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Question-120322




Question Number 120322 by sdfg last updated on 30/Oct/20
Commented by TANMAY PANACEA last updated on 30/Oct/20
do x′ means =(dx/dt) pls...
$${do}\:{x}'\:{means}\:=\frac{{dx}}{{dt}}\:{pls}… \\ $$
Commented by Dwaipayan Shikari last updated on 30/Oct/20
(dx/dt)=x−2y−t^2   ⇒x=xt−2yt−(t^3 /3)+C  Or  (dx/dy)=x−2y−t^2 ⇒x=xy−y^2 −t^2 y+C
$$\frac{{dx}}{{dt}}={x}−\mathrm{2}{y}−{t}^{\mathrm{2}} \\ $$$$\Rightarrow{x}={xt}−\mathrm{2}{yt}−\frac{{t}^{\mathrm{3}} }{\mathrm{3}}+{C} \\ $$$${Or} \\ $$$$\frac{{dx}}{{dy}}={x}−\mathrm{2}{y}−{t}^{\mathrm{2}} \Rightarrow{x}={xy}−{y}^{\mathrm{2}} −{t}^{\mathrm{2}} {y}+{C} \\ $$
Commented by prakash jain last updated on 31/Oct/20
Second Step:  y is a constant or a function of t?
$$\mathrm{Second}\:\mathrm{Step}: \\ $$$${y}\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{or}\:\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:{t}? \\ $$
Answered by mr W last updated on 31/Oct/20
x=at^2 +bt+c  y=pt^2 +qt+r  z=ut^2 +vt+w    2at+b=at^2 +bt+c−2pt^2 −2qt−2r−t^2   (a−2p−1)t^2 +(b−2a−2q)t−b+c−2r=0  ⇒a−2p−1=0   (1)  ⇒−2a+b−2q=0   (2)  ⇒−b+c−2r=0   (3)  ......
$${x}={at}^{\mathrm{2}} +{bt}+{c} \\ $$$${y}={pt}^{\mathrm{2}} +{qt}+{r} \\ $$$${z}={ut}^{\mathrm{2}} +{vt}+{w} \\ $$$$ \\ $$$$\mathrm{2}{at}+{b}={at}^{\mathrm{2}} +{bt}+{c}−\mathrm{2}{pt}^{\mathrm{2}} −\mathrm{2}{qt}−\mathrm{2}{r}−{t}^{\mathrm{2}} \\ $$$$\left({a}−\mathrm{2}{p}−\mathrm{1}\right){t}^{\mathrm{2}} +\left({b}−\mathrm{2}{a}−\mathrm{2}{q}\right){t}−{b}+{c}−\mathrm{2}{r}=\mathrm{0} \\ $$$$\Rightarrow{a}−\mathrm{2}{p}−\mathrm{1}=\mathrm{0}\:\:\:\left(\mathrm{1}\right) \\ $$$$\Rightarrow−\mathrm{2}{a}+{b}−\mathrm{2}{q}=\mathrm{0}\:\:\:\left(\mathrm{2}\right) \\ $$$$\Rightarrow−{b}+{c}−\mathrm{2}{r}=\mathrm{0}\:\:\:\left(\mathrm{3}\right) \\ $$$$…… \\ $$

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