v^3-6v^2+2v}{v^4-2v^3+v^2}\;dv=-4\int\dfrac{dx}{x}}$

$\textsf{In L.H.S, Numerator is the actual derivative of denominator}$

$\mathsf{log(v^4-2v^3+v^2)=-4\;log\,x+logC}$

$\mathsf{log(v^4-2v^3+v^2)=log\,x^{-4}+logC}$

$\mathsf{log(v^4-2v^3+v^2)=log\,x^{-4}C}$

$\implies\mathsf{v^4-2v^3+v^2=\dfrac{C}{x^4}}$

$\implies\mathsf{\dfrac{y^4}{x^4}-2\dfrac{y^3}{x^3}+\dfrac{y^2}{x^2}=\dfrac{C}{x^4}}$

$\implies\mathsf{\dfrac{y^4-2xy^3+x^2y^2}{x^4}=\dfrac{C}{x^4}}$

$\implies\boxed{\mathsf{y^4-2xy^3+x^2y^2=C}}$

$\textsf{which is the required solution}$

$\underline{\textbf{Find more:}}$

Solve x(x-y)dy+y^2dx=0​

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d​

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