Exams Joint Entrance Examination Mathematics Differential Equations Question 464 Question 464 of 490 Medium MCQ
Let x = x ( t ) x=x(\mathrm{t}) x = x ( t ) and y = y ( t ) y=y(\mathrm{t}) y = y ( t ) be solutions of the differential equations d x d t + a x = 0 \frac{\mathrm{d} x}{\mathrm{dt}}+\mathrm{a} x=0 dt d x + a x = 0 and d y d t + b y = 0 \frac{\mathrm{d} y}{\mathrm{dt}}+\mathrm{by}=0 dt d y + by = 0 respectively, a , b ∈ R \mathrm{a}, \mathrm{b} \in \mathbf{R} a , b ∈ R . Given that x ( 0 ) = 2 ; y ( 0 ) = 1 x(0)=2 ; y(0)=1 x ( 0 ) = 2 ; y ( 0 ) = 1 and 3 y ( 1 ) = 2 x ( 1 ) 3 y(1)=2 x(1) 3 y ( 1 ) = 2 x ( 1 ) , the value of t \mathrm{t} t , for which x ( t ) = y ( t ) x(\mathrm{t})=y(\mathrm{t}) x ( t ) = y ( t ) , is :
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\[ d x d t + a x = 0 d x x = − a d t ∫ d x x = − a ∫ d t ln ∣ x ∣ = − a t + c a t t = 0 , x = 2 ln 2 = 0 + c ln x = − a t + ln 2 x 2 = e − a t x = 2 e − a t .... (i) \begin{aligned}
& \frac{\mathrm{dx}}{\mathrm{dt}}+\mathrm{ax}=0 \\
& \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{adt} \\
& \int \frac{\mathrm{dx}}{\mathrm{x}}=-\mathrm{a} \int \mathrm{dt} \\
& \ln |\mathrm{x}|=-\mathrm{at}+\mathrm{c} \\
& \mathrm{at} \mathrm{t}=0, \mathrm{x}=2 \\
& \ln 2=0+\mathrm{c} \\
& \ln \mathrm{x}=-\mathrm{at}+\ln 2 \\
& \frac{\mathrm{x}}{2}=\mathrm{e}^{-\mathrm{at}} \\
& \mathrm{x}=2 \mathrm{e}^{-\mathrm{at}} \quad \text{.... (i)}
\end{aligned} dt dx + ax = 0 x dx = − adt ∫ x dx = − a ∫ dt ln ∣ x ∣ = − at + c at t = 0 , x = 2 ln 2 = 0 + c ln x = − at + ln 2 2 x = e − at x = 2 e − at .... (i) \]
\[ d y d t + b y = 0 d y y = − b d t ln ∣ y ∣ = − b t + λ t = 0 , y = 1 0 = 0 + λ y = e − b t ..... (ii) \begin{aligned}
& \frac{d y}{d t}+b y=0 \\
& \frac{d y}{y}=-b d t \\
& \ln |y|=-b t+\lambda \\
& t=0, y=1 \\
& 0=0+\lambda \\
& y=e^{-b t} \quad \text{..... (ii)}
\end{aligned} d t d y + b y = 0 y d y = − b d t ln ∣ y ∣ = − b t + λ t = 0 , y = 1 0 = 0 + λ y = e − b t ..... (ii) \]
According to question
\[ 3 y ( 1 ) = 2 x ( 1 ) 3 e − b = 2 ( 2 e − a ) e a − b = 4 3 \begin{aligned}
& 3 \mathrm{y}(1)=2 \mathrm{x}(1) \\
& 3 \mathrm{e}^{-\mathrm{b}}=2\left(2 \mathrm{e}^{-\mathrm{a}}\right) \\
& \mathrm{e}^{\mathrm{a}-\mathrm{b}}=\frac{4}{3}
\end{aligned} 3 y ( 1 ) = 2 x ( 1 ) 3 e − b = 2 ( 2 e − a ) e a − b = 3 4 \]
\[ For x ( t ) = y ( t ) \[ 2 e − a t = e − b t 2 = e ( a − b ) t 2 = ( 4 3 ) t log 4 3 2 = t \begin{aligned}
& \text { For } x(t)=y(t) \\
& \[\begin{gathered} 2 \mathrm{e}^{-a t}=e^{-b t} \\ 2=e^{(a-b) t} \\ 2=\left(\frac{4}{3}\right)^t \\ \log _{\frac{4}{3}} 2=t \end{gathered}\end{aligned} For x ( t ) = y ( t ) \[ 2 e − a t = e − b t 2 = e ( a − b ) t 2 = ( 3 4 ) t log 3 4 2 = t \]
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