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

`x=1+(3)/(1-Ke^(3t))`

Explanation:

Given that

`(dx)/(dt)=x^2-5x+4`

This is a differential equation.

Now by separating variables

`(dx)/(x^2-5x+4)=dt`

`(dx)/((x-1)(x-4))=dt`

`(1)/(3)\left((1)/(x-4)-(1)/(x-1)\right)dx=dt`

Now by integrating both side

`\int(1)/(3)\left((1)/(x-4)-(1)/(x-1)\right)dx=\int dt`

`(1)/(3)\left(\ln(x-4)-\ln(x-1)\right )=t+C`

Where C is the constant

`(1)/(3)\ln(x-4)/(x-1)=t+C`

`(x-4)/(x-1)=Ke^(3t)` K is the constant.

`x=1+(3)/(1-Ke^(3t))`

So the solution of above differential equation is

`x=1+(3)/(1-Ke^(3t))`