The Method of Variation of Parameters

This page is about second order differential equations of this type:

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

where P(x), Q(x) and f(x) are functions of x.

The simpler case where f(x) = 0:

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

is "homogeneous" and is explained on Introduction to Second Order Differential Equations. Please learn that method first to help you understand this page.

Two Methods

There are two main methods to solve equations like

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

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

Variation of Parameters (that we will learn here) which works on a wide range of functions but is a little messy to use.

Variation of Parameters

To keep things simple, we are only going to look at the case:

d2ydx2 + pdydx + qy = f(x)

where p and q are constants and f(x) is a non-zero function of x.

The complete solution to such an equation can be found by combining two types of solution:

  1. The general solution of the homogeneous equation d2ydx2 + pdydx + qy = 0
  2. Particular solutions of the non-homogeneous equation d2ydx2 + pdydx + qy = f(x)

Note that f(x) could be a single function or a sum of two or more functions.

Once we have found the general solution and all the particular solutions, then the final complete solution is found by adding all the solutions together.

This method relies on integration.

The problem with this method is that, although it may yield a solution, in some cases the solution has to be left as an integral.

Start with the General Solution

On Introduction to Second Order Differential Equations we learn how to find the general solution.

Basically we take the equation

d2ydx2 + pdydx + qy = 0

and reduce it to the "characteristic equation":

r2 + pr + q = 0

Which is a quadratic equation that has three possible solution types depending on the discriminant p2 - 4q. When p2 - 4q 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) )

The Fundamental Solutions of The Equation

In all three cases above "y" is made of two parts:

y1 and y2 are known as the fundamental solutions of the equation

And y1 and y2 are said to be linearly independent because neither function is a constant multiple of the other.

The Wronskian

When y1 and y2 are the two fundamental solutions of the homogeneous equation

d2ydx2 + pdydx + qy = 0

then the Wronskian W(y1, y2) is the determinant of the matrix

  matrix for the Wronskian 

So

W(y1, y2) = y1y2' − y2y1'

The Wronskian is named after the Polish mathematician and philosopher Józef Hoene-Wronski (1776-1853).

Since y1 and y2 are linearly independent, the value of the Wronskian cannot equal zero.

The Particular Solution

Using the Wronskian we can now find the particular solution of the differential equation

d2ydx2 + pdydx + qy = f(x)

using the formula:

yp(x) = −y1(x)y2(x)f(x)W(y1,y2)dx + y2(x)y1(x)f(x)W(y1,y2)dx


Example 1: Solve d2ydx2 − 3dydx + 2y = e3x

1. Find the general solution of d2ydx2 − 3dydx + 2y = 0

The characteristic equation is: r2 − 3r + 2 = 0

Factor: (r − 1)(r − 2) = 0

r = 1 or 2

So the general solution of the differential equation is y = Aex+Be2x

So in this case the fundamental solutions and their derivatives are:

y1(x) = ex

y1'(x) = ex

y2(x) = e2x

y2'(x) = 2e2x

2. Find the Wronskian:

W(y1, y2) = y1y2' − y2y1' = 2e3x − e3x = e3x

3. Find the particular solution using the formula:

yp(x) = −y1(x)y2(x)f(x)W(y1,y2)dx + y2(x)y1(x)f(x)W(y1,y2)dx

4. First we solve the integrals:

y2(x)f(x)W(y1,y2)dx


= e2xe3xe3xdx

= e2xdx

= 12e2x

So:

−y1(x)y2(x)f(x)W(y1,y2)dx = −(ex)(12e2x) = −12e3x

And also:

y1(x)f(x)W(y1,y2)dx


= exe3xe3xdx

= exdx

= ex

So:

y2(x)y1(x)f(x)W(y1,y2)dx = (e2x)(ex) = e3x

Finally:

yp(x) = −y1(x)y2(x)f(x)W(y1,y2)dx + y2(x)y1(x)f(x)W(y1,y2)dx

= −12e3x + e3x

= 12e3x

and the complete solution of the differential equation d2ydx2 − 3dydx + 2y = e3x is

y = Aex + Be2x + 12e3x

Which looks like this (example values of A and B):

