Second Order Differential Equations

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 = e^x

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

Example 1: Solve

d²ydx² + dydx − 6y = 0

Let y = e^rx so we get:

• dydx = re^rx
• d²ydx² = r²e^rx

Substitute these into the equation above:

r²e^rx + re^rx − 6e^rx = 0

Simplify:

e^rx(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 = e^2x

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 = Ae^2x + Be^−3x

Check

Let us check that answer. First take derivatives:

y = Ae^2x + Be^−3x

dydx = 2Ae^2x − 3Be^−3x

d²ydx² = 4Ae^2x + 9Be^−3x

Now substitute into the original equation:

d²ydx² + dydx − 6y = 0

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

4Ae^2x + 9Be^−3x + 2Ae^2x − 3Be^−3x − 6Ae^2x − 6Be^−3x = 0

4Ae^2x + 2Ae^2x − 6Ae^2x+ 9Be^−3x− 3Be^−3x − 6Be^−3x = 0

0 = 0

It worked!

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 = Ae^4x + Be^5x

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}

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 = e^5x

 

BUT when e^5x is a solution, then xe^5x is also a solution!

So, in this case our solution is:

y = Ae^5x + Bxe^5x

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

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 e^2x is a common factor:

y = e^2x( Ae^3ix + Be^−3ix )

But we haven't finished yet ... !

Euler's formula tells us that:

e^ix = 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:

Ae^3ix + 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 = e^2x( Ccos(3x) + iDsin(3x) )

 

Check

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

y = e^2x( Ccos(3x) + iDsin(3x) )

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

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

Substitute:

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

Example 8: Solve

d²ydx² − 6dydx + 25y = 0

The characteristic equation is:

r² − 6r + 25 = 0

Use the quadratic equation formula:

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 = e^3x(Ccos(4x) + iDsin(4x))

Example 9: Solve

9d²ydx² + 12dydx + 29y = 0

The characteristic equation is:

9r² + 12r + 29 = 0

Use the quadratic equation formula:

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))