You should not be trying to do non-homogeneous differential equations until you can easily handle homogenous ones!
y"- 5y'+ 4y= 0 is a homogeneous linear differential equation
with constant coefficients. The standard way of solving such an equation is just what you started with- but without the "V":
Illusionist said:
y''-5y'+4y = 0, y1= e^mx
y=y(1)=(e^mx)
y'=[m(e^mx)]
y''=[m^2(e^mx)]
Sub. back in I get:
m^2e^mx- 5me^mx+ 4e^mx= 0
Since e^mx is never 0, divide the entire equation by it to get m^2- 5m+ 4= 0, the "characteristic equation" for your differential equation. After you have done a few of those you should be able to write down the characteristic equation for such a differential equation without having to do the substitution!
Yes, it is true that m^2- 5m+ 4= (m-4)(m-1) so m= 4 and m= 1 are solutions to that equation which tells you that both e^(x) and e^(4x) are solutions to y"- 5y'+ 4y= 0. You should know enough o fthe theory to know that the set of all solutions of a linear homogeneous
second order equation form a
2 dimensional vector space and so you know how to form the general solution to the equation.
The y(x)= Ve^mx where V is a function of x (the "variation of parameters" method) requires that you already know the solutions to the homogeneous equation: you need, for these problems, y(x)= u(x)e^x+ v(x)e^(4x).
A simpler method, that, however, only works when the right hand side is an exponential, sine or cosine, polynomial, or combination of those, is the "undetermined coefficients" method. In that case you can make a good "guess" as to what the specific solution must be and just try to find coefficients that will work.
For example, for f(x)= 8x^2- 25 try y(x)= Ax^2+ Bx+ C. Put that into your equation and determine what A, B, C must be to make the equation true. For f(x)= 9e^x, try y= Axe^x. (Do you see
why you need that "x" multiplying e^x? If not review your textbook.) For the last, you want to combine those previous answers.