Aex + Be2x + 12e3x


Example 2:  Solve d2ydx2 − y = 2x2 − x − 3


1. Find the general solution of d2ydx2 − y = 0

The characteristic equation is: r2 − 1 = 0

Factor: (r − 1)(r + 1) = 0

r = 1 or −1

So the general solution of the differential equation is y = Aex+Be−x

So in this case the fundamental solutions and their derivatives are:

y1(x) = ex

y1'(x) = ex

y2(x) = e−x

y2'(x) = −e−x

2. Find the Wronskian:

W(y1, y2) = y1y2' − y2y1' = −exe−x − exe−x = −2

3. Find the particular solution using the formula:

yp(x) = −y1(x)y2(x)f(x)W(y1,y2)dx + y2(x)y1(x)f(x)W(y1,y2)dx

4. Solve the integrals:

Each of the integrals can be obtained by using Integration by Parts twice:

y2(x)f(x)W(y1,y2)dx


= e−x (2x2−x−3)−2dx

= −12 (2x2−x−3)e−xdx

= −12[ −(2x2−x−3)e−x + (4x−1)e−x dx ]

= −12[ −(2x2−x−3)e−x − (4x − 1)e−x + 4e−xdx ]

= −12[ −(2x2−x−3)e−x − (4x − 1)e−x − 4e−x ]

= e−x2[ 2x2 − x − 3 + 4x −1 + 4 ]

= e−x2[ 2x2 + 3x ]

So:

−y1(x)y2(x)f(x)W(y1,y2)dx = (−ex)[e−x2( 2x2 + 3x )] = −12(2x2 + 3x)

And this one:

y1(x)f(x)W(y1,y2)dx


= ex (2x2−x−3)−2dx

= −12 (2x2−x−3)exdx

= −12[ (2x2−x−3)ex (4x−1)ex dx ]

= −12[ (2x2−x−3)ex − (4x − 1)ex + 4exdx ]

= −12[ (2x2−x−3)ex − (4x − 1)ex + 4ex ]

= −ex2[ 2x2 − x − 3 − 4x + 1 + 4 ]

= −ex2[ 2x2 − 5x + 2 ]
 

So:

y2(x)y1(x)f(x)W(y1,y2)dx = (e−x)[−ex2( 2x2 − 5x + 2 ) ] = −12( 2x2 − 5x + 2 ) 

Finally:

yp(x) = −y1(x)y2(x)f(x)W(y1,y2)dx + y2(x)y1(x)f(x)W(y1,y2)dx

= −12( 2x2 + 3x ) − 12( 2x2 − 5x + 2 ) 

= −12( 4x2 − 2x + 2 )

= −2x2 + x − 1

and the complete solution of the differential equation d2ydx2 − y = 2x2 − x − 3 is

y = Aex + Be−x − 2x2 + x − 1

(This is the same answer that we got in Example 1 on the page Method of undetermined coefficients.)

Example 3:  Solve d2ydx2 − 6dydx + 9y =1x


1. Find the general solution of d2ydx2 − 6dydx + 9y = 0

The characteristic equation is: r2 − 6r + 9 = 0

Factor: (r − 3)(r − 3) = 0

r = 3

So the general solution of the differential equation is y = Ae3x + Bxe3x

And so in this case the fundamental solutions and their derivatives are:

y1(x) = e3x

y1'(x) = 3e3x

y2(x) = xe3x

y2'(x) = (3x + 1)e3x

2. Find the Wronskian:

W(y1, y2) = y1y2' − y2y1' = (3x + 1)e3xe3x − 3xe3xe3x = e6x

3. Find the particular solution using the formula:

yp(x) = −y1(x)y2(x)f(x)W(y1,y2)dx + y2(x)y1(x)f(x)W(y1,y2)dx

4. Solve the integrals:

y2(x)f(x)W(y1,y2)dx


= (xe3x)x−1 e6xdx   (Note: 1x = x−1)

= e−3xdx

= −13e−3x

So:

−y1(x)y2(x)f(x)W(y1,y2)dx = −(e3x)(−13e−3x) = 13

And this one:

y1(x)f(x)W(y1,y2)dx


= e3xx−1e6xdx

= e−3xx−1dx

This cannot be integrated, so this is an example where the answer has to be left as an integral.

