4.6 (#ThrowbackThursday)

Section 4.6 is entitled “Variation of Parameters.”

We’ve seen this method in the past, except we used it only for first-order differential equations. In this section, we’re interested in solving the equation y” + p(t)y’ + q(t)y = g(t). Notice that p and q are functions of t, instead of restricting them to be constants.

If we first look at the homogeneous equation y” + p(t)y’ + q(t)y = 0, then the general solution to that equation would be y_h = C₁y₁ + C₂y₂. As usual, C₁ and C₂ are arbitrary constants.

Like our previous process, the method of variation of parameters makes us think of C₁ and C₂ are unknown functions v₁(t) and v₂(t). Then we’ll look for a particular solution to our original inhomogeneous equation, in which our particular solution will have the form y_p = v₁y₁ + v₂y₂. In other words, we have a lot of similarity between this method of variation of parameters and our previous one we used to solved linear equations back in Chapter 2 (i.e. http://differentialequationsjourney.blogspot.com/2013/09/super-serious-and-not-so-funny-linear.html).

When we compute the derivative of y_p,

What we have in the brackets, we will set equal to zero.

Differentiating again, we get

When we plug y_p, y_p’ and y_p” into our original inhomogeneous equation y” + p(t)y’ + q(t)y = g(t), we get

Since we defined y₁ and y₂ to be solutions to the homogeneous equation, this simplifies to be

In other words, y_p will be a solution to our original inhomogeneous equation provided that 

At this point, we have two equations for v₁ and v₂, which are

Since this is a system of linear equations, we can observe that our coefficient matrix would be

This means that this system can be solved provided that the determinant of this matrix is nonzero. Something to observe about the determinant is that is also the Wronskian of y₁ and y₂.

Since y₁ and y₂ form a fundamental set of solutions for our equation, this means they are linearly independent and that means the Wronskian will be nonzero. Hooray!

Back to our system of equations for v₁ and v₂: We can solve these equations to obtain

Something to note about these two equations is that the denominators will always be nonzero since they are both the Wronskian. Double hooray!

When we integrate these equations, we get

Now we’ll have a particular solution once we plug these back into y_p = v₁y₁ + v₂y₂.

And now you have a method to solve inhomogeneous differential equations of higher orders! Of course, this method requires for you to first find a fundamental set of solutions to the homogeneous equations, and also to be able to compute those scary-looking integrals for v₁ and v₂, but this method is still important.

That’s it for 4.6! I’ll see you in 4.7!