DIFFYQS Second order linear ODEs

The proof becomes even simpler to state if we use the operator notation. An operator is an object that eats functions and spits out functions (kind of like what a function is, but a function eats numbers and spits out numbers). Define the operator

L

L y = y^{″} + p y^{'} + q y .

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The differential equation now becomes

L y = 0 .

The operator (and the equation)

L

being linear means that

L (C_{1} y_{1} + C_{2} y_{2}) = C_{1} L y_{1} + C_{2} L y_{2} .

It is almost as if we were “multiplying” by

L .

The proof above becomes

L y = L (C_{1} y_{1} + C_{2} y_{2}) = C_{1} L y_{1} + C_{2} L y_{2} = C_{1} \cdot 0 + C_{2} \cdot 0 = 0 .

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Two different solutions to the second equation

y^{″} - k^{2} y = 0

are

y_{1} = \cosh (k x)

and

y_{2} = \sinh (k x) .

Recalling the definition of

\sinh

and

\cosh,

we note that these are solutions by superposition as they are linear combinations of the two exponential solutions:

\cosh (k x) = \frac{e^{k x} + e^{- k x}}{2} = (^{1} /_{2}) e^{k x} + (^{1} /_{2}) e^{- k x}

and

\sinh (k x) = \frac{e^{k x} - e^{- k x}}{2} = (^{1} /_{2}) e^{k x} - (^{1} /_{2}) e^{- k x} .

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The functions

\sinh

and

\cosh

are sometimes more convenient to use than the exponential. Let us review some of their properties:

\begin{aligned} \cosh 0 = 1, & \sinh 0 = 0, \\ \frac{d}{d x} [\cosh x] = \sinh x, & \frac{d}{d x} [\sinh x] = \cosh x, \\ \cosh^{2} x - \sinh^{2} x = 1 . \end{aligned}

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Exercise 2.1.1.

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Derive these properties using the definitions of

\sinh

and

\cosh

in terms of exponentials.

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Linear equations have nice and simple answers to the existence and uniqueness question.

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Theorem 2.1.2. Existence and uniqueness.

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Suppose

p, q, f

are continuous functions on some interval

I,

a

is a number in

I,

and

b_{0}, b_{1}

are constants. Then the equation

y^{″} + p (x) y^{'} + q (x) y = f (x),

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has exactly one solution

y (x)

defined on the interval

I

satisfying the initial conditions

y (a) = b_{0}, y^{'} (a) = b_{1} .

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For example, the equation

y^{″} + k^{2} y = 0

with

y (0) = b_{0}

and

y^{'} (0) = b_{1}

has the solution

y (x) = b_{0} \cos (k x) + \frac{b_{1}}{k} \sin (k x) .

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The equation

y^{″} - k^{2} y = 0

with

y (0) = b_{0}

and

y^{'} (0) = b_{1}

has the solution

y (x) = b_{0} \cosh (k x) + \frac{b_{1}}{k} \sinh (k x) .

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Using

\cosh

and

\sinh

in this solution allows us to solve for the initial conditions in a cleaner way than if we have used the exponentials.

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The initial conditions for a second order ODE consist of two equations. Common sense tells us that if we have two arbitrary constants and two equations, then we should be able to solve for the constants and find a solution to the differential equation satisfying the initial conditions.

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Question: Suppose we find two different solutions

y_{1}

and

y_{2}

to the homogeneous equation (2.2). Can every solution be written (using superposition) in the form

y = C_{1} y_{1} + C_{2} y_{2} ?

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Answer is affirmative! Provided that

y_{1}

and

y_{2}

are different enough in the following sense. We say

y_{1}

and

y_{2}

are linearly independent if one is not a constant multiple of the other.

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Theorem 2.1.3.

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Let

p, q

be continuous functions. Let

y_{1}

and

y_{2}

be two linearly independent solutions to the homogeneous equation (2.2). Then every other solution is of the form

y = C_{1} y_{1} + C_{2} y_{2} .

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That is,

y = C_{1} y_{1} + C_{2} y_{2}

is the general solution.

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For example, we found the solutions

y_{1} = \sin x

and

y_{2} = \cos x

for the equation

y^{″} + y = 0 .

It is not hard to see that sine and cosine are not constant multiples of each other. Indeed, if

\sin x = A \cos x

for some constant

A,

plugging in

x = 0

would imply

A = 0 .

But then

\sin x = 0

for all

x,

which is preposterous. So

y_{1}

and

y_{2}

are linearly independent. Hence,

y = C_{1} \cos x + C_{2} \sin x

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is the general solution to

y^{″} + y = 0 .

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For two functions, checking linear independence is rather simple. Let us see another example. Consider

y^{″} - 2 x^{- 2} y = 0 .

Then

y_{1} = x^{2}

and

y_{2} =^{1} /_{x}

are solutions. To see that they are linearly independent, suppose one is a multiple of the other:

y_{1} = A y_{2},

we just have to find out that

A

cannot be a constant. In this case we have

A =^{y_{1}} /_{y_{2}} = x^{3},

this most decidedly not a constant. So

y = C_{1} x^{2} + C_{2}^{1} /_{x}

is the general solution.

