JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

Let

I

be the purchase value of an equipment and

V(t)

be the value after it has been used for

t

years. The value

V(t)

depreciates at a rate given by the differential equation

\frac{d V}{d t}=-k(T-t)

, where

k>0

is a constant and

T

is the total life in years of the equipment. Then the scrap value

V(T)

of the equipment is:

View answer and explanation

Correct option: B

I-\frac{k T^2}{2}

Option B (

I-\frac{{k T^2}}{{2}}

) is correct because it represents the scrap value V(T) obtained by solving the given differential equation with the initial condition V(0) = I.

More Questions from DIFFERENTIAL EQUATIONS

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y=y(x)

\left(3 y^{2}-5 x^{2}\right) y d x+2 x\left(x^{2}-y^{2}\right) d y=0

y(1)=1

\left|(y(2))^{3}-12 y(2)\right|

x

{x^2} + {y^2} = 1

\left( {y' = \frac{{dy}}{{dx}},y'' = \frac{{{d^2}y}}{{d{x^2}}}} \right)

x = A\cos 4t + B\sin 4t,

\frac{{{d^2}x}}{{d{t^2}}}

y=y_{1}(x)

y=y_{2}(x)

\frac{d y}{d x}=y+7

y_{1}(0)=0

y_{2}(0)=1

y=y_{1}(x)

y=y_{2}(x)

y=y(t)

\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}

\alpha>0, \beta>0

\gamma>0

\lim \limits_{{x \rightarrow \infty}} y(t)