Answer to Question #301669 in Differential Equations for Madhumita

Question

The given differential equation (1-6y^2-3x^2y)

Answer 1

Given problem is incomplete.

Assume, we need to solve: (4x -6y-1) dx + (3x –2y-2) dy = 0

Solution:

Let us solve the differential equation $(4x -6y-1) dx + (3x -2y-2) dy = 0.$

Let us use the transformation $x=u+1,\ y=v+\frac{1}{2}.$ Then we have the equation

$(4u -6v) du + (3u -2v) dv = 0.$

Let $u=tv,$ then $t=\frac{u}{v}=\frac{x-1}{y-\frac{1}2}=\frac{2x-2}{2y-1}.$

It follows that $du=tdv+vdt,$ and hence

$(4tv -6v) (tdv+vdt) + (3tv –2v) dv = 0.$

Then after dividing by $v$ we get the equation

$(4t -6) (tdv+vdt) + (3t –2) dv = 0$ or $(4t -6) vdt + (4t^2-3t -2) dv = 0.$

it follows that

$\frac{dv}{v}=-\frac{(4t -6)dt}{4t^2-3t -2} =-\frac{1}2\frac{(8t - 12)dt}{4t^2-3t -2} =-\frac{1}2\frac{(8t - 3)dt}{4t^2-3t -2} +\frac{9}2\frac{dt}{4(t^2-\frac{3}4t -\frac{1}2)} =-\frac{1}2\frac{d(4t^2-3t -2)}{4t^2-3t -2} +\frac{9}8\frac{dt}{(t-\frac{3}8)^2 -\frac{41}{64}}.$

Therefore,

$\int\frac{dv}{v}=-\frac{1}2\int\frac{d(4t^2-3t -2)}{4t^2-3t -2} +\frac{9}8\int\frac{dt}{(t-\frac{3}8)^2 -\frac{41}{64}}$

We conclude that

$\ln|v|=-\frac{1}2\ln|4t^2-3t-2|+\frac{9}8\frac{1}{2\frac{\sqrt{41}}{8}}\ln|\frac{t-\frac{3}8-\frac{\sqrt{41}}{8}}{t-\frac{3}8+\frac{\sqrt{41}}{8}}|+C,$

and hence the general solution is

$\ln|y-\frac{1}2|=-\frac{1}2\ln|4(\frac{2x-2}{2y-1})^2-3(\frac{2x-2}{2y-1})-2|+\frac{9}{2\sqrt{41}}\ln|\frac{\frac{2x-2}{2y-1}-\frac{3}8-\frac{\sqrt{41}}{8}}{\frac{2x-2}{2y-1}-\frac{3}8+\frac{\sqrt{41}}{8}}|+C.$