Find the solution of the initial-value problem $\frac { d y } { d t } = 3 y + 1$ , $y ( 0 ) = 2$ .
A) $y = \frac { 1 } { 3 } \left( 7 e ^ { t } - 1 \right)$
B) $y = \left( 7 e ^ { t } - 5 \right)$
C) $y = \frac { 1 } { 3 } \left( e ^ { 3 t } - 1 \right)$
D) $y = \left( 3 e ^ { 3 t } - 1 \right)$
E) $y = \frac { 1 } { 3 } \left( 5 e ^ { 3 t } + 1 \right)$
F) $y = \left( 5 e ^ { 3 t } + 1 \right)$
G) $y = \frac { 1 } { 3 } \left( 7 e ^ { 3 t } - 1 \right)$
H) $y = \left( 7 e ^ { 3 t } - 1 \right)$

Question

Find the solution of the initial-value problem $\frac { d y } { d t } = 3 y + 1$ , $y ( 0 ) = 2$ .
A) $y = \frac { 1 } { 3 } \left( 7 e ^ { t } - 1 \right)$ 
B) $y = \left( 7 e ^ { t } - 5 \right)$ 
C) $y = \frac { 1 } { 3 } \left( e ^ { 3 t } - 1 \right)$ 
D) $y = \left( 3 e ^ { 3 t } - 1 \right)$ 
E) $y = \frac { 1 } { 3 } \left( 5 e ^ { 3 t } + 1 \right)$ 
F) $y = \left( 5 e ^ { 3 t } + 1 \right)$ 
G) $y = \frac { 1 } { 3 } \left( 7 e ^ { 3 t } - 1 \right)$ 
H) $y = \left( 7 e ^ { 3 t } - 1 \right)$

Quizplus · Accepted Answer

The differential equation is separable and linear. Solving it gives $y = Ce^{3t} - \frac{1}{3}$. Using the initial condition $y(0) = 2$, we find $C = \frac{7}{3}$. Thus, the solution is $y = \frac{1}{3}(7e^{3t} - 1)$.

Find the Solution of the Initial-Value Problem dydt=3y+1\frac { d y } { d t } = 3 y + 1dtdy​=3y+1

Find the Solution of the Initial-Value Problem $\frac { d y } { d t } = 3 y + 1$