not separable

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \frac{dy}{dx} = \frac{x + 2y}{3y - 2x}\)

 

[logistic_guy]

Let us try to write the differential equation in a different form.

\(\displaystyle (x + 2y) \ dx + (2x - 3y) \ dy = 0\)

 

[logistic_guy]

Let \(\displaystyle M = x + 2y\) and \(\displaystyle N = 2x - 3y\)

Then,

\(\displaystyle \frac{\partial M}{\partial y} = 2 = \frac{\partial N}{\partial x}\)



This means that the differential equation is exact and we can solve it by the method we have learnt in previous posts.

 

[logistic_guy]

You will be surprised that everything will go smoothly from now and on.

\(\displaystyle \frac{\partial f}{\partial x} = M\)

Or

\(\displaystyle \frac{\partial f}{\partial x} = x + 2y\)


\(\displaystyle \int \frac{\partial f}{\partial x} \ dx = \int (x + 2y) \ dx\)


\(\displaystyle f(x,y) = \frac{x^2}{2} + 2xy + g(y)\)

 

[logistic_guy]

We continue like math heroes.



\(\displaystyle \frac{\partial f}{\partial y} = 2x + g'(y)\)

 

[logistic_guy]

\(\displaystyle \frac{\partial f}{\partial y} = 2x + g'(y)\)

\(\displaystyle \int \frac{\partial f}{\partial y} \ dy = \int [\ 2x + g'(y) \ ] \ dy\)


\(\displaystyle \int N \ dy = \int [\ 2x + g'(y) \ ] \ dy\)


\(\displaystyle \int [\ 2x - 3y \ ] \ dy = \int [\ 2x + g'(y) \ ] \ dy\)


\(\displaystyle 2xy - \frac{3y^2}{2} = 2xy + g(y)\)


\(\displaystyle g(y) = -\frac{3y^2}{2}\)