assume that y1t and y2t are two solutions to 1

Question

Assume that Y₁(t) and Y₂(t) are two solutions to (1) and y₁(t) and y₂(t) are a fundamental set of solutions to the associated homogeneous differential equation (2) so,

Y₁ (t) - Y₂ (t) is a solution to (2) and it can be written as

Y₁ (t ) - Y₂ (t ) =  c₁ y₁ (t ) + c₂ y₂ (t)

Note the notation used now. Capital letters considered as solutions to (1) while lower case letters considered as to solutions to (2. It is a fairly common convention while dealing with non-homogeneous differential equations.

This theorem is simple enough to prove thus let's do that. To prove this Y₁(t) - Y₂(t) is a solution to (2) all we require to do is plug this in the differential equation and check this.

(Y₁ (t ) - Y₂ (t ))'' + p(t) (Y₁ (t ) - Y₂ (t ))' + q(t) (Y₁ (t ) - Y₂ (t )) = 0

Y''₁ + p(t) Y'₁ + q(t) Y₁ - (Y''₂ + p(t) Y'₂ + q(t) Y₂) = 0

g(t) - g(t) = 0

0 = 0

We utilized the fact that Y₁(t) and Y₂(t) are two solutions to (1) into the third step. Since they are solutions to (1) we know as

Y''₁ + p(t) Y'₁ + q(t) Y₁ = g(t)

Y''₂ + p(t) Y'₂ + q(t) Y₂ = g(t)

Therefore, we were capable to prove that the difference of the two solutions is a solution to (2).

Proving as,

Y₁ (t) - Y₂ (t) = c₁ y₁ (t ) + c₂ y₂ (t) is even easier.

As y₁(t) and y₂(t) are a fundamental set of solutions to (2) we identify that they form a general solution and thus any solution to (2) can be written as,

Y (t) = c₁ y₁ (t ) + c₂ y₂ (t)

Well, Y₁(t) - Y₂(t) is a solution to (2), as we've illustrated above, thus it can be written as,

Y₁ (t) - Y₂ (t) = c₁ y₁ (t ) + c₂ y₂ (t)

Thus, what does this theorem do for us? We can utilize this theorem to write down the type of the general solution to (1). Let's assume that y(t) is the general solution to (1) and that Y_P(t) is any solution to (1) which we can get our hands on. After that using the second part of our theorem as,

y(t) - Y_p(t) = c₁ y₁ (t) + c₂ y₂(t)

Here y1(t) and y2(t) are a fundamental set of solutions for (2). So solving for y(t) provides,

y(t) = c₁ y₁ (t) + c₂ y₂(t) + Y_p(t)

We can here call,

y_c= c₁ y₁ (t ) + c₂ y₂ (t)

The complementary solution and Y_P(t) a specific solution. The general solution to a differential equation can after that be written as,

y(t) = y_c + Y_p(t)

Here, to solve a nonhomogeneous differential equation, we will require solving the homogeneous differential equation, (2), that for constant coefficient differential equations is pretty simple to do, and we'll require a solution to (1).

It seems to be a circular argument. So as to write down a solution to (1) we require a solution. Though, this isn't the problem that this seems to be. There are ways to get a solution to (1).

They just won't, in common, be the general solution. Actually, the next two sections are devoted to accurately that, finding a particular solution to a non-homogeneous differential equation.

There are two general methods for determining particular solutions: Undetermined Coefficients and Variation of Parameters. Both have their disadvantages and advantages as you will see in the subsequent couple of sections.