So:

y2(x)y1(x)f(x)W(y1,y2)dx = ( xe3x )( e−3xx−1dx ) = xe3xe−3xx−1dx

Finally:

yp(x) = −y1(x)y2(x)f(x)W(y1,y2)dx + y2(x)y1(x)f(x)W(y1,y2)dx

= 13 + xe3xe−3xx−1dx

So the complete solution of the differential equation d2ydx2 − 6dydx + 9y = 1x is

y = Ae3x + Bxe3x + 13 + xe3xe−3xx−1dx

Example 4 (Harder example):  Solve d2ydx2 − 6dydx + 13y = 195cos(4x)


This example uses the following trigonometric identities

sin2(θ) + cos2(θ) = 1

sin⁡(θ ± φ) = sin(θ)cos(φ) ± cos(θ)sin(φ)

cos⁡(θ ± φ) = cos(θ)cos(φ) minus/plus sin(θ)sin(φ)

sin(θ)cos(φ) = 12[sin⁡(θ + φ) + sin⁡(θ − φ)]
cos(θ)cos(φ) = 12[cos⁡(θ − φ) + cos⁡(θ + φ)]


1. Find the general solution of d2ydx2 − 6dydx + 13y = 0

The characteristic equation is: r2 − 6r + 13 = 0

Use the quadratic equation formula

x = −b ± √(b2 − 4ac)2a

with a = 1, b = −6 and c = 13

So:

r = −(−6) ± √[(−6)2 − 4(1)(13)] 2(1)

= 6 ± √[36−52] 2

= 6 ± √[−16] 2

= 6 ± 4i 2

= 3 ± 2i

So α = 3 and β = 2

y = e3x[Acos(2x) + iBsin(2x)]

So in this case we have:

y1(x) = e3xcos(2x)

y1'(x) = e3x[3cos(2x) − 2sin(2x)]

y2(x) = e3xsin(2x)

y2'(x) = e3x[3sin(2x) + 2cos(2x)]

2. Find the Wronskian:

W(y1, y2) = y1y2' − y2y1'

= e6xcos(2x)[3sin(2x) + 2cos(2x)] − e6xsin(2x)[3cos(2x) − 2sin(2x)]

= e6x[3cos(2x)sin(2x) +2cos2(2x) − 3sin(2x)cos(2x) + 2sin2(2x)]

=2e6x


3. Find the particular solution using the formula:

yp(x) = −y1(x)y2(x)f(x)W(y1,y2)dx + y2(x)y1(x)f(x)W(y1,y2)dx


4. Solve the integrals:

y2(x)f(x)W(y1,y2)dx


= e3xsin⁡(2x)[195cos⁡(4x)] 2e6xdx 

= 1952e−3xsin(2x)cos(4x)dx

= 1954e−3x[sin(6x) − sin(2x)]dx    ...  (1)

In this case, we won’t do the integration yet, for reasons that will become clear in a moment.

The other integral is:

y1(x)f(x)W(y1,y2)dx

= e3xcos(2x)[195cos(4x)]2e6xdx

= 1952e−3xcos(2x)cos(4x)dx

= 1954e−3x[cos(6x) + cos(2x)]dx    ...  (2)


  
From equations (1) and (2) we see that there are four very similar integrations that we need to perform:

I1 = e−3xsin(6x)dx
I2 = e−3xsin(2x)dx
I3 = e−3xcos(6x)dx
I4 = e−3xcos(2x)dx

Each of these could be obtained by using Integration by Parts twice, but there’s an easier method:

I1 = e−3xsin(6x)dx = −16e−3xcos(6x) − 36e−3xcos(6x)dx = − 16e−3xcos(6x) − 12I3

2I1 + I3 = − 13e−3xcos(6x)    ...  (3)

I2 = e−3xsin(2x)dx = −12e−3xcos(2x) − 32e−3xcos(2x)dx = − 12e−3xcos(2x) − 32I4

2I2 + 3I4 = − e−3xcos(2x)    ...  (4)

I3 = e−3xcos(6x)dx = 16e−3xsin(6x) + 36e−3xsin(6x)dx = 16e−3xsin(6x) + 12I1
2I3 I1 = 13e−3xsin(6x)    ...  (5)
I4 = e−3xcos(2x)dx = 12e−3xsin(2x) + 32e−3xsin(2x)dx = 12e−3xsin(2x) + 32I2

