Question

Find and classify any equilibrium solutions and then sketch typical solution curves to the differential equation: dx /dt = x 2 − 5x + 4.

Answer 1

Answer:

$x=1+\dfrac{3}{1-Ke^{3t}}$

Step-by-step explanation:

Given that

$\dfrac{dx}{dt}=x^2-5x+4$

This is a differential equation.

Now by separating variables

$\dfrac{dx}{x^2-5x+4}=dt$

$\dfrac{dx}{(x-1)(x-4)}=dt$

$\dfrac{1}{3}\left(\dfrac{1}{x-4}-\dfrac{1}{x-1}\right)dx=dt$

Now by integrating both side

$\int\dfrac{1}{3}\left(\dfrac{1}{x-4}-\dfrac{1}{x-1}\right)dx=\int dt$

$\dfrac{1}{3}\left(\ln(x-4)-\ln(x-1)\right )=t+C$

Where C is the constant

$\dfrac{1}{3}\ln\dfrac{x-4}{x-1}=t+C$

$\dfrac{x-4}{x-1}=Ke^{3t}$    K is the constant.

$x=1+\dfrac{3}{1-Ke^{3t}}$

So the solution of above differential equation is

$x=1+\dfrac{3}{1-Ke^{3t}}$