Finding General Solution of Second Order Odes

Second Order Differential Equations

Here we learn how to solve equations of this type:

d²y dx² + p dy dx + 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:

dy dx + y² = 5x

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

Example:

d²y dx² + xy = sin(x)

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

Example:

d³y dx³ + x dy dx + y = e^x

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

Before tackling second order differential equations, make sure you are familiar with the various methods for solving first order differential equations.

Second Order Differential Equations

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

d²y dx² + P(x) dy dx + Q(x)y = f(x)

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

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

Variation of Parameters 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"):

d²y dx² + P(x) dy dx + Q(x)y = 0

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

d²y dx² + p dy dx + 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 e ^x equals the value of e ^x :

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

f(x) = e^rx

We find:

• the first derivative is f'(x) = re^rx
• the second derivative is f''(x) = r²e^rx

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

d²y dx² + dy dx − 6y = 0

Let y = e^rx so we get:

• dy dx = re^rx
• d²y dx² = 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

dy dx = 2Ae^2x − 3Be^−3x

d²y dx² = 4Ae^2x + 9Be^−3x

Now substitute into the original equation:

d²y dx² + dy dx − 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!

So, does this method work generally?

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

With y = e^rx as a solution of the differential equation:

d²y dx² + p dy dx + qy = 0

we get:

r²e^rx + pre^rx + qe^rx = 0

e^rx(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

d²y dx² + p dy dx + qy = 0

through the "characteristic equation":

r² + pr + q = 0

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

y = Ae^r₁x + Be^r₂x

Example 2: Solve

d²y dx² − 9 dy dx + 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

6 d²y dx² + 5 dy dx − 6y = 0

The characteristic equation is:

6r² + 5r − 6 = 0

Factor:

(3r − 2)(2r + 3) = 0

r = 2 3 or −3 2

So the general solution of our differential equation is:

y = Ae^{( 2 3 x)} + Be^{( −3 2 x)}

Example 4: Solve

9 d²y dx² − 6 dy dx − 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

d²y dx² − 10 dy dx + 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!

Why? I can show you:

y = xe^5x

dy dx = e^5x + 5xe^5x

d²y dx² = 5e^5x + 5e^5x + 25xe^5x

So

d²y dx² − 10 dy dx + 25y

= 5e^5x + 5e^5x + 25xe^5x − 10(e^5x + 5xe^5x) + 25xe^5x

= (5e^5x + 5e^5x − 10e^5x) + (25xe^5x − 50xe^5x + 25xe^5x) = 0

So, in this case our solution is:

y = Ae^5x + Bxe^5x

How does this work in the general case?

With y = xe^rx we get the derivatives:

• dy dx = e^rx + rxe^rx
• d²y dx² = re^rx + re^rx + r²xe^rx

So

d²y dx² + p dy dx + qy

= (re^rx + re^rx + r²xe^rx) + p( e^rx + rxe^rx ) + q( xe^rx )

= e^rx(r + r + r²x + p + prx + qx)

= e^rx(2r + p + x(r² + pr + q))

= e^rx(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 = Ae^rx + Bxe^rx

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

Example 6: Solve

4 d²y dx² + 4 dy dx + y = 0

The characteristic equation is:

4r² + 4r + 1 = 0

Then:

(2r + 1)² = 0

r = − 1 2

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

d²y dx² − 4 dy dx + 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) )

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

d²y dx² = e^2x( −(6C+9iD)sin(3x) + (−9C+6iD)cos(3x)) + 2e^2x(2C+3iD)cos(3x) + (−3C+2iD)sin(3x) )

Substitute:

d²y dx² − 4 dy dx + 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?

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 = e^vx ( Ccos(wx) + iDsin(wx) )

Example 8: Solve

d²y dx² − 6 dy dx + 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

9 d²y dx² + 12 dy dx + 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 = − 2 3 ± 5 3 i

And we get the solution:

y = e^{(− 2 3 )x}(Ccos( 5 3 x) + iDsin( 5 3 x))

Summary

To solve a linear second order differential equation of the form

d²y dx² + p dy dx + 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 = Ae^r₁x + Be^r₂x

zero we get one real root, and the solution is

y = Ae^rx + Bxe^rx

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

y = e^vx ( Ccos(wx) + iDsin(wx) )

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Finding General Solution of Second Order Odes

Source: https://www.mathsisfun.com/calculus/differential-equations-second-order.html

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