2I4 − 3I2 = e−3xsin(2x)    ...  (6)

Solve equations (3) and (5) simultaneously:

2I1 + I3 = − 13e−3xcos(6x)    ...  (3)

2I3 I1 = 13e−3xsin(6x)    ...  (5)

Multiply equation (5) by 2 and add them together (term I1 will neutralize):

5I= − 13e−3xcos(6x) + 23e−3xsin(6x)

            = 13e−3x[2sin(6x) − cos(6x)]

I= 115e−3x[2sin(6x) − cos(6x)]

Multiply equation (3) by 2 and subtract (term I3 will neutralize):

5I= − 23e−3xcos(6x) − 13e−3xsin(6x)

            = − 13e−3x[2cos(6x) + sin(6x)]

I= − 115e−3x[2cos(6x) + sin(6x)]

Solve equations (4) and (6) simultaneously:

2I2 + 3I4 = − e−3xcos(2x)    ...  (4)

2I4 − 3I2 = e−3xsin(2x)    ...  (6)

Multiply equation (4) by 3 and equation (6) by 2 and add (term I2 will neutralize):

13I= − 3e−3xcos(2x) + 2e−3xsin(2x)

            =e−3x[2sin(2x) − 3 cos(2x)]

I= 113e−3x[2sin(2x) − 3cos(2x)]

Multiply equation (4) by 2 and equation (6) by 3 and subtract (term I4 will neutralize):

13I= − 2e−3xcos(2x) − 3e−3xsin(2x)

            =− e−3x[2cos(2x) + 3 sin(2x)]

I= − 113e−3x[2cos(2x) + 3sin(2x)]

Substitute into (1) and (2):

y2(x)f(x)W(y1,y2)dx

= 1954e−3x[sin(6x) − sin(2x)]dx   ... (1)  

= 1954[ 115e−3x[2cos(6x) + sin(6x)] − [−113e−3x[2cos(2x) + 3sin(2x)]]]

= e−3x4[−13(2cos(6x)+sin(6x))+15(2 cos⁡(2x)+3sin(2x))]

y1(x)f(x)W(y1,y2)dx

= 1954e−3x[cos(6x) + cos(2x)]dx    ...  (2)

= 1954[115e−3x[2sin(6x) − cos(6x)] + 113e−3x[2sin(2x) − 3cos(2x)]]

= e−3x4[13(2sin(6x) − cos(6x)) + 15(2sin⁡(2x) − 3cos(2x))]

So yp(x) = −y1(x)y2(x)f(x)W(y1,y2)dx + y2(x)y1(x)f(x)W(y1,y2)dx

= − e3xcos(2x)e−3x4[−13(2cos(6x)+sin(6x)) + 15(2 cos⁡(2x)+3sin(2x))] + e3xsin(2x)e−3x4[13(2sin(6x) − cos(6x)) + 15(2sin⁡(2x) − 3cos(2x))]

= − 14cos(2x) [−13(2cos(6x) − sin(6x)) + 15(2 cos⁡(2x) + 3sin(2x))] +14 sin⁡(2x)[13(2sin(6x) − cos(6x)) + 15(2 sin⁡(2x) − 3cos(2x))]  

= 14[26cos(2x)cos(6x) + 13cos(2x)sin(6x) − 30cos2(2x) − 45cos(2x)sin(2x) + 26sin(2x)sin(6x) − 13sin(2x)cos(6x) + 30sin2(2x) − 45sin(2x)cos(2x)]

= 14[26[cos(2x)cos(6x) + sin(2x)sin(6x)] + 13[cos(2x)sin(6x) − sin(2x)cos(6x)] − 30[cos2(2x) − sin2(2x)] − 45[cos(2x)sin(2x) + sin(2x)cos(2x)]]

= 14[26cos(4x) + 13sin(4x) − 30cos(4x) − 45sin(4x)]

= 14[−4cos(4x) − 32sin(4x)]

= −cos⁡(4x) − 8 sin⁡(4x)

So the complete solution of the differential equation d2ydx2 − 6dydx + 13y = 195cos(4x) is

y = e3x(Acos(2x) + iBsin(2x)) − cos(4x) − 8sin(4x)