# Second Order Differential Equations

Here we learn how to solve equations of this type:

d2ydx2 + pdydx + qy = 0

## Differential Equation

A Differential Equation is an equation with a function and one or more of its derivatives:

Example: an equation with the function y and its derivative dy dx

## Order

The Order is the highest derivative (is it a first derivative? a second derivative? etc):

### Example:

dydx + y2 = 5x

It has only the first derivative dy dx , so is "First Order"

### Example:

d2ydx2 + xy = sin(x)

This has a second derivative d2y dx2 , so is "Second Order" or "Order 2"

### Example:

d3ydx3 + xdydx + y = ex

This has a third derivative d3y dx3 which outranks the dy dx , so is "Third Order" or "Order 3"

## Second Order Differential Equations

We can solve a second order differential equation of the type:

d2ydx2 + P(x)dydx + Q(x)y = f(x)

where P(x), Q(x) and f(x) are functions of x, by using:

Variation of Parameters which only works when f(x) is a polynomial, exponential, sine, cosine or a linear combination of those.

Undetermined Coefficients which is a little messier but works on a wider range of functions.

But here we begin by learning the case where f(x) = 0 (this makes it "homogeneous"):

d2ydx2 + P(x)dydx + Q(x)y = 0

and also where the functions P(X) and Q(x) are constants p and q:

d2ydx2 + pdydx + qy = 0

Let's learn to solve them!

### e to the rescue

We are going to use a special property of the derivative of the exponential function:

At any point the slope (derivative) of ex equals the value of ex :

And when we introduce a value "r" like this:

f(x) = erx

We find:

• the first derivative is f'(x) = rerx
• the second derivative is f''(x) = r2erx

In other words, the first and second derivatives of f(x) are both multiples of f(x)

This is going to help us a lot!

### Example 1: Solve

d2ydx2 + dydx − 6y = 0

Let y = erx so we get:

• dydx = rerx
• d2ydx2 = r2erx

Substitute these into the equation above:

r2erx + rerx − 6erx = 0

Simplify:

erx(r2 + r − 6) = 0

r2 + r − 6 = 0

We have reduced the differential equation to an ordinary quadratic equation!

This quadratic equation is given the special name of characteristic equation.

We can factor this one to:

(r − 2)(r + 3) = 0

So r = 2 or −3

And so we have two solutions:

y = e2x

y = e−3x

But that’s not the final answer because we can combine different multiples of these two answers to get a more general solution:

y = Ae2x + Be−3x

### Check

Let us check that answer. First take derivatives:

y = Ae2x + Be−3x

dydx = 2Ae2x − 3Be−3x

d2ydx2 = 4Ae2x + 9Be−3x

Now substitute into the original equation:

d2ydx2 + dydx − 6y = 0

(4Ae2x + 9Be−3x) + (2Ae2x − 3Be−3x) − 6(Ae2x + Be−3x) = 0

4Ae2x + 9Be−3x + 2Ae2x − 3Be−3x − 6Ae2x − 6Be−3x = 0

4Ae2x + 2Ae2x − 6Ae2x+ 9Be−3x− 3Be−3x − 6Be−3x = 0

0 = 0

It worked!

## So, does this method work generally?

Well, yes and no. The answer to this question depends on the constants p and q.

With y = erx as a solution of the differential equation:

d2ydx2 + pdydx + qy = 0

we get:

r2erx + prerx + qerx = 0

erx(r2 + pr + q) = 0

r2 + pr + q = 0

This is a quadratic equation, and there can be three types of answer:

• two real roots
• one real root (i.e. both real roots are the same)
• two complex roots

How we solve it depends which type!

