Question

(x-y-2)dx+(x-2y-3)dy=0. solve this differential equations

Answer 1

Equations of First Order and First Degree

Given: $(x-y-2)dx+(x-2y-3)dy=0$

To find: the solution

Solution:

Here, $(x-y-2)dx+(x-2y-3)dy=0$

$\Rightarrow \frac{dy}{dx}=-\frac{x-y-2}{x-2y-3}$

$\Rightarrow \frac{dy}{dx}=\frac{-x+y+2}{x-2y-3}$

The two lines $-x+y+2=0$ and $x-2y-3=0$ intersect at the point $(1,\: -1)$.

The substitution $x=X+1,\:y=Y-1$ gives

$\quad\frac{dY}{dX}=\frac{-X+Y}{X-2Y}$

$\Rightarrow \frac{dY}{dX}=\frac{-1+Y/X}{1-2Y/X}$

Put $Y=vX$. Then the equation becomes

$\quad v+X\frac{dv}{dX}=\frac{-1+v}{1-2v}$

$\Rightarrow X\frac{dv}{dX}=\frac{-1+v}{1-2v}-v$

$\Rightarrow X\frac{dv}{dX}=\frac{-1+v-v+2v^{2}}{1-2v}$

$\Rightarrow X\frac{dv}{dX}=\frac{-1+2v^{2}}{1-2v}$

$\Rightarrow X\frac{dv}{dX}=-\frac{2v^{2}-1}{2v-1}$

$\Rightarrow \frac{dX}{X}+\frac{2v-1}{2v^{2}-1}dv=0$

$\Rightarrow \frac{dX}{X}+\frac{1}{2}\frac{4v-2}{2v^{2}-1}dv=0$

$\Rightarrow \frac{dX}{X}+\frac{1}{2}\frac{4v}{2v^{2}-1}dv-\frac{1}{2}\frac{dv}{v^{2}-(\frac{1}{\sqrt{2}})^{2}}=1$

On integration, we get

$\quad \int \frac{dX}{X}+\frac{1}{2}\int\frac{4v}{2v^{2}-1}dv-\int\frac{1}{2}\frac{dv}{v^{2}-(\frac{1}{\sqrt{2}})^{2}}=1$

$\Rightarrow logX+\frac{1}{2}.log(2v^{2}-1)-\frac{1}{2}.\frac{1}{2\frac{1}{\sqrt{2}}}.log(\frac{v-\frac{1}{\sqrt{2}}}{v+\frac{1}{\sqrt{2}}})=c$ where $c$ is constant of integration

$\Rightarrow log(x-1)+\frac{1}{2}.log[2(\frac{y+1}{x-1})^{2}-1]-\frac{1}{2\sqrt{2}}.log(\frac{\frac{y+1}{x-1}-\frac{1}{\sqrt{2}}}{\frac{y+1}{x-1}+\frac{1}{\sqrt{2}}})=c$

• Putting $v=\frac{Y}{X}=\frac{y+1}{x-1}$

This is the required solution.