Solve: $(3xy-2ay^2)dx+(x^2-2axy)dy=0$

Question

Solve: $(3xy-2ay^2)dx+(x^2-2axy)dy=0$

The given differential equation is not exact.

The given differential equation is homogeneous.

Therefore,

$I.F=\frac{1}{Mx+Ny}$

$=\frac{1}{(3xy-2ay^2)x+(x^2-2axy)y}$

$=\frac{1}{xy(4x-4ay)}$

With Integrating Factor of this form (ugly value), it is difficult to compute further integration, that is involved in solving the equation.

So, can we not solve the equation using the rule:

If in the differential eqn $Mdx+Ndy=0$, provided $ Mx+Ny \ne 0$, andM and N ar eboth homogeneous then

$I.F=\frac{1}{Mx+Ny}$

Answer 1

Hint: $$(3xy-2ay^2)dx+(x^2-2axy)dy=0$$ With $$M=3xy-2ay^2~~~~~~,~~~~~~N=x^2-2axy$$ then $$M_y=3x-4ay~~~~~~,~~~~~~N_x=2x-2ay$$ so the equation is not exact. For integrating factor $$\mu=\dfrac{M_y-N_x}{N}=\dfrac{x-2ay}{x(x-2ay)}=\dfrac1x$$ therefore $I=e^{\int \mu\ dx}=x$ is integrating factor.

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Can you proceed?

Answer 2

Put y=vx

$\frac{dy}{dx} = \frac{2ay^2-3xy}{x^2-2axy} \\ v+x\frac{dv}{dx} = \frac{2av^2-3v}{1-2av}\\x\frac{dv}{dx} =\frac{4v(av-1)}{1-2av} \\ \frac{1-2av}{4v(av-1)}dv=\frac{dx}{x} \\(\frac{-1}{4v}-\frac{a}{4(av-1)})dv=\frac{dx}{x}\\v(av-1)x^4=C\\x^2y(ay-x)=C, \space some \space constant$

Other simpler way would be multiply with x

Given differential equation becomes:

$d(x^3 y)-ad(x^2y^2)=0$

U can directly integrate from here