For the given differential equation, find the particular solution $y = - 9 \ln 2$ when $x = - \ln 2$ . $\frac { \mathrm { d } y } { \mathrm {~d} x } = 9 - e ^ { - x }$
A) $y = 9 x + e ^ { x } - 2$
B) $y = 9 x + e ^ { - x } - 2$
C) $y = - 9 x - e ^ { - x } + 2$
D) $y = 9 x + e ^ { - x }$
E) $y = e ^ { - x } - 2$

Question

For the given differential equation, find the particular solution $y = - 9 \ln 2$ when $x = - \ln 2$ .   $\frac { \mathrm { d } y } { \mathrm {~d} x } = 9 - e ^ { - x }$  
A) $y = 9 x + e ^ { x } - 2$
B)   $y = 9 x + e ^ { - x } - 2$
C)   $y = - 9 x - e ^ { - x } + 2$
D)   $y = 9 x + e ^ { - x }$
E)   $y = e ^ { - x } - 2$

Quizplus · Accepted Answer

To solve the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } = 9 - e ^ { - x }$, we integrate the right side with respect to $x$ to get $y = 9x + e^{-x} + C$, where $C$ is the constant of integration. To find $C$, we use the given condition $y = - 9 \ln 2$ when $x = - \ln 2$. Substituting these values into the solution gives $- 9 \ln 2 = 9(-\ln 2) + e^{\ln 2} + C$. Simplifying, we get $- 9 \ln 2 = -9\ln 2 + 2 + C$, leading to $C = -2$. Therefore, the particular solution is $y = 9x + e^{-x} - 2$.

For the Given Differential Equation, Find the Particular Solution y=−9ln⁡2y = - 9 \ln 2y=−9ln2

For the Given Differential Equation, Find the Particular Solution $y = - 9 \ln 2$