The Method of Variation of Parameters (2024)

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.

Please read Introduction to Second Order Differential Equations first, it shows how to solve the simpler "hom*ogeneous" case where f(x)=0

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:

The general solution of the hom*ogeneous equation d²ydx² + pdydx + qy = 0
Particular solutions of the non-hom*ogeneous 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 = 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) )

The Fundamental Solutions of The Equation

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

y = Ae^r₁x + Be^r₂x is made of y₁ = Ae^r₁x and y₂ = Be^r₂x
y = Ae^rx + Bxe^rx is made of y₁ = Ae^rx and y₂ = Bxe^rx
y = e^vx ( Ccos(wx) + iDsin(wx) ) is made of y₁ = e^vxCcos(wx) and y₂ = e^vxiDsin(wx)

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 hom*ogeneous equation

d²ydx² + pdydx + qy = 0

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

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:

y_p(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 = e^3x

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 = Ae^x+Be^2x

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

y₁(x) = e^x

y₁'(x) = e^x

y₂(x) = e^2x

y₂'(x) = 2e^2x

2. Find the Wronskian:

W(y₁, y₂) = y₁y₂' − y₂y₁' = 2e^3x − e^3x = e^3x

3. Find the particular solution using the formula:

y_p(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

= ∫e^2xe^3xe^3xdx

= ∫e^2xdx

= 12e^2x

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 = Ae^x+Be^−x

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

y₁(x) = e^x

y₁'(x) = e^x

y₂(x) = e^−x

y₂'(x) = −e^−x

2. Find the Wronskian:

W(y₁, y₂) = y₁y₂' − y₂y₁' = −e^xe^−x − e^xe^−x = −2

3. Find the particular solution using the formula:

y_p(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^−xdx ]

= −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 = (−e^x)[e^−x2( 2x² + 3x )] =−12(2x² + 3x)

And this one:

∫y₁(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^xdx ]

= −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 − 5x + 2 ]

So:

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

Finally:

y_p(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 = Ae^x + 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 = Ae^3x + Bxe^3x

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

y₁(x) = e^3x

y₁'(x) = 3e^3x

y₂(x) = xe^3x

y₂'(x) = (3x + 1)e^3x

2. Find the Wronskian:

W(y₁, y₂) = y₁y₂' − y₂y₁' = (3x + 1)e^3xe^3x − 3xe^3xe^3x = e^6x

3. Find the particular solution using the formula:

y_p(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

= ∫(xe^3x)x⁻¹ e^6xdx (Note: 1x = x⁻¹)

= ∫e^−3xdx

= −13e^−3x

So:

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

And this one:

∫y₁(x)f(x)W(y₁, y₂)dx

= ∫e^3xx⁻¹e^6xdx

= ∫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 = ( xe^3x)( ∫e^−3xx⁻¹dx ) = xe^3x∫e^−3xx⁻¹dx

Finally:

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

= 13 + xe^3x∫e^−3xx⁻¹dx

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

y = Ae^3x + Bxe^3x + 13 + xe^3x∫e^−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) 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 = e^3x[Acos(2x) + iBsin(2x)]

So in this case we have:

y₁(x) = e^3xcos(2x)

y₁'(x) = e^3x[3cos(2x) − 2sin(2x)]

y₂(x) = e^3xsin(2x)

y₂'(x) = e^3x[3sin(2x) + 2cos(2x)]

2. Find the Wronskian:

W(y₁, y₂) = y₁y₂' − y₂y₁'

= e^6xcos(2x)[3sin(2x) + 2cos(2x)] − e^6xsin(2x)[3cos(2x) − 2sin(2x)]

= e^6x[3cos(2x)sin(2x) +2cos²(2x) − 3sin(2x)cos(2x) + 2sin²(2x)]

= 2e^6x

3. Find the particular solution using the formula:

y_p(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

= ∫e^3xsin⁡(2x)[195cos⁡(4x)] 2e^6xdx

= 1952∫e^−3xsin(2x)cos(4x)dx

= 1954∫e^−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

= ∫e^3xcos(2x)[195cos(4x)]2e^6xdx

= 1952∫e^−3xcos(2x)cos(4x)dx

= 1954∫e^−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) − 36∫e^−3xcos(6x)dx = − 16e^−3xcos(6x) − 12I₃

⇒ 2I₁+ I₃ = − 13e^−3xcos(6x) ... (3)

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

⇒ 2I₂+ 3I₄ = − e^−3xcos(2x) ... (4)

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

⇒ 5I₃= − 13e^−3xcos(6x) + 23e^−3xsin(6x)

= 13e^−3x[2sin(6x) − cos(6x)]

⇒ I₃= 115e^−3x[2sin(6x) − cos(6x)]

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

⇒ 5I₁= − 23e^−3xcos(6x) − 13e^−3xsin(6x)

= − 13e^−3x[2cos(6x) + sin(6x)]

⇒ I₁= − 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):

⇒ 13I₄= − 3e^−3xcos(2x) + 2e^−3xsin(2x)

=e^−3x[2sin(2x) − 3 cos(2x)]

⇒ I₄= 113e^−3x[2sin(2x) − 3cos(2x)]

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

⇒ 13I₂= − 2e^−3xcos(2x) − 3e^−3xsin(2x)

=− e^−3x[2cos(2x) + 3 sin(2x)]

⇒ I₂= − 113e^−3x[2cos(2x) + 3sin(2x)]

Substitute into (1) and (2):

∫y₂(x)f(x)W(y₁, y₂)dx

= 1954∫e^−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

= 1954∫e^−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 y_p(x) = −y₁(x)∫y₂(x)f(x)W(y₁, y₂)dx + y₂(x)∫y₁(x)f(x)W(y₁, y₂)dx

= − e^3xcos(2x)e^−3x4[−13(2cos(6x)+sin(6x)) + 15(2 cos⁡(2x)+3sin(2x))] + e^3xsin(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 = e^3x(Acos(2x) + iBsin(2x)) − cos(4x) − 8sin(4x)

9529, 9530, 9531, 9532, 9533, 9534, 9535, 9536, 9537, 9538

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The Method of Variation of Parameters (2024)

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Two Methods in Variation of Parameters

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The Method of Variation of Parameters (2024)

Two Methods

Variation of Parameters

Start with the General Solution

The Fundamental Solutions of The Equation

The Wronskian

The Particular Solution

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

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

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

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

FAQs

What are the methods of variation of parameters? ›

What is the variation method in math? ›

Example 1: Solve d²ydx² − 3dydx + 2y = e^3x

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

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

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