suchonosdy

2022-07-17

Solve this separable differential equation

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

$\frac{dx(t)}{dt}=8-3x$ with $x(0)=4$?

My solution thus far is the following:

$\int \frac{dx(t)}{8-3x}=\int dt+C$

$\Rightarrow \text{ln}|8-3x(t)|=-3(t+C)$

Now using the fact that $x(0)=4$ we can solve for $C=-\frac{1}{3}\text{ln}|4|$. From this, I would seem to get

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

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

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

$\frac{dx(t)}{dt}=8-3x$ with $x(0)=4$?

My solution thus far is the following:

$\int \frac{dx(t)}{8-3x}=\int dt+C$

$\Rightarrow \text{ln}|8-3x(t)|=-3(t+C)$

Now using the fact that $x(0)=4$ we can solve for $C=-\frac{1}{3}\text{ln}|4|$. From this, I would seem to get

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

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

taguetzbo

Beginner2022-07-18Added 16 answers

$\frac{dx(t)}{dt}=8-3x$

$\int \frac{dx(t)}{8-3x}=\int dt$

$\frac{ln(|8-3x|)}{-3}=t+C$

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

When t=0, x=4 therefore

$ln4=3C$ and $C=\frac{ln4}{3}$

$(ln|8-3x|)=-3t+ln4$

$|8-3x|={e}^{-3t+ln4}={e}^{-3t}.{e}^{ln4}=4.{e}^{-3t}$

therefore required solution

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

$\int \frac{dx(t)}{8-3x}=\int dt$

$\frac{ln(|8-3x|)}{-3}=t+C$

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

When t=0, x=4 therefore

$ln4=3C$ and $C=\frac{ln4}{3}$

$(ln|8-3x|)=-3t+ln4$

$|8-3x|={e}^{-3t+ln4}={e}^{-3t}.{e}^{ln4}=4.{e}^{-3t}$

therefore required solution

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

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