Solve the diferential equatuion (dy/dx)=((2x+y+1)/(x−2y+3))


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Question Number 37913 by gunawan last updated on 19/Jun/18

$Solve the diferential equatuion$ $\frac{d y}{d x} = \frac{2 x + y + 1}{x - 2 y + 3}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jun/18
	$let x=X+h y=Y+k (dx/dX)=1 (dy/dY)=1 so(dy/dx)=(dY/dX) (dY/dX)=((2(X+h)+Y+k+1)/((X+h)−2(Y+k)+3)) (dY/dX)=((2X+Y+2h+k+1)/(X−2Y+h−2k+3)) put 2h+k+1=0 h−2k+3=0 2h+k+1=0 2h−4k+6=0 substructing 5k−5=0 k=1 h−2k+3=0 h−2+3=0 h=−1 (dY/dX)=((2X+Y)/(X−2Y)) (dY/dX)=((2+(Y/X))/(1−2(Y/X))) put (Y/X)=V Y=VX (dY/dX)=V+X(dV/dX) V+X(dV/dX)=((2+V)/(1−2V)) X(dV/dX)=((2+V)/(1−2V))−V ((X(dV/dX)=((2+V−V+2V^2 )/(1−2V)))/) ((1−2V)/(2+2V^2 ))dV=(dX/X) (1/2)∫(dV/(1+V^2 ))−(1/2)∫((2V)/(1+V^2 ))=∫(dX/X) (1/2)tan^(−1) V−(1/2)ln(1+V^2 )=lnX+lnc (1/2)tan^(−1) ((Y/X))−(1/2)ln(1+(Y^2 /X^2 ))=lnX+lnc (1/2)tan^(−1) (((y−k)/(x−h)))−(1/2)ln{1+(((y−k)/(x−h)))^2 }=ln(x−h) +lnc (1/2)tan^(−1) (((y−1)/(x+1)))−(1/2)ln{1+(((y−1)/(x+1)))^2 }=ln(x+1) +lnc Ans$
	$l e t x = X + h$ $y = Y + k$ $\frac{d x}{d X} = 1 \frac{d y}{d Y} = 1$ $s o \frac{d y}{d x} = \frac{d Y}{d X}$ $\frac{d Y}{d X} = \frac{2 (X + h) + Y + k + 1}{(X + h) - 2 (Y + k) + 3}$ $\frac{d Y}{d X} = \frac{2 X + Y + 2 h + k + 1}{X - 2 Y + h - 2 k + 3}$ $p u t 2 h + k + 1 = 0$ $h - 2 k + 3 = 0$ $2 h + k + 1 = 0$ $2 h - 4 k + 6 = 0$ $s u b s t r u c t i n g$ $5 k - 5 = 0 k = 1$ $h - 2 k + 3 = 0$ $h - 2 + 3 = 0 h = - 1$ $\frac{d Y}{d X} = \frac{2 X + Y}{X - 2 Y}$ $\frac{d Y}{d X} = \frac{2 + \frac{Y}{X}}{1 - 2 \frac{Y}{X}} p u t \frac{Y}{X} = V$ $Y = V X$ $\frac{d Y}{d X} = V + X \frac{d V}{d X}$ $V + X \frac{d V}{d X} = \frac{2 + V}{1 - 2 V}$ $X \frac{d V}{d X} = \frac{2 + V}{1 - 2 V} - V$ $\frac{X \frac{d V}{d X} = \frac{2 + V - V + 2 V^{2}}{1 - 2 V}}{}$ $\frac{1 - 2 V}{2 + 2 V^{2}} d V = \frac{d X}{X}$ $\frac{1}{2} \int \frac{d V}{1 + V^{2}} - \frac{1}{2} \int \frac{2 V}{1 + V^{2}} = \int \frac{d X}{X}$ $\frac{1}{2} {t a n}^{- 1} V - \frac{1}{2} l n (1 + V^{2}) = l n X + l n c$ $\frac{1}{2} {t a n}^{- 1} (\frac{Y}{X}) - \frac{1}{2} l n (1 + \frac{Y^{2}}{X^{2}}) = l n X + l n c$ $\frac{1}{2} {t a n}^{- 1} (\frac{y - k}{x - h}) - \frac{1}{2} l n {1 + {(\frac{y - k}{x - h})}^{2}} = l n (x - h)$ $+ l n c$ $\frac{1}{2} {t a n}^{- 1} (\frac{y - 1}{x + 1}) - \frac{1}{2} l n {1 + {(\frac{y - 1}{x + 1})}^{2}} = l n (x + 1)$ $+ l n c A n s$
	Commented by gunawan last updated on 19/Jun/18

	$Wow, Nice Sir$
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