Question

solve dy/ dx= x+y+4/x-y-6​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\dfrac{dy}{dx}=\dfrac{x+y+4}{x-y-6}}$

$\sf{Let\,\,x=X+h\,\,\,\,\,\,and\,\,\,\,\,\,y=Y+k}$

$\sf{\dfrac{dy}{dx}=\dfrac{d(Y+k)}{d(X+h)}=\dfrac{dY}{dX}}$

So,

$\tt{\dfrac{dY}{dX}=\dfrac{(X+h)+(Y+k)+4}{(X+h)-(Y+k)-6}}$

$\tt{\implies\dfrac{dY}{dX}=\dfrac{X+Y+h+k+4}{X-Y+h-k-6}}$

$\sf{Put\,\,\,\,h+k+4=0\,\,\,\,\&\,\,\,\,h-k-6=0}$

Add the above equations

$\sf{h+k+4+h-k-6=0}$

$\sf{\implies\,2h-2=0}$

$\implies\boxed{\sf{h=1}}\,\,\,....(1)$

Put the value of h in any of the equations, we get,

$\boxed{\sf{k=-5}}\,\,\,\,....(2)$

Now,

$\tt{\implies\dfrac{dY}{dX}=\dfrac{X+Y+0}{X-Y+0}}$

$\tt{\implies\dfrac{dY}{dX}=\dfrac{X+Y}{X-Y}}$

Put  $\sf{Y=VX\,\,\,\,...(3)}$

$\sf{\implies\,\dfrac{dY}{dX}=V+X\dfrac{dV}{dX}}$

So,

$\tt{\implies\,V+X\dfrac{dV}{dX}=\dfrac{X+VX}{X-VX}}$

$\tt{\implies\,V+X\dfrac{dV}{dX}=\dfrac{1+V}{1-V}}$

$\tt{\implies\,X\dfrac{dV}{dX}=\dfrac{1+V}{1-V}-V}$

$\tt{\implies\,X\dfrac{dV}{dX}=\dfrac{1+V-V+V^2}{1-V}}$

$\tt{\implies\,X\dfrac{dV}{dX}=\dfrac{1+V^2}{1-V}}$

$\tt{\implies\,\dfrac{1-V}{1+V^2}dV=\dfrac{dX}{X}$

$\tt{\implies\,\displaystyle\int\dfrac{1-V}{1+V^2}dV=\int\dfrac{dX}{X}$

$\tt{\implies\,\displaystyle\int\dfrac{1}{1+V^2}dV-\int\dfrac{V}{1+V^2}\,dV=\int\dfrac{dX}{X}$

$\tt{\implies\,\displaystyle\int\dfrac{1}{1+V^2}dV-\dfrac{1}{2}\int\dfrac{2V}{1+V^2}\,dV=\int\dfrac{dX}{X}$

$\tt{\implies\,\displaystyle\,tan^{-1}\,V-\dfrac{1}{2}\ln|1+V^2|=\ln|X|+C$

From (3), we have,

$\tt{\implies\,\displaystyle\,tan^{-1}\bigg(\dfrac{Y}{X}\bigg)-\ln\sqrt{1+\bigg(\dfrac{Y}{X}\bigg)^2}=\ln|X|+C$

$\tt{\implies\,\displaystyle\,tan^{-1}\bigg(\dfrac{Y}{X}\bigg)-\ln\sqrt{1+\dfrac{Y^2}{X^2}}=\ln|X|+C$

$\tt{\implies\,\displaystyle\,tan^{-1}\bigg(\dfrac{Y}{X}\bigg)-\ln\sqrt{\dfrac{X^2+Y^2}{X^2}}=\ln|X|+C$

$\tt{\implies\,\displaystyle\,tan^{-1}\bigg(\dfrac{Y}{X}\bigg)-\ln\dfrac{\sqrt{X^2+Y^2}}{X}=\ln|X|+C$

$\tt{\implies\,\displaystyle\,tan^{-1}\bigg(\dfrac{Y}{X}\bigg)-\ln\sqrt{X^2+Y^2}+\ln|X|=\ln|X|+C$

$\tt{\implies\,\displaystyle\,tan^{-1}\bigg(\dfrac{Y}{X}\bigg)-\ln\sqrt{X^2+Y^2}=C$

Recall that, we have taken x=X+h and y=Y+k

From (1) and (2), we get,

x=X+1 and y=Y-5

So,

X=x-1 and Y=y+5

$\tt{\implies\,\displaystyle\,tan^{-1}\bigg(\dfrac{y+5}{x-1}\bigg)-\ln\sqrt{(x-1)^2+(y+5)^2}=C$

$\tt{\implies\,\displaystyle\,tan^{-1}\bigg(\dfrac{y+5}{x-1}\bigg)-\ln\sqrt{x^2-2x+1+y^2+10y+25}=C$

$\tt{\implies\,\displaystyle\,tan^{-1}\bigg(\dfrac{y+5}{x-1}\bigg)-\ln\sqrt{x^2-2x+y^2+10y+26}=C$

$\tt{\implies\,\displaystyle\,tan^{-1}\bigg(\dfrac{y+5}{x-1}\bigg)-\ln\sqrt{x^2+y^2-2x+10y+26}=C$

Answer 2

Given:

