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dy-dt-dx-dt-t-dy-dt-y-t-dx-dt-x-




Question Number 1030 by 123456 last updated on 20/May/15
(dy/dt)−(dx/dt)=((t(dy/dt)+y)/(t(dx/dt)+x))
$$\frac{{dy}}{{dt}}−\frac{{dx}}{{dt}}=\frac{{t}\frac{{dy}}{{dt}}+{y}}{{t}\frac{{dx}}{{dt}}+{x}} \\ $$
Answered by prakash jain last updated on 21/May/15
y=k_1 t,x=k_2 t  (dy/dt)=k_1 ,(dx/dt)=k_2   k_1 −k_2 =((k_1 t+k_1 t)/(k_2 t+k_2 t))=(k_1 /k_2 )  k_1 −k_2 =(k_1 /k_2 )  k_2 ^2 −k_1 k_2 +k_1 =0  k_2 =((k_1 ±(√(k_1 ^2 −4k_1 )))/2)  k_1 ^2 −4k_1 ≥0⇒k_1 (k_1 −4)≥0⇒k_1 ≤0∨k_1 ≥4
$${y}={k}_{\mathrm{1}} {t},{x}={k}_{\mathrm{2}} {t} \\ $$$$\frac{{dy}}{{dt}}={k}_{\mathrm{1}} ,\frac{{dx}}{{dt}}={k}_{\mathrm{2}} \\ $$$${k}_{\mathrm{1}} −{k}_{\mathrm{2}} =\frac{{k}_{\mathrm{1}} {t}+{k}_{\mathrm{1}} {t}}{{k}_{\mathrm{2}} {t}+{k}_{\mathrm{2}} {t}}=\frac{{k}_{\mathrm{1}} }{{k}_{\mathrm{2}} } \\ $$$${k}_{\mathrm{1}} −{k}_{\mathrm{2}} =\frac{{k}_{\mathrm{1}} }{{k}_{\mathrm{2}} } \\ $$$${k}_{\mathrm{2}} ^{\mathrm{2}} −{k}_{\mathrm{1}} {k}_{\mathrm{2}} +{k}_{\mathrm{1}} =\mathrm{0} \\ $$$${k}_{\mathrm{2}} =\frac{{k}_{\mathrm{1}} \pm\sqrt{{k}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{4}{k}_{\mathrm{1}} }}{\mathrm{2}} \\ $$$${k}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{4}{k}_{\mathrm{1}} \geqslant\mathrm{0}\Rightarrow{k}_{\mathrm{1}} \left({k}_{\mathrm{1}} −\mathrm{4}\right)\geqslant\mathrm{0}\Rightarrow{k}_{\mathrm{1}} \leqslant\mathrm{0}\vee{k}_{\mathrm{1}} \geqslant\mathrm{4} \\ $$$$ \\ $$

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