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Question Number 42012 by rahul 19 last updated on 16/Aug/18
Solve : (du/dt) = ((3u−7t)/(−7u+3t))
$$\mathrm{Solve}\:: \\ $$$$\frac{{d}\mathrm{u}}{{d}\mathrm{t}}\:=\:\frac{\mathrm{3u}−\mathrm{7t}}{−\mathrm{7u}+\mathrm{3t}} \\ $$
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
$${method}\:\mathrm{1} \\ $$$${this}\:{method}\:{for}\:{homogeneous}\:{equation} \\ $$$${put}\:{v}=\frac{{y}}{{x}}\:\:\:{here}\:{i}\:{have}\:{put}\:{y}=\frac{{u}}{{t}} \\ $$$$ \\ $$$$\frac{{du}}{{dt}}=\frac{\mathrm{3}\left(\frac{{u}}{{t}}\right)−\mathrm{7}}{−\mathrm{7}\left(\frac{{u}}{{t}}\right)+\mathrm{3}} \\ $$$${y}=\frac{{u}}{{t}}\:\:\:{u}={yt} \\ $$$$\frac{{du}}{{dt}}={y}+{t}\frac{{dy}}{{dt}} \\ $$$${y}+{t}\frac{{dy}}{{dt}}=\frac{\mathrm{3}{y}−\mathrm{7}}{−\mathrm{7}{y}+\mathrm{3}} \\ $$$${t}\frac{{dy}}{{dt}}=\frac{\mathrm{3}{y}−\mathrm{7}}{−\mathrm{7}{y}+\mathrm{3}}−{y} \\ $$$${t}\frac{{dy}}{{dt}}=\frac{\mathrm{3}{y}−\mathrm{7}+\mathrm{7}{y}^{\mathrm{2}} −\mathrm{3}{y}}{−\mathrm{7}{y}+\mathrm{3}} \\ $$$$\frac{−\mathrm{7}{y}+\mathrm{3}}{\mathrm{7}{y}^{\mathrm{2}} −\mathrm{7}}{dy}=\frac{{dt}}{{t}} \\ $$$$−\frac{{ydy}}{{y}^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{3}{dy}}{\mathrm{7}{y}^{\mathrm{2}} −\mathrm{7}}=\frac{{dt}}{{t}} \\ $$$$\frac{\mathrm{3}}{\mathrm{14}}\int\frac{\left({y}+\mathrm{1}\right)−\left({y}−\mathrm{1}\right)}{\left({y}+\mathrm{1}\right)\left({y}−\mathrm{1}\right)}{dy}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({y}^{\mathrm{2}} −\mathrm{1}\right)}{{y}^{\mathrm{2}} −\mathrm{1}}=\int\frac{{dt}}{{t}} \\ $$$$\frac{\mathrm{3}}{\mathrm{14}}{ln}\mid\left(\frac{{y}−\mathrm{1}}{{y}+\mathrm{1}}\right)\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({y}^{\mathrm{2}} −\mathrm{1}\right)={lnt}+{lnc} \\ $$$${ln}\left\{\left(\frac{{y}−\mathrm{1}}{{y}+\mathrm{1}}\right)^{\frac{\mathrm{3}}{\mathrm{14}}} \right\}−{ln}\left({y}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} ={ln}\left({tc}\right) \\ $$$${ln}\left\{\left(\frac{{y}−\mathrm{1}}{\left.{y}+\mathrm{1}\right)}\right)^{\frac{\mathrm{3}}{\mathrm{14}}} ×\frac{\mathrm{1}}{\left({y}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\right\}={lntc} \\ $$$$\frac{\left({y}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{14}}−\frac{\mathrm{1}}{\mathrm{2}}} }{\left({y}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{14}}+\frac{\mathrm{1}}{\mathrm{2}}} }.={tc} \\ $$$$\frac{\left({y}−\mathrm{1}\right)^{\frac{\mathrm{3}−\mathrm{7}}{\mathrm{14}}} }{\left({y}+\mathrm{1}\right)^{\frac{\mathrm{10}}{\mathrm{14}}} }={tc} \\ $$$$\frac{\left({y}−\mathrm{1}\right)^{\frac{−\mathrm{4}}{\mathrm{14}}} }{\left({y}+\mathrm{1}\right)^{\frac{\mathrm{10}}{\mathrm{14}}} }={tc} \\ $$$$\frac{\left(\frac{{u}}{{t}}−\mathrm{1}\right)^{\frac{−\mathrm{4}}{\mathrm{14}}} }{\left(\frac{{u}}{{t}}+\mathrm{1}\right)^{\frac{\mathrm{10}}{\mathrm{14}}} }={tc} \\ $$$$ \\ $$$$ \\ $$
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
