Consider the market model
$\begin{array} { l }
Q _ { S } = 4 P - 3 \
Q _ { D } = - 2 P + 13 \
\frac { d P } { d t } = 0.4 \left( Q _ { D } - Q _ { S } \right)
\end{array}$
Find an expression for $Q _ { D } ( t )$ when $P ( 0 ) = 2$.
A) $\frac { 2 } { 3 } \left( 4 - e ^ { - 2.4 t } \right)$

B) $\frac { 1 } { 3 } \left( 23 + 4 e ^ { - 2.4 t } \right)$

C) $\frac { 1 } { 3 } \left( 4 + e ^ { - 2.4 t } \right)$

D) $\frac { 2 } { 3 } \left( 4 + e ^ { - 2.4 t } \right)$

E) $\frac { 1 } { 3 } \left( 23 - 8 e ^ { - 2.4 t } \right)$

Question

Consider the market model 
$\begin{array} { l } 
Q _ { S } = 4 P - 3 \
Q _ { D } = - 2 P + 13 \
\frac { d P } { d t } = 0.4 \left( Q _ { D } - Q _ { S } \right)
\end{array}$
Find an expression for $Q _ { D } ( t )$ when $P ( 0 ) = 2$.
A) $\frac { 2 } { 3 } \left( 4 - e ^ { - 2.4 t } \right)$

B) $\frac { 1 } { 3 } \left( 23 + 4 e ^ { - 2.4 t } \right)$

C) $\frac { 1 } { 3 } \left( 4 + e ^ { - 2.4 t } \right)$

D) $\frac { 2 } { 3 } \left( 4 + e ^ { - 2.4 t } \right)$

E) $\frac { 1 } { 3 } \left( 23 - 8 e ^ { - 2.4 t } \right)$

Quizplus · Accepted Answer

We start by substituting the equation for Qs and Qd into the third equation and solve for dP/dt.

\begin{aligned} \frac { d P } { d t } &= 0.4 \left( Q _ { D } - Q _ { S } ight) \ &= 0.4 \left( ( - 2 P + 13 ) - ( 4 P - 3 ) ight) \ &= - 0.8 P + 4.4 \end{aligned}

This is a separable differential equation that we can solve using integration.

\begin{aligned} \frac { d P } { d t } &= - 0.8 P + 4.4 \ \frac { d P } { - 0.8 P + 4.4 } &= d t \ \int \frac { 1 } { - 0.8 P + 4.4 } d P &= \int d t \ - \frac { 1 } { 0.8 } \ln | - 0.8 P + 4.4 | &= t + C \ \ln | 4.4 - 0.8 P | &= - 0.8 t + C \ | 4.4 - 0.8 P | &= e ^ { - 0.8 t + C } \ 4.4 - 0.8 P &= \pm e ^ { - 0.8 t + C } \ P &= \frac { 23 } { 8 } - \frac { 1 } { 2 } e ^ { - 0.8 t + C } \end{aligned}

To determine the value of the constant C, we use the initial condition P(0) = 2.

\begin{aligned} P ( 0 ) &= \frac { 23 } { 8 } - \frac { 1 } { 2 } e ^ { C } \ 2 &= \frac { 23 } { 8 } - \frac { 1 } { 2 } e ^ { C } \ e ^ { C } &= \frac { 9 } { 4 } \ C &= \ln \frac { 9 } { 4 } \end{aligned}

Substituting this value of C back into our equation for P, we get

\begin{aligned} P &= \frac { 23 } { 8 } - \frac { 1 } { 2 } e ^ { - 0.8 t + \ln \frac { 9 } { 4 } } \ &= \frac { 23 } { 8 } - \frac { 1 } { 2 } \cdot \frac { 9 } { 4 } e ^ { - 0.8 t } \ &= \frac { 23 } { 8 } - \frac { 9 } { 8 } e ^ { - 0.8 t } \end{aligned}

To find Qd, we substitute this expression for P into the equation for Qd.

\begin{aligned} Q _ { D } &= - 2 P + 13 \ &= - 2 \left( \frac { 23 } { 8 } - \frac { 9 } { 8 } e ^ { - 0.8 t } ight) + 13 \ &= \frac { 1 } { 4 } e ^ { - 0.8 t } + \frac { 11 } { 2 } \end{aligned}

Thus the final answer is (B).

Consider the Market Model QS=4P−3QD=−2P+13dPdt=0.4(QD−QS)\begin{array} { l } Q _ { S } = 4 P - 3 \\Q _ { D } = - 2 P + 13 \\\frac { d P } { d t } = 0.4 \left( Q _ { D } - Q _ { S } \right)\end{array}QS​=4P−3QD​=−2P+13dtdP​=0.4(QD​−QS​)​

Consider the Market Model $\begin{array} { l } Q _ { S } = 4 P - 3 \\Q _ { D } = - 2 P + 13 \\\frac { d P } { d t } = 0.4 \left( Q _ { D } - Q _ { S } \right)\end{array}$