Solve this separable differential equation How would I go about solving the following separable differential equation? (dx(t))/(dt)=8-3x with x(0)=4?

Question

Solve this separable differential equation
How would I go about solving the following separable differential equation?

\frac{d x (t)}{d t} = 8 - 3 x

with

x (0) = 4

?
My solution thus far is the following:

\int \frac{d x (t)}{8 - 3 x} = \int d t + C

\Rightarrow ln | 8 - 3 x (t) | = - 3 (t + C)

Now using the fact that

x (0) = 4

we can solve for

C = - \frac{1}{3} ln | 4 |

. From this, I would seem to get

| 8 - 3 x (t) | = e^{- 3 t - \frac{1}{3} ln 4}

but something seems to go wrong. Am I making a mistake somewhere and how could I solve this entirely?

Answer 1

\frac{d x (t)}{d t} = 8 - 3 x

\int \frac{d x (t)}{8 - 3 x} = \int d t

\frac{l n (| 8 - 3 x |)}{- 3} = t + C

(l n | 8 - 3 x |) = - 3 t + 3 C

When t=0, x=4 therefore

l n 4 = 3 C

and

C = \frac{l n 4}{3}

(l n | 8 - 3 x |) = - 3 t + l n 4

| 8 - 3 x | = e^{- 3 t + l n 4} = e^{- 3 t} . e^{l n 4} = 4. e^{- 3 t}

therefore required solution

| 8 - 3 x | = 4. e^{- 3 t}

Answers (1)