We can easily find which type by calculating the discriminant p2 − 4q. When it is

• positive we get two real roots
• zero we get one real root
• negative we get two complex roots

## Two Real Roots

When the discriminant p2 − 4q is positive we can go straight from the differential equation

d2ydx2 + pdydx + qy = 0

through the "characteristic equation":

r2 + pr + q = 0

to the general solution with two real roots r1 and r2:

y = Aer1x + Ber2x

### Example 2: Solve

d2ydx2 − 9dydx + 20y = 0

The characteristic equation is:

r2 − 9r + 20 = 0

Factor:

(r − 4)(r − 5) = 0

r = 4 or 5

So the general solution of our differential equation is:

y = Ae4x + Be5x

And here are some sample values:

### Example 3: Solve

6d2ydx2 + 5dydx − 6y = 0

The characteristic equation is:

6r2 + 5r − 6 = 0

Factor:

(3r − 2)(2r + 3) = 0

r = 23 or −32

So the general solution of our differential equation is:

y = Ae(23x) + Be(−32x)

### Example 4: Solve

9d2ydx2 − 6dydx − y = 0

The characteristic equation is:

9r2 − 6r − 1 = 0

This does not factor easily, so we use the quadratic equation formula:

x = −b ± √(b2 − 4ac) 2a

with a = 9, b = −6 and c = −1

x = −(−6) ± √((−6)2 − 4×9×(−1)) 2×9

x = 6 ± √(36 + 36) 18

x = 6 ± 6√2 18

x = 1 ± √2 3

So the general solution of the differential equation is

y = Ae(1 + √2 3)x + Be(1 − √2 3)x

## One Real Root

When the discriminant p2 − 4q is zero we get one real root (i.e. both real roots are equal).

Here are some examples:

### Example 5: Solve

d2ydx2 − 10dydx + 25y = 0

The characteristic equation is:

r2 − 10r + 25 = 0

Factor:

(r − 5)(r − 5) = 0

r = 5

So we have one solution: y = e5x

BUT when e5x is a solution, then xe5x is also a solution!

Why? I can show you:

y = xe5x

dydx = e5x + 5xe5x

d2ydx2 = 5e5x + 5e5x + 25xe5x

So

d2ydx2 − 10dydx + 25y

= 5e5x + 5e5x + 25xe5x − 10(e5x + 5xe5x) + 25xe5x

= (5e5x + 5e5x − 10e5x) + (25xe5x − 50xe5x + 25xe5x) = 0

So, in this case our solution is:

y = Ae5x + Bxe5x

### How does this work in the general case?

With y = xerx we get the derivatives:

• dydx = erx + rxerx
• d2ydx2 = rerx + rerx + r2xerx

So

d2ydx2 + p dydx + qy

= (rerx + rerx + r2xerx) + p( erx + rxerx ) + q( xerx )

= erx(r + r + r2x + p + prx + qx)

= erx(2r + p + x(r2 + pr + q))

= erx(2r + p) because we already know that r2 + pr + q = 0

And when r2 + pr + q has a repeated root, then r = −p2 and 2r + p = 0

So if r is a repeated root of the characteristic equation, then the general solution is

y = Aerx + Bxerx

Let's try another example to see how quickly we can get a solution:

### Example 6: Solve

4d2ydx2 + 4dydx + y = 0

The characteristic equation is:

4r2 + 4r + 1 = 0

Then:

(2r + 1)2 = 0

r = −12

So the solution of the differential equation is:

y = Ae(−½)x + Bxe(−½)x

## Complex roots

When the discriminant p2 − 4q is negative we get complex roots.

Let’s try an example to help us work out how to do this type:

### Example 7: Solve

d2ydx2 − 4dydx + 13y = 0

The characteristic equation is:

r2 − 4r + 13 = 0

This does not factor, so we use the quadratic equation formula:

x = −b ± √(b2 − 4ac) 2a

with a = 1, b = −4 and c = 13

x = −(−4) ± √((−4)2 − 4×1×13) 2×1

x = 4 ± √(16 − 52) 2

x = 4 ± √(−36) 2

x = 4 ± 6i 2

x = 2 ± 3i

If we follow the method used for two real roots, then we can try the solution:

y = Ae(2+3i)x + Be(2−3i)x

We can simplify this since e2x is a common factor:

y = e2x( Ae3ix + Be−3ix )

But we haven't finished yet ... !