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If you have one nonzero solution to a second order linear homogeneous equation, then you can find another one. This is the reduction of order method. The idea is that if we somehow found

y_{1}

as a solution of

y^{″} + p (x) y^{'} + q (x) y = 0,

then we try a second solution of the form

y_{2} (x) = y_{1} (x) v (x) .

We just need to find

v .

We plug

y_{2}

into the equation:

\begin{aligned} 0 = y_{2}^{″} + p (x) y_{2}^{'} + q (x) y_{2} & = \underset{y_{2}^{″}}{\underset{⏟}{y_{1}^{″} v + 2 y_{1}^{'} v^{'} + y_{1} v^{″}}} + p (x) \underset{y_{2}^{'}}{\underset{⏟}{(y_{1}^{'} v + y_{1} v^{'})}} + q (x) \underset{y_{2}}{\underset{⏟}{y_{1} v}} \\ = y_{1} v^{″} + (2 y_{1}^{'} + p (x) y_{1}) v^{'} + {(y_{1}^{″} + p (x) y_{1}^{'} + q (x) y_{1})}^{0} v . \end{aligned}

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In other words,

y_{1} v^{″} + (2 y_{1}^{'} + p (x) y_{1}) v^{'} = 0 .

Using

w = v^{'},

we have the first order linear equation

y_{1} w^{'} + (2 y_{1}^{'} + p (x) y_{1}) w = 0 .

After solving this equation for

w

(integrating factor), we find

v

by antidifferentiating

w .

We then form

y_{2}

by computing

y_{1} v .

For example, suppose we somehow know

y_{1} = x

is a solution to

y^{″} + x^{- 1} y^{'} - x^{- 2} y = 0 .

The equation for

w

is then

x w^{'} + 3 w = 0 .

We find a solution,

w = C x^{- 3},

and we find an antiderivative

v = \frac{- C}{2 x^{2}} .

Hence

y_{2} = y_{1} v = \frac{- C}{2 x} .

Any

C

works and so

C = - 2

makes

y_{2} =^{1} /_{x} .

Thus, the general solution is

y = C_{1} x + C_{2}^{1} /_{x} .

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Since we have a formula for the solution to the first order linear equation, we can write a formula for

y_{2} :

y_{2} (x) = y_{1} (x) \int \frac{e^{- \int p (x) d x}}{{(y_{1} (x))}^{2}} d x

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However, it is much easier to remember that we just need to try

y_{2} (x) = y_{1} (x) v (x)

and find

v (x)

as we did above. The technique works for higher order equations too: You get to reduce the order by one for each solution you find. So it is better to remember how to do it rather than a specific formula.

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We will study the solution of nonhomogeneous equations in Section 2.5. We will first focus on finding general solutions to homogeneous equations.

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Exercises Exercises

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2.1.2.

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Show that

y = e^{x}

and

y = e^{2 x}

are linearly independent.

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2.1.3.

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Take

y^{″} + 5 y = 10 x + 5 .

Find (guess!) a solution.

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2.1.4.

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Prove the superposition principle for nonhomogeneous equations. Suppose that

y_{1}

is a solution to

L y_{1} = f (x)

and

y_{2}

is a solution to

L y_{2} = g (x)

(same linear operator

L

). Show that

y = y_{1} + y_{2}

solves

L y = f (x) + g (x) .

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2.1.5.

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For the equation

x^{2} y^{″} - x y^{'} = 0,

find two solutions, show that they are linearly independent and find the general solution. Hint: Try

y = x^{r} .

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Equations of the form

a x^{2} y^{″} + b x y^{'} + c y = 0

are called Euler’s equations or Cauchy–Euler equations. They are solved by trying

y = x^{r}

and solving for

r

(assume

x \geq 0

for simplicity).

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2.1.6.

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Suppose that

{(b - a)}^{2} - 4 a c > 0 .

Find a formula for the general solution of Euler’s equation (see above) $a x^{2} y^{″} + b x y^{'} + c y = 0 .$ Hint: Try $y = x^{r}$ and find a formula for $r .$
What happens when ${(b - a)}^{2} - 4 a c = 0$ or ${(b - a)}^{2} - 4 a c < 0 ?$

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We will revisit the case when

{(b - a)}^{2} - 4 a c < 0

later.

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2.1.7.

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Same equation as in Exercise 2.1.6. Suppose

{(b - a)}^{2} - 4 a c = 0 .

Find a formula for the general solution of

a x^{2} y^{″} + b x y^{'} + c y = 0 .

Hint: Try

y = x^{r} \ln x

for the second solution.

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2.1.8. reduction of order.

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Suppose

y_{1}

is a solution to

y^{″} + p (x) y^{'} + q (x) y = 0 .

By directly plugging into the equation, show that

y_{2} (x) = y_{1} (x) \int \frac{e^{- \int p (x) d x}}{{(y_{1} (x))}^{2}} d x

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is also a solution.

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2.1.9. Chebyshev’s equation of order 1.

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Take

(1 - x^{2}) y^{″} - x y^{'} + y = 0 .

Show that $y = x$ is a solution.
Use reduction of order to find a second linearly independent solution.
Write down the general solution.

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2.1.10. Hermite’s equation of order 2.

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Take

y^{″} - 2 x y^{'} + 4 y = 0 .

Show that $y = 1 - 2 x^{2}$ is a solution.
Use reduction of order to find a second linearly independent solution. (It’s OK to leave a definite integral in the formula.)
Write down the general solution.