Math, asked by mohitparmar, 1 year ago

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

Answers

Answered by Swarup1998
28

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.

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