$\frac{d y}{d x}=\frac{x+y+4}{x-y-6}$

$\text { Let } x=X+h \text { and } y=Y+k$

$\frac{d y}{d x}=\frac{d(Y+k)}{d(X+h)}=\frac{d Y}{d X}$

$\frac{d Y}{d X}=\frac{(X+h)+(Y+k)+4}{(X+h)-(Y+k)-6}$

$\begin{array}{l}\frac{d Y}{d X}=\frac{X+Y+h+k+4}{X-Y+h-k-6} \\\end{array}$

$put\ h+k+4=0 \& h-k-6=0$

$\text { Add the above equations }$

$\begin{array}{l}\mathrm{h}+\mathrm{k}+4+\mathrm{h}-\mathrm{k}-6=0 \\ 2 \mathrm{~h}-2=0 \\ \mathrm{h}=1\end{array}$-(1)

$\text { Put the value of } h \text { in any of the equations, we get, }$

$\mathrm{k}=-5$                            -(2)

$\begin{array}{l}\frac{d Y}{d X}=\frac{X+Y+0}{X-Y+0} \\\end{array}$

$\frac{d Y}{d X}=\frac{X+Y}{X-Y}$

$\begin{array}{l}\text { Put } Y=V X \ldots(3) \\ \frac{d Y}{d X}=V+X \frac{d V}{d X}\end{array}$

$\begin{aligned}\mathrm{V}+\mathrm{X} \frac{\mathrm{dV}}{\mathrm{dX}} &=\frac{\mathrm{X}+\mathrm{VX}}{\mathrm{X}-\mathrm{VX}} \\\mathrm{V}+\mathrm{X} \frac{\mathrm{dV}}{\mathrm{dX}} &=\frac{1+\mathrm{V}}{1-\mathrm{V}}\end{aligned}$

$\begin{aligned}\mathrm{X} \frac{\mathrm{dV}}{\mathrm{dX}} &=\frac{1+\mathrm{V}}{1-\mathrm{V}}-\mathrm{V} \\\mathrm{X} \frac{\mathrm{dV}}{\mathrm{dX}} &=\frac{1+\mathrm{V}-\mathrm{V}+\mathrm{V}^{2}}{1-\mathrm{V}}\end{aligned}$

$\begin{array}{l}\frac{d V}{d X}=\frac{1+V^{2}}{1-V} \\\frac{1-V}{1+V^{2}} d V=\frac{d X}{X}\end{array}$

$\begin{array}{l}\int \frac{1-\mathrm{V}}{1+\mathrm{V}^{2}} \mathrm{dV}=\int \frac{\mathrm{dX}}{\mathrm{X}} \\\int \frac{1}{1+\mathrm{V}^{2}} \mathrm{dV}-\int \frac{\mathrm{V}}{1+\mathrm{V}^{2}} \mathrm{dV}=\int \frac{\mathrm{d} \mathrm{X}}{\mathrm{X}} \\\int \frac{1}{1+\mathrm{V}^{2}} \mathrm{dV}-\frac{1}{2} \int \frac{2 \mathrm{~V}}{1+\mathrm{V}^{2}} \mathrm{dV}=\int \frac{\mathrm{d} \mathrm{X}}{\mathrm{X}}\end{array}$

$\tan ^{-1} \mathrm{~V}-\frac{1}{2} \ln \left|1+\mathrm{V}^{2}\right|=\ln |\mathrm{X}|+\mathrm{C}$

From (3) we have

$\begin{array}{l}\tan ^{-1}\left(\frac{\mathrm{Y}}{\mathrm{X}}\right)-\ln \sqrt{1+\left(\frac{\mathrm{Y}}{\mathrm{X}}\right)^{2}}=\ln |\mathrm{X}|+\mathrm{C} \\\tan ^{-1}\left(\frac{\mathrm{Y}}{\mathrm{X}}\right)-\ln \sqrt{1+\frac{\mathrm{Y}^{2}}{\mathrm{X}^{2}}}=\ln |\mathrm{X}|+\mathrm{C}\end{array}$

$\begin{array}{l}\tan ^{-1}\left(\frac{Y}{X}\right)-\ln \sqrt{\frac{X^{2}+Y^{2}}{X^{2}}}=\ln |X|+C \\\tan ^{-1}\left(\frac{Y}{X}\right)-\ln \frac{\sqrt{X^{2}+Y^{2}}}{X}=\ln |X|+C\end{array}$

$\begin{array}{l}\tan ^{-1}\left(\frac{Y}{X}\right)-\ln \sqrt{X^{2}+Y^{2}}+\ln |X|=\ln |X|+C \\\tan ^{-1}\left(\frac{Y}{X}\right)-\ln \sqrt{X^{2}+Y^{2}}=C\end{array}$

$\text { Recall that, we have taken } X=X+h \text { and } y=Y+k$

$X=X+1 \text { and } y=Y-5\\X=x-1 \text { and } Y=y+5$

$\begin{array}{l}\tan ^{-1}\left(\frac{y+5}{x-1}\right)-\ln \sqrt{(x-1)^{2}+(y+5)^{2}}=C \\\tan ^{-1}\left(\frac{y+5}{x-1}\right)-\ln \sqrt{x^{2}-2 x+1+y^{2}+10 y+25}=C \\\tan ^{-1}\left(\frac{y+5}{x-1}\right)-\ln \sqrt{x^{2}-2 x+y^{2}+10 y+26}=C \\\tan ^{-1}\left(\frac{y+5}{x-1}\right)-\ln \sqrt{x^{2}+y^{2}-2 x+10 y+26}=C\end{array}$