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likoan [24]
3 years ago
7

Find the general solution of the equation: (y^2-2xy)dx + (2xy-x^2)dy = 0

Mathematics
1 answer:
lys-0071 [83]3 years ago
6 0

The ODE is exact, since

\dfrac{\partial(y^2-2xy)}{\partial y}=2y-2x

\dfrac{\partial(2xy-x^2)}{\partial x}=2y-2x

so there is solution f(x,y)=C such that

\dfrac{\partial f}{\partial x}=y^2-2xy

\dfrac{\partial f}{\partial y}=2xy-x^2

Integrating both sides of the first PDE wrt x gives

f(x,y)=xy^2-x^2y+g(y)

Differentiating both sides wrt y gives

\dfrac{\partial f}{\partial y}=2xy-x^2+\dfrac{\mathrm dg}{\mathrm dy}=2xy-x^2

\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C

Then the solution to the ODE is

f(x,y)=\boxed{xy^2-x^2y=C}

# # #

Alternatively, we can see that the ODE is homogeneous, since replacing x\to tx and y\to ty reduces to the same ODE:

((ty)^2-2(tx)(ty))\,\mathrm d(tx)+(2(tx)(ty)-(tx)^2)\,\mathrm d(ty)=0

t^3(y^2-2xy)\,\mathrm dx+t^3(2xy-x^2)\,\mathrm dy=0

(y^2-2xy)\,\mathrm dx+(2xy-x^2)\,\mathrm dy=0

This tells us we can solve by substituting y(x)=xv(x), so that \mathrm dy(x)=x\,\mathrm dv(x)+v(x)\,\mathrm dx, and the ODE becomes

(x^2v^2-2x^2v)\,\mathrm dx+(2x^2v-x^2)(x\,\mathrm dv+v\,\mathrm dx)=0

v(v-2)\,\mathrm dx+(2v-1)(x\,\mathrm dv+v\,\mathrm dx)=0

3v(v-1)\,\mathrm dx+x(2v-1)\,\mathrm dv=0

which is separable as

\dfrac{1-2v}{3v(v-1)}\,\mathrm dv=\dfrac{\mathrm dx}x

Integrating both sides gives

-\dfrac13\ln|v(1-v)|=\ln|x|+C

v(1-v)=\dfrac C{x^3}

and solving in terms of y(x),

\dfrac yx\left(1-\dfrac yx\right)=\dfrac C{x^3}

xy(x-y)=C

\boxed{x^2y-xy^2=C}

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