$${method}\mathrm{2} \\ $$$$−\mathrm{7}{udu}+\mathrm{3}{tdu}=\mathrm{3}{udt}−\mathrm{7}{tdt} \\ $$$$\mathrm{3}{tdu}−\mathrm{3}{udt}=\mathrm{7}{udu}−\mathrm{7}{tdt} \\ $$$$\mathrm{3}\left(\frac{{tdu}−{udt}}{{t}^{\mathrm{2}} }\right){t}^{\mathrm{2}} \:=\frac{\mathrm{7}}{\mathrm{2}}\left\{\left({du}^{\mathrm{2}} \right)−\left({dt}^{\mathrm{2}} \right)\right\} \\ $$$$\mathrm{3}{t}^{\mathrm{2}} {d}\left(\frac{{u}}{{t}}\right)=\frac{\mathrm{7}}{\mathrm{2}}{d}\left({u}^{\mathrm{2}} −{t}^{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{3}{t}^{\mathrm{2}} {d}\left(\frac{{u}}{{t}}\right)}{{u}^{\mathrm{2}} −{t}^{\mathrm{2}} }=\frac{\mathrm{7}}{\mathrm{2}}×\frac{{d}\left({u}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)}{{u}^{\mathrm{2}} −{t}^{\mathrm{2}} } \\ $$$$\mathrm{3}\frac{{d}\left(\frac{{u}}{{t}}\right)}{\left(\frac{{u}^{\mathrm{2}} }{{t}^{\mathrm{2}} }\right)−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}}\frac{{d}\left({u}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)}{{u}^{\mathrm{2}} −{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}.\int\frac{\left(\frac{{u}}{{t}}+\mathrm{1}\right)−\left(\frac{{u}}{{t}}−\mathrm{1}\right)}{\left(\frac{{u}}{{t}}+\mathrm{1}\right)\left(\frac{{u}}{{t}}−\mathrm{1}\right)}{d}\left(\frac{{u}}{{t}}\right)=\frac{\mathrm{7}}{\mathrm{2}}\int\frac{{d}\left({u}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)}{{u}^{\mathrm{2}} −{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}{ln}\mid\left(\frac{\frac{{u}}{{t}}−\mathrm{1}}{\frac{{u}}{{t}}+\mathrm{1}}\right)\mid=\frac{\mathrm{7}}{\mathrm{2}}{ln}\mid\left({u}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)\mid+{lnc} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by rahul 19 last updated on 17/Aug/18
thanks sir��
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18
its ok...
$${its}\:{ok}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18
d(xy)=xdy+ydx d(x^2 +y^2 )=2(xdx+ydy) d((y/x))=((xdy−ydx)/x^2 ) d((x/y))=((ydx−xdy)/y^2 ) some times above fail then put x=rcosα y=rsinα r^2 =x^2 +y^2 2rdr=2xdx+2ydy rdr=xdx+ydy tanα=(y/x) sec^2 α(dα/dx)=((x(dy/dx)−y)/x^2 ) sec^2 αdα=((xdy−ydx)/x^2 ) use them
$${d}\left({xy}\right)={xdy}+{ydx} \\ $$$${d}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{2}\left({xdx}+{ydy}\right) \\ $$$${d}\left(\frac{{y}}{{x}}\right)=\frac{{xdy}−{ydx}}{{x}^{\mathrm{2}} } \\ $$$${d}\left(\frac{{x}}{{y}}\right)=\frac{{ydx}−{xdy}}{{y}^{\mathrm{2}} } \\ $$$${some}\:{times}\:{above}\:{fail}\:{then}\:{put} \\ $$$${x}={rcos}\alpha\:\:{y}={rsin}\alpha \\ $$$${r}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\:\:\:\mathrm{2}{rdr}=\mathrm{2}{xdx}+\mathrm{2}{ydy} \\ $$$${rdr}={xdx}+{ydy} \\ $$$${tan}\alpha=\frac{{y}}{{x}} \\ $$$${sec}^{\mathrm{2}} \alpha\frac{{d}\alpha}{{dx}}=\frac{{x}\frac{{dy}}{{dx}}−{y}}{{x}^{\mathrm{2}} } \\ $$$${sec}^{\mathrm{2}} \alpha{d}\alpha=\frac{{xdy}−{ydx}}{{x}^{\mathrm{2}} } \\ $$$${use}\:{them}\: \\ $$
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.
$$\mathrm{Sir},\:\mathrm{don}'\mathrm{t}\:\mathrm{you}\:\mathrm{feel}\:\mathrm{quite}\:\mathrm{bad}\:\mathrm{when} \\ $$$$\mathrm{you}\:\mathrm{came}\:\mathrm{to}\:\mathrm{know}\:\mathrm{our}\:\mathrm{beloved}\:\mathrm{Atalji} \\ $$$$\mathrm{is}\:\mathrm{no}\:\mathrm{more}\:\mathrm{with}\:\mathrm{us}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 17/Aug/18
yes...he was a true mass leader...good orator.. excellent writer...above all true replica of human being...
$${yes}…{he}\:{was}\:{a}\:{true}\:{mass}\:{leader}…{good}\:{orator}.. \\ $$$${excellent}\:{writer}…{above}\:{all}\:{true}\:{replica}\:{of}\:{human} \\ $$$${being}… \\ $$

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