Question

What is the general solution of the differential equation `y'' - 10y' +25 = 0`?

Answer 1

`y = Axe^(5x) + Be^(5x)`

Explanation:

We have:

`y'' - 10y' +25 = 0` ..... [A]

This is a Second order linear Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution, `y_c` of the homogeneous equation by looking at the Auxiliary Equation, which is the polynomial equation with the coefficients of the derivatives

Complimentary Function

The associated Auxiliary equation is:

`m^2-10m+25 = 0`
`(m-5)^2`

Which has repeated real solutions `m=5`

Thus the solution of the homogeneous equation is:

`y_c = (Ax+B)e^(5x)`
`\ \ \ = Axe^(5x) + Be^(5x)`

Confirming the quoted solution