# Solve the differential equation dy/dx = (x + 2y - 3)/(2x + y - 3)

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Solve the differential equation dy/dx = (x + 2y - 3)/(2x + y - 3)

by (53.4k points)
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Take x = X + l; y = Y + m (l and m are constants). So,

dy/dx = dY/dX

Therefore, the equation becomes (in X, Y)

If l, m are chosen to satisfy

In X, Y the equation is homogeneous and of the first degree. Put Y = VX. Then

Separating the variables (X, V) and integrating,

where A is an arbitrary constant.

ago by (51.0k points)

Take $x$ = X + l; y = Y + m  ($\because$ l and m are constants)

So, $\frac{dy}{dx}$ = $\frac{dY}{dX}$

Therefore, the equations becomes (in X, Y)

$\frac{dY}{dX}$ = $\frac{X+2Y+ \ell+2m-3}{2X+Y+2\ell+m-3}$ = $\frac{X+2Y}{2X+y}$

If l, m are chosen to satisfy

$\ell$ + 2m - 3 = 0

2$\ell$ + m - 3 = 0

(where $\ell$ = 1 and m = 1)

In X, Y the equation is homogeneous and of the first degree. Put Y = VX. Then

V + X$\frac{dV}{dX}$ = $\frac{X+2VX}{2X+VX}$ = $\frac{1+2V}{2+V}$

X$\frac{dV}{dX}$ = $\frac{1+2V-(2+V)V}{2+V}$ = $\frac{1-V^2}{2+V}$

Separating the variables (X,V) and integrating,

$\int\frac{2+V}{1-V^2}dV$ = $\int{\frac{dX}{X}}$+A

where A is an arbitrary constant.

$\int\frac{2+V}{1-V^2}dV$ = $\int{\frac{dX}{X}}$ + A

$\Rightarrow$ $\int\frac{2}{1-V^2}dV$ + $\frac{1}{-2}\int\frac{-2V}{1-V^2}dV$ = $\int{\frac{dX}{X}}$ + A

$\Rightarrow$ 2$\big(\frac{1}{2}log|\frac{1+V}{1-V}|\big)$ - $\frac{1}{2}$log(1-V2) = logX + logC

$\big(\because\int\frac{1}{a^2-x^2}dx$ = $\frac{1}{2a}log|\frac{a+x}{a-x}|$ and here a = 1 and putting A = logC $\big)$

$\Rightarrow$ log$|\frac{1+V}{1-V}|$ - log$\sqrt{1-V^2}$ = logXC  ($\because$ logan = nloga)

$\Rightarrow$ log $\Big(\big(\frac{1+V}{1-V}\big)\frac{1}{\sqrt{1-v^2}}\Big)$ = logXC  $\big(log\frac{A}{B}= logA-logB\big)$

$\Rightarrow$ $\frac{1+V}{1-V}$ $\frac{1}{\sqrt{1-V^2}}$ = XC

Now putting V = $\frac{Y}{X}$ ($\because$ Y = VX)

$\frac{1+\frac{Y}{X}}{1-\frac{Y}{X}}$ $\frac{1}{\sqrt{1-\frac{Y^2}{X^2}}}$ = XC

$\Rightarrow$ $\frac{X+Y}{X-Y}$ $\frac{X}{\sqrt{X^2-Y^2}}$ = XC

$\Rightarrow$

 $\frac{X+Y}{X-Y}$ $\frac{1}{\sqrt{X^2-Y^2}}$ = C

where C is constant

$\Rightarrow$ X+Y = C (X-Y) $\sqrt{X^2-Y^2}$

Now

$x= X+\ell$ = X + 1 $\Rightarrow$ X = $x-1$

y = Y + m = Y + 1 $\Rightarrow$ Y = y -1

then, $x-1$ + y - 1 = C ( $x-1$ - y + 1) $\sqrt{(x-1)^2-(y-1)^2}$

$\Rightarrow$

 $x+y-2$ = C ($x-y$) $\sqrt{(x-1)^2-(y-1)^2}$