Euler's formula tells us that:

eix = cos(x) + i sin(x)

So now we can follow a whole new avenue to (eventually) make things simpler.

Looking just at the "A plus B" part:

Ae3ix + Be−3ix

A(cos(3x) + i sin(3x)) + B(cos(−3x) + i sin(−3x))

Acos(3x) + Bcos(−3x) + i(Asin(3x) + Bsin(−3x))

Now apply the Trigonometric Identities: cos(−θ)=cos(θ) and sin(−θ)=−sin(θ):

Acos(3x) + Bcos(3x) + i(Asin(3x) − Bsin(3x)

(A+B)cos(3x) + i(A−B)sin(3x)

Replace A+B by C, and A−B by D:

Ccos(3x) + iDsin(3x)

And we get the solution:

y = e2x( Ccos(3x) + iDsin(3x) )

### Check

We have our answer, but maybe we should check that it does indeed satisfy the original equation:

y = e2x( Ccos(3x) + iDsin(3x) )

dydx = e2x( −3Csin(3x)+3iDcos(3x) ) + 2e2x( Ccos(3x)+iDsin(3x) )

d2ydx2 = e2x( −(6C+9iD)sin(3x) + (−9C+6iD)cos(3x)) + 2e2x(2C+3iD)cos(3x) + (−3C+2iD)sin(3x) )

Substitute:

d2ydx2 − 4dydx + 13y = e2x( −(6C+9iD)sin(3x) + (−9C+6iD)cos(3x)) + 2e2x(2C+3iD)cos(3x) + (−3C+2iD)sin(3x) ) − 4( e2x( −3Csin(3x)+3iDcos(3x) ) + 2e2x( Ccos(3x)+iDsin(3x) ) ) + 13( e2x(Ccos(3x) + iDsin(3x)) )

... hey, why don't YOU try adding up all the terms to see if they equal zero ... if not please let me know, OK?

### How do we generalize this?

Generally, when we solve the characteristic equation with complex roots, we will get two solutions r1 = v + wi and r2 = v − wi

So the general solution of the differential equation is

y = evx ( Ccos(wx) + iDsin(wx) )

### Example 8: Solve

d2ydx2 − 6dydx + 25y = 0

The characteristic equation is:

r2 − 6r + 25 = 0

x = −b ± √(b2 − 4ac) 2a

with a = 1, b = −6 and c = 25

x = −(−6) ± √((−6)2 − 4×1×25) 2×1

x = 6 ± √(36 − 100) 2

x = 6 ± √(−64) 2

x = 6 ± 8i 2

x = 3 ± 4i

And we get the solution:

y = e3x(Ccos(4x) + iDsin(4x))

### Example 9: Solve

9d2ydx2 + 12dydx + 29y = 0

The characteristic equation is:

9r2 + 12r + 29 = 0

x = −b ± √(b2 − 4ac) 2a

with a = 9, b = 12 and c = 29

x = −12 ± √(122 − 4×9×29) 2×9

x = −12 ± √(144 − 1044) 18

x = −12 ± √(−900) 18

x = −12 ± 30i 18

x = −23 ± 53i

And we get the solution:

y = e(−23)x(Ccos(53x) + iDsin(53x))

## Summary

To solve a linear second order differential equation of the form

d2ydx2 + pdydx + qy = 0

where p and q are constants, we must find the roots of the characteristic equation

r2 + pr + q = 0

There are three cases, depending on the discriminant p2 - 4q. When it is

positive we get two real roots, and the solution is

y = Aer1x + Ber2x

zero we get one real root, and the solution is

y = Aerx + Bxerx

negative we get two complex roots r1 = v + wi and r2 = v − wi, and the solution is

y = evx ( Ccos(wx) + iDsin(wx) )