Solve the initial value problem.
-$$\frac { d y } { d t } + 5 y = 3 ; y ( 0 ) = 1$$
A) $y = \frac { 3 } { 5 } e ^ { - 5 t } + \frac { 2 } { 5 }$
B) $y = \frac { 2 } { 5 } e ^ { 5 t } + \frac { 3 } { 5 }$
C) $y = \frac { 2 } { 5 } e ^ { - 5 t } + \frac { 3 } { 5 }$
D) $y = \frac { 3 } { 5 } e ^ { 5 t } + \frac { 2 } { 5 }$

Question

Quizplus · Accepted Answer

First, we need to find the general solution of the differential equation:
$$\frac { d y } { d t } + 5 y = 3 $$
The associated homogeneous equation is:
$$\frac { d y } { d t } + 5 y = 0 $$
whose solution is:
$$y_h = Ce^{-5t}$$
where C is a constant.

To find the particular solution, we can use the method of undetermined coefficients. We guess a solution of the form:
$$y_p = A$$
where A is a constant. Substituting this into the differential equation, we get:
$$\frac { d y_p } { d t } + 5 y_p = 0$$
$$0 + 5A = 3$$
$$A = \frac {3}{5}$$
Therefore, the general solution of the differential equation is:
$$y = y_h + y_p = Ce^{-5t} + \frac {3}{5}$$

Using the initial condition $y(0) = 1$, we can find the value of C:
$$y(0) = C \times e^{0} + \frac {3}{5}= 1$$
$$C = \frac {2}{5}$$
Therefore, the solution of the initial value problem is:
$$y = \frac {2}{5}e^{-5t} + \frac {3}{5}$$

Solve the Initial Value Problem dydt+5y=3;y(0)=1\frac { d y } { d t } + 5 y = 3 ; y ( 0 ) = 1dtdy​+5y=3;y(0)=1

Solve the Initial Value Problem $\frac { d y } { d t } + 5 y = 3 ; y ( 0 ) = 1$