suchonosdy

2022-07-17

Solve this separable differential equation
How would I go about solving the following separable differential equation?
$\frac{dx\left(t\right)}{dt}=8-3x$ with $x\left(0\right)=4$?
My solution thus far is the following:
$\int \frac{dx\left(t\right)}{8-3x}=\int dt+C$
$⇒\text{ln}|8-3x\left(t\right)|=-3\left(t+C\right)$
Now using the fact that $x\left(0\right)=4$ we can solve for $C=-\frac{1}{3}\text{ln}|4|$. From this, I would seem to get
$|8-3x\left(t\right)|={e}^{-3t-\frac{1}{3}\text{ln}4}$
but something seems to go wrong. Am I making a mistake somewhere and how could I solve this entirely?

taguetzbo

$\frac{dx\left(t\right)}{dt}=8-3x$
$\int \frac{dx\left(t\right)}{8-3x}=\int dt$
$\frac{ln\left(|8-3x|\right)}{-3}=t+C$
$\left(ln|8-3x|\right)=-3t+3C$
When t=0, x=4 therefore
$ln4=3C$ and $C=\frac{ln4}{3}$
$\left(ln|8-3x|\right)=-3t+ln4$
$|8-3x|={e}^{-3t+ln4}={e}^{-3t}.{e}^{ln4}=4.{e}^{-3t}$
therefore required solution
$|8-3x|=4.{e}^{-3t}$

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