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Differential Equations question

2011 · Shift 0 · Q38
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  5. /2011 · Shift 0 · Q38

Differential Equations question

2011 · Shift 0 · Q38

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let III be the purchase value of an equipment and V(t)V(t)V(t) be the value after it has been used for ttt years. The value V(t)V(t)V(t) depreciates at a rate given by differential equation dv(t)dt=−k(T−t),{{dv\left( t \right)} \over {dt}} = - k\left( {T - t} \right),dtdv(t)​=−k(T−t), where k>0k\gt 0k>0 is a constant and TTT is the total life in years of the equipment. Then the scrap value V(T)V(T)V(T) of the equipment is
  1. A
    I−kT22I - {{k{T^2}} \over 2}I−2kT2​
  2. B
    I−k(T−t)22I - {{k{{\left( {T - t} \right)}^2}} \over 2}I−2k(T−t)2​
  3. C
    e−kT{e^{ - kT}}e−kT
  4. D
    T2−1k{T^2} - {1 \over k}T2−k1​
View written solutionFree

Correct answer: A

  1. Given differential equation

    The value of the equipment satisfies dVdt=−k(T−t),k>0.\frac{dV}{dt}=-k(T-t), \qquad k>0.dtdV​=−k(T−t),k>0.

    Also, at time t=0t=0t=0, the purchase value is V(0)=I.V(0)=I.V(0)=I.

  2. Integrate the differential equation

    We have dVdt=−k(T−t).\frac{dV}{dt}=-k(T-t).dtdV​=−k(T−t).

    Integrating with respect to ttt: V(t)=∫−k(T−t) dt=−k(Tt−t22)+C.V(t)=\int -k(T-t)\,dt = -k\left(Tt-\frac{t^2}{2}\right)+C.V(t)=∫−k(T−t)dt=−k(Tt−2t2​)+C.

    So, V(t)=−kTt+kt22+C.V(t)=-kTt+\frac{k t^2}{2}+C.V(t)=−kTt+2kt2​+C.

  3. Use the initial condition

    Since V(0)=IV(0)=IV(0)=I, I=−kT(0)+k⋅022+C=C.I = -kT(0)+\frac{k\cdot 0^2}{2}+C = C.I=−kT(0)+2k⋅02​+C=C.

    Hence, V(t)=I−kTt+kt22.V(t)=I-kTt+\frac{k t^2}{2}. V(t)=I−kTt+2kt2​.

  4. Find the scrap value V(T)V(T)V(T)

    Put t=Tt=Tt=T: V(T)=I−kT(T)+kT22V(T)=I-kT(T)+\frac{kT^2}{2}V(T)=I−kT(T)+2kT2​ =I−kT2+kT22=I-kT^2+\frac{kT^2}{2}=I−kT2+2kT2​ =I−kT22.=I-\frac{kT^2}{2}. =I−2kT2​.

  5. Match with the options

    V(T)=I−kT22V(T)=I-\frac{kT^2}{2}V(T)=I−2kT2​ which is Option A.

  6. Check other options briefly

    • B depends on ttt, so it cannot represent the scrap value at t=Tt=Tt=T unless simplified.
    • C is unrelated to the linear differential equation here.
    • D has incorrect form and dimensions.

Therefore, the correct answer is A.

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