Solve the differential equation. $4 \frac { d w } { d t } + 5 e ^ { t + w } = 0$
A) $w = \frac { 5 } { 4 } e ^ { t } - C$
B) $w = \ln \left( \frac { 4 } { 5 } t \right) - t$
C) $w = - \ln \left( \frac { 5 e ^ { z } } { 4 } - C \right)$
D) $w = \ln \left( \frac { 4 } { 5 } + \frac { 4 } { 5 } C e ^ { t } \right) - t$
E) $\text { none of these }$

Question

Quizplus · Accepted Answer

Separating variables and integrating, we get$$4 dw = -5e^{t+w}dt$$$$\Rightarrow \int 4 dw = -\int 5e^{t+w}dt$$$$\Rightarrow 4w = -5e^{t+w}+C$$$$\Rightarrow w = -\frac{5}{4}e^{t+w}+\frac{C}{4}$$$$\Rightarrow w+\frac{5}{4}e^{t+w}=\frac{C}{4}$$$$\Rightarrow e^{t+w}=\frac{4}{5}\left(w-\frac{C}{4}\right)$$$$\Rightarrow w=\ln\left(\frac{4}{5}\left(w-\frac{C}{4}\right)\right)-t$$$$\Rightarrow w=-\ln\left(\frac{5e^z}{4}-C\right)$$

Solve the Differential Equation 4dwdt+5et+w=04 \frac { d w } { d t } + 5 e ^ { t + w } = 04dtdw​+5et+w=0

Solve the Differential Equation $4 \frac { d w } { d t } + 5 e ^ { t + w } = 0$