# Prove that if the functions `y_1` and `y_2` are independnt solutios of the DE `y'' + p(x)y' + q(x)y = 0` , then `y_3=y_1+y_2` and `y_4=y_1-y_2` also form a fundamental set of solutions.

txmedteach | High School Teacher | (Level 3) Associate Educator

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We start by acknowledging certain assumptions implied by the problem before we start tackling `y_3` and `y_4`.  Let's figure out what it really means for `y_1` and `y_2` to be solutions to the differential equation.

Well, the following equations must hold true if those two functions are solutions:

`y_1'' + p(x)y_1' + q(x)y_1 = 0`

`y_2'' + p(x)y_2'+q(x)y_2 = 0`

Alright, now, we need to figure out whether the following hold true as a result:

`y_3'' + p(x)y_3'+q(x)y_3 = 0`

`y_4'' + p(x)y_4' + q(x) y_4 = 0`

Well, let's start with `y_3`. We'll try to show it's a solution first determining what its derivatives are. Now, to do this, all we need to do is recognze that the derivative is a linear operator. In other words, given `f(x)` and `g(x)`, their derivatives `f'(x)` and `g'(x)` , and constants `a` and `b` :

`d/dx (af + bg) = af' + b g'`

In other words, the derivative distributes its operation much like a constant factor.

Using this property, let's find the first and second derivative of `y_3`:

`y_3' = d/dx(y_1 + y_2)`

`y_3'= y_1' + y_2'`

`y_3'' = d/dx (y_1' + y_2')`

`y_3'' = y_1'' + y_2''`

So the first and second derivatives of `y_3` are simply the sum of the first derivatives and second derivatives of `y_1` and `y_2`! Now, we can substitute these two into our differential equation for `y_3`:

`y_3'' + p(x)y_3' + q(x)y_3 = 0`

`y_2'' + y_1'' + p(x)(y_2' + y_1') + q(x)(y_2+y_1) = 0`

`y_2'' + y_1'' + p(x)y_2' + p(x)y_1' + q(x)y_2 + q(x)y_1 = 0`

Now, let's do some rearranging and grouping:
`(y_2'' + p(x)y_2' + q(x)y_2) + (y_1'' + p(x)y_1' + q(x)y_1) = 0`

Well, referencing our assumption:

`y_1'' + p(x) y_1' + q(x)y_1 = 0`

which, remember also holds for `y_2`, we know that our first group of three terms and our second group of three terms both become 0!

`0 + 0 = 0`

Therefore, `y_3` must also be a solution to the differential equation!

Now, to prove that `y_4` is a solution, we're going to do the same thing. First, let's start by finding the first and second derivatives of `y_4`:

`y_4' = d/dx(y_1 - y_2) = y_1' - y_2'`

`y_4'' = d/dx(y_1' -y_2') = y_1'' - y_2''`

Now, we can substitute these two expressions into our differential equation for `y_4`:

`y_4'' + p(x)y_4' + q(x)y_4 = 0`

`(y_1'' - y_2'') + p(x)(y_1' - y_2') + q(x)(y_1 - y_2) = 0`

Now, we do some distributing and rearranging;

`y_1'' - y_2'' + p(x)y_1' - p(x)y_2' + q(x)y_1 - q(x)y_2 = 0`

`y_1'' + p(x)y_1' + q(x)y_1 - y_2'' - p(x)y_2' - q(x)y_2=0`

Now, remember, when we use the distributive property on subtraction (or more correctly, -1), we change the sign for each term inside the parentheses:

`(y_1'' + p(x)y_1' +q(x)y_1) - (y_2'' + p(x)y_2' + q(x)y_2) = 0`

Well, again, based on our first assumption, we can substitute 0 for each term inside the parentheses!

`0 - 0 = 0`

Well, that is certainly true! Therefore, `y_4` is also a solution to the differential equation.

Note that in general, the solutions to the homogeneous linear second-order differential equation (which is pretty much defined by the equation `y'' + p(x)y' + q(x)y = 0`) are going to include any linear combination of `y_1` and `y_2`. You could say that the solution set to the equation will be the following:

`{y | a,b in RR, y = ay_1 + by_2}`

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