Here is one single example.

My claim is one, e to the x and e to the negative 2x is a fundamental set of

solutions of this constant coefficient second order homogeneous differential equation.

Knowing that, solve the initial value problem,

y double prime plus y prime minus 2y is equal to four.

So now we have the second order constant coefficient

for nonhomogeneous differential equation.

Satisfy the two initial conditions say,

y of zero is equal to zero,

and the y prime of zero is equal to negative one.

The first part is easy computation.

You can show very easily there,

e to the x and e to the negative 2x are solutions of this differential equation.

Moreover, computed their Wronskian and see that the Wronskian is never vanishing.

That means these two are linearly independent.

That means this is our fundamental set of a solution. That's the easy part.

So I would leave it to you.

Now, let's consider this nonhomogeneous problem.

By the theory I've just introduced here,

the general solution of this nonhomogeneous equation is

complementary solution plus any one particular solution of this problem.

What is the complementary solution?

In other words, what is the general solution of a corresponding homogeneous problem.

Since you already have a fundamental set of solutions,

general solution of a corresponding homogeneous equation

must be linear combination of these two.

That's this complementary solution.

Right here, c1e to the x plus c2e to the negative 2x.

That's the complementary resolution.

Now, it's time to find anyone of the particular solution of this nonhomogeneous problem.

You can find the bias,

just a simple inspection.

Y is equal to negative two.

Then it's derivative, first derivative,

second derivative i equal to zero,

and y is equal to negative two,

then negative two times negative two,

that is equal to positive four.

What does that mean?

Negative two is a particular solution of this differential equation.

Is it really a particular solution?

So by the theorem,

complementary solution plus a particular solution

is a general solution of this differential equation.

So this is a general solution.

And they represents all possible solutions with arbitrary two constants,

c sub one and the c sub two.

Now it's time to check the initial condition.

Those two initial conditions.

You see, y of zero is equal to zero gives you when x is equal to zero in this expression,

you get c1 plus c2 minus two.

That must be equal to zero.

Second initial condition, derivative of y at x is equal to zero equal to negative one.

And that gives you,

c1 minus 2c2 is equal to negative one.

So with this very simple simultaneous equation for c1 and c2,

you will get c1 and the c2 are both equal to one.

So finally, the solution of this given initial value problem

will be e to the x plus e to the minus 2x minus 2.