Question

Consider the following differential equation` :
$\frac{dx}{dt}=3x$ initial condition :$x=5$ at $t=0$. The value of $x$ at $t=4$ is

• A. $-2e^{10}$
• B. $5e^{12}$
• C. $2e^5$
• D. $-5e^{12}$

Answer 1

The correct option is B $5e^{12}$

$\frac{dx}{dt}=3x$
$\int \frac{dx}{x}=\int 3dt$
$lnx=3t+C$
$\text{at}t=0,x=5$
$ln5=C$
So $lnx=3t+ln5$
$ln\frac{x}{5}=3t$
$\frac{x}{5}=e^{3t}$
$x=5e^{3t}$
At $t=4$ $x=5e^{12}$