Welcome to a lesson
on the derivation of the variation of parameters formula
to solve linear second order
nonhomogeneous differential equations.
Which means we'll be solving differential equations
in this form here, where g of x is a non-zero function.
So we'll call the nonhomogeneous differential equation
equation one, and then the differential equation given
in this form here where the right side is equal to zero,
is called the homogeneous differential equation
corresponding to equation one,
or the associated homogeneous differential equation.
So we'll call the homogeneous differential equation
equation two.
And the variation of parameters method
states that if we're given a differential equation
in this form here, where p, q and g are continuous
on an open interval and y sub one and y sub two
are linearly independent solutions to equation two,
the homogeneous differential equation,
then a particular solution
to the nonhomogeneous differential equation
is given by big y sub p and this formula here.
And therefore, the general solution
to the nonhomogeneous differential equation
would be y of x.
So to begin the derivation, we'll assume that y sub c,
given by this sum, is a complementary function
of equation one and the general solution to equation two,
again where equation one is
the nonhomogeneous differential equation
and equation two is the homogeneous differential equation,
and y sub one and y sub two
are a fundamental set of solutions.
So our goal here is to find two solutions,
u sub one and u sub two, such that y sub p is
a particular solution to equation one,
so what we'll do next is find
the first and second derivatives of big y sub p
and then perform substitution into equation one.
But before we do this,
we are going to make one important assumption.
We're going to assume that u sub one prime
times y sub one, plus u sub two prime times y sub two
is equal to zero,
and this assumption will simplify our derivatives
as well as help us find u sub one and u sub two.
So to find our first derivative,
we'll have to find the derivative of the first term
and the second term in big y sub p.
To do this, we'll have to apply
the product rule to each term.
So the derivative of this term
is equal to the sum of these two terms,
and the derivative of the second term
is equal to the sum of the second two terms.
But now if we take a look at our assumption,
notice that u sub one prime times y sub one,
which is the same as this term,
plus u sub two prime times y sub two,
which is this term, is equal to zero,
and therefore the first derivative
of big y sub p simplifies to the sum of these two terms.
Which does make it easier to find our second derivative.
The derivative of this first term requires our product rule,
giving us the sum of these two terms,
and the derivative of the second term
gives us the sum of these two terms.
So now that we have y, y prime, and y double prime,
we'll now perform substitution into equation one.
So here's equation one,
and now notice that y double prime is here,
then we have p of x times y prime,
here's y prime.
And then we have q of x times y,
and here's y.
And here's where the algebra gets a little tricky,
what we're going to do now is eliminate the parentheses,
we'll distribute p of x and distribute q of x
and then rearrange the terms, forming this equation here.
Notice how all of these terms here
had a common factor of u sub one,
and all these terms here had a common factor of u sub two.
But remember that y sub one and y sub two
were particular solutions to the
homogeneous differential equation,
which means all of this
is equal to zero and so is all of this.
And therefore, this simplifies very nicely to
the sum of these first two terms
must equal g of x.
So we're going to use this equation,
and then our assumption,
to solve this as a system of equations,
and we'll solve this for u sub one and u sub two,
the two functions we're looking for.
So we'll solve this as a system using Cramer's Rule,
and then integrate to find u sub one and u sub two.
Notice when solving the system,
we'll be able to solve for u sub one prime
and u sub two prime,
but we still have to find the interderivatives to find
u sub one and u sub two.
So let's go ahead and set this up using Cramer's Rule.
Notice how I did change the order of these equations,
equation one is now the assumption,
and equation two is the equation that we derived.
And now when we set this up using Cramer's Rule,
notice how the two by two determinants in the denominator
end up being the Wronskians of y sub one and y sub two.
Then looking at the two by two determinants
in the numerator, notice for y sub one prime,
we would have zero minus y sub two times g of x,
given here,
and then for y sub two prime,
we would have y sub one times g of x minus zero,
which we see here.
So notice here on the left
we have u sub one prime, and u sub two prime
as this quotient,
but because we're after u sub one and u sub two,
we'd have to integrate both sides of the equation,
which would give us this formula here for u sub one,
and this formula here for u sub two.
Remember these are the two functions that we were after
in order to find our particular solution
to the nonhomogeneous differential equation.
So now we have everything we need,
here's big y sub p,
and now we know a formula for u sub one and u sub two,
using the method of variation of parameters.
So to summarize, one more time,
to use the variation of parameters method
to solve a linear, second order
nonhomogenous differential equation,
we can find a particular solution
using this formula here for big y sub p,
and then once we have big y sub p,
we can form the general solution using y of x given here.
Now that we've derived this formula,
we'll now look at several examples.
I hope you found this helpful.