Solve : (du/dt) = ((3u−7t)/(−7u+3t))


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Question Number 42012 by rahul 19 last updated on 16/Aug/18

$Solve :$ $\frac{d u}{d t} = \frac{3 u - 7 t}{- 7 u + 3 t}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
	$method 1 this method for homogeneous equation put v=(y/x) here i have put y=(u/t) (du/dt)=((3((u/t))−7)/(−7((u/t))+3)) y=(u/t) u=yt (du/dt)=y+t(dy/dt) y+t(dy/dt)=((3y−7)/(−7y+3)) t(dy/dt)=((3y−7)/(−7y+3))−y t(dy/dt)=((3y−7+7y^2 −3y)/(−7y+3)) ((−7y+3)/(7y^2 −7))dy=(dt/t) −((ydy)/(y^2 −1))+((3dy)/(7y^2 −7))=(dt/t) (3/(14))∫(((y+1)−(y−1))/((y+1)(y−1)))dy−(1/2)∫((d(y^2 −1))/(y^2 −1))=∫(dt/t) (3/(14))ln∣(((y−1)/(y+1)))∣−(1/2)ln(y^2 −1)=lnt+lnc ln{(((y−1)/(y+1)))^(3/(14)) }−ln(y^2 −1)^(1/2) =ln(tc) ln{(((y−1)/(y+1))))^(3/(14)) ×(1/((y^2 −1)^(1/2) ))}=lntc (((y−1)^((3/(14))−(1/2)) )/((y+1)^((3/(14))+(1/2)) )).=tc (((y−1)^((3−7)/(14)) )/((y+1)^((10)/(14)) ))=tc (((y−1)^((−4)/(14)) )/((y+1)^((10)/(14)) ))=tc ((((u/t)−1)^((−4)/(14)) )/(((u/t)+1)^((10)/(14)) ))=tc$
	$m e t h o d 1$ $t h i s m e t h o d f o r h o m o g e n e o u s e q u a t i o n$ $p u t v = \frac{y}{x} h e r e i h a v e p u t y = \frac{u}{t}$ $\frac{d u}{d t} = \frac{3 (\frac{u}{t}) - 7}{- 7 (\frac{u}{t}) + 3}$ $y = \frac{u}{t} u = y t$ $\frac{d u}{d t} = y + t \frac{d y}{d t}$ $y + t \frac{d y}{d t} = \frac{3 y - 7}{- 7 y + 3}$ $t \frac{d y}{d t} = \frac{3 y - 7}{- 7 y + 3} - y$ $t \frac{d y}{d t} = \frac{3 y - 7 + 7 y^{2} - 3 y}{- 7 y + 3}$ $\frac{- 7 y + 3}{7 y^{2} - 7} d y = \frac{d t}{t}$ $- \frac{y d y}{y^{2} - 1} + \frac{3 d y}{7 y^{2} - 7} = \frac{d t}{t}$ $\frac{3}{14} \int \frac{(y + 1) - (y - 1)}{(y + 1) (y - 1)} d y - \frac{1}{2} \int \frac{d (y^{2} - 1)}{y^{2} - 1} = \int \frac{d t}{t}$ $\frac{3}{14} l n ∣ (\frac{y - 1}{y + 1}) ∣ - \frac{1}{2} l n (y^{2} - 1) = l n t + l n c$ $l n {{(\frac{y - 1}{y + 1})}^{\frac{3}{14}}} - l n {(y^{2} - 1)}^{\frac{1}{2}} = l n (t c)$ $l n {{(\frac{y - 1}{y + 1)})}^{\frac{3}{14}} \times \frac{1}{{(y^{2} - 1)}^{\frac{1}{2}}}} = l n t c$ $\frac{{(y - 1)}^{\frac{3}{14} - \frac{1}{2}}}{{(y + 1)}^{\frac{3}{14} + \frac{1}{2}}} . = t c$ $\frac{{(y - 1)}^{\frac{3 - 7}{14}}}{{(y + 1)}^{\frac{10}{14}}} = t c$ $\frac{{(y - 1)}^{\frac{- 4}{14}}}{{(y + 1)}^{\frac{10}{14}}} = t c$ $\frac{{(\frac{u}{t} - 1)}^{\frac{- 4}{14}}}{{(\frac{u}{t} + 1)}^{\frac{10}{14}}} = t c$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
	$method2 −7udu+3tdu=3udt−7tdt 3tdu−3udt=7udu−7tdt 3(((tdu−udt)/t^2 ))t^2 =(7/2){(du^2 )−(dt^2 )} 3t^2 d((u/t))=(7/2)d(u^2 −t^2 ) ((3t^2 d((u/t)))/(u^2 −t^2 ))=(7/2)×((d(u^2 −t^2 ))/(u^2 −t^2 )) 3((d((u/t)))/(((u^2 /t^2 ))−1))=(7/2)((d(u^2 −t^2 ))/(u^2 −t^2 )) (3/2).∫((((u/t)+1)−((u/t)−1))/(((u/t)+1)((u/t)−1)))d((u/t))=(7/2)∫((d(u^2 −t^2 ))/(u^2 −t^2 )) (3/2)ln∣((((u/t)−1)/((u/t)+1)))∣=(7/2)ln∣(u^2 −t^2 )∣+lnc$
	$m e t h o d 2$ $- 7 u d u + 3 t d u = 3 u d t - 7 t d t$ $3 t d u - 3 u d t = 7 u d u - 7 t d t$ $3 (\frac{t d u - u d t}{t^{2}}) t^{2} = \frac{7}{2} {({d u}^{2}) - ({d t}^{2})}$ $3 t^{2} d (\frac{u}{t}) = \frac{7}{2} d (u^{2} - t^{2})$ $\frac{3 t^{2} d (\frac{u}{t})}{u^{2} - t^{2}} = \frac{7}{2} \times \frac{d (u^{2} - t^{2})}{u^{2} - t^{2}}$ $3 \frac{d (\frac{u}{t})}{(\frac{u^{2}}{t^{2}}) - 1} = \frac{7}{2} \frac{d (u^{2} - t^{2})}{u^{2} - t^{2}}$ $\frac{3}{2} . \int \frac{(\frac{u}{t} + 1) - (\frac{u}{t} - 1)}{(\frac{u}{t} + 1) (\frac{u}{t} - 1)} d (\frac{u}{t}) = \frac{7}{2} \int \frac{d (u^{2} - t^{2})}{u^{2} - t^{2}}$ $\frac{3}{2} l n ∣ (\frac{\frac{u}{t} - 1}{\frac{u}{t} + 1}) ∣= \frac{7}{2} l n ∣ (u^{2} - t^{2}) ∣ + l n c$
	Commented by rahul 19 last updated on 17/Aug/18
	thanks sir��
	Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18

