# General solution to differential equations

I’m reviewing differential equations after taking the course about 5-6 years ago and I have a couple of questions about the solutions of differential equations.
1) First why is the general form of the solution to linear homogenous differential equations, with non-equal and real roots to the complementary equation, of the forum
$$y = {e^{{\alpha _1}x}} + {e^{{\alpha _2}x}} + \cdot \cdot \cdot + {e^{{\alpha _n}x}}$$
And real and equal roots to the complementary equation
$$y = {e^{{\alpha _1}x}} + x{e^{{\alpha _2}x}} + \cdot \cdot \cdot + {x^{n - 1}}{e^{{\alpha _n}x}}$$

It seems that the resources on this subjects simply tell you this is the form of the solution with very little explanation of why this is known to be the form of the solution. If anyone can give me some insight into this I would appreciate it. Thanks

2) The second question is how do we know that the general solution to the nth order non-homogenous differential equation is: $$y(t) = {y_p}(t) + {y_h}(t)$$

Simon Bridge
Homework Helper
1) derive it - you'll see why.

2) ... a particular solution to the non-homogeneous and the general solution of the homogeneous part? Starting from 1 and because any superposition of solutions is itself a solution. It's like the homogeneous solution provides a set of lines and the particular solution selects one of them.

It is difficult to answer why questions in math. All the manipulations are exercises in tautology so the thing is because the thing is. The basic method to finding solutions to DEs is to guess at them ... what you are learning is rules of thumb for the guesswork. For instance, a homogeneous DE means that the derivative is proportional to the function ... and the only function that has that property is the exponential or functions derived from exponentials like trig functions, so you guess exponentials. The rest follows from there.

HallsofIvy