	$i t s o k . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18

	$d (x y) = x d y + y d x$ $d (x^{2} + y^{2}) = 2 (x d x + y d y)$ $d (\frac{y}{x}) = \frac{x d y - y d x}{x^{2}}$ $d (\frac{x}{y}) = \frac{y d x - x d y}{y^{2}}$ $s o m e t i m e s a b o v e f a i l t h e n p u t$ $x = r c o s α y = r s i n α$ $r^{2} = x^{2} + y^{2} 2 r d r = 2 x d x + 2 y d y$ $r d r = x d x + y d y$ $t a n α = \frac{y}{x}$ ${s e c}^{2} α \frac{d α}{d x} = \frac{x \frac{d y}{d x} - y}{x^{2}}$ ${s e c}^{2} α d α = \frac{x d y - y d x}{x^{2}}$ $u s e t h e m$
	Commented by rahul 19 last updated on 17/Aug/18

	$Sir, {don}^{'} t you feel quite bad when$ $you came to know our beloved Atalji$ $is no more with us .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18

	$y e s . . . h e w a s a t r u e m a s s l e a d e r . . . g o o d o r a t o r . .$ $e x c e l l e n t w r i t e r . . . a b o v e a l l t r u e r e p l i c a o f h u m a n$ $b e i n g . . .$
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