Find a Particular Solution of Y 3y 2y Sin T

The Method of Variation of Parameters

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

d²y dx² + P(x) dy dx + Q(x)y = f(x)

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

Two Methods

There are two main methods to solve equations like

d²y dx² + P(x) dy dx + 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:

d²y dx² + p dy dx + 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 d²y dx² + p dy dx + qy = 0
2. Particular solutions of the non-homogeneous equation d²y dx² + p dy dx + 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

d²y dx² + p dy dx + 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 homogeneous equation

d²y dx² + p dy dx + qy = 0

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

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

d²y dx² + p dy dx + 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 d²y dx² − 3 dy dx + 2y = e^3x

1. Find the general solution of d²y dx² − 3 dy dx + 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^3x e^3x dx

= ∫e^2xdx

= 12e^2x

So:

−y₁(x)∫ y₂(x)f(x) W(y₁, y₂) dx = −(e^x)( 12e^2x) = − 12e^3x

And also:

∫ y₁(x)f(x) W(y₁, y₂) dx

= ∫ e^xe^3x e^3x dx

= ∫e^xdx

= e^x

So:

y₂(x)∫ y₁(x)f(x) W(y₁, y₂) dx = (e^2x)(e^x) = e^3x

Finally:

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

= − 12e^3x + e^3x

= 12e^3x

and the complete solution of the differential equation d²y dx² − 3 dy dx + 2y = e^3x is

y = Ae^x + Be^2x + 12e^3x

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

Example 2:  Solve d²y dx² − y = 2x² − x − 3

1. Find the general solution of d²y dx² − 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:

∫ y₂(x)f(x) W(y₁, y₂) dx

= ∫ e^−x (2x²−x−3) −2 dx

= − 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^−x 2[ 2x² − x − 3 + 4x −1 + 4 ]

= e^−x 2[ 2x² + 3x ]

So:

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

And this one:

∫ y₁(x)f(x) W(y₁, y₂) dx

= ∫ e^x (2x²−x−3) −2 dx

= − 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^x 2[ 2x² − x − 3 − 4x + 1 + 4 ]

= −e^x 2[ 2x² − 5x + 2 ]

So:

y₂(x)∫ y₁(x)f(x) W(y₁, y₂) dx = (e^−x)[ −e^x 2( 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 d²y dx² − 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 d²y dx² − 6 dy dx + 9y = 1 x

1. Find the general solution of d²y dx² − 6 dy dx + 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^6x dx   (Note: 1 x = 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⁻¹ e^6x dx

= ∫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 d²y dx² − 6 dy dx + 9y = 1x is

y = Ae^3x + Bxe^3x + 13 + xe^3x ∫e^−3xx⁻¹dx

Example 4 (Harder example):  Solve d²y dx² − 6 dy dx + 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) = 1 2 [sin⁡(θ + φ) + sin⁡(θ − φ)]
cos(θ)cos(φ) = 1 2 [cos⁡(θ − φ) + cos⁡(θ + φ)]

1. Find the general solution of d²y dx² − 6 dy dx + 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^6x dx

= 195 2 ∫e^−3xsin(2x)cos(4x)dx

= 195 4 ∫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^6x dx

= 195 2 ∫e^−3xcos(2x)cos(4x)dx

= 195 4 ∫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:

I ₁ = ∫e^−3xsin(6x)dx
I ₂ = ∫e^−3xsin(2x)dx
I ₃ = ∫e^−3xcos(6x)dx
I ₄ = ∫e^−3xcos(2x)dx

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

I ₁ = ∫e^−3xsin(6x)dx = − 1 6 e^−3xcos(6x) − 3 6 ∫e^−3xcos(6x)dx = − 1 6 e^−3xcos(6x) − 12 I ₃

⇒ 2I ₁ + I ₃ = − 13e^−3xcos(6x)    ...  (3)

I ₂ = ∫e^−3xsin(2x)dx = − 1 2 e^−3xcos(2x) − 32 ∫e^−3xcos(2x)dx = − 12e^−3xcos(2x) − 32 I ₄

⇒ 2I ₂ + 3I ₄ = − e^−3xcos(2x)    ...  (4)

I ₃ = ∫e^−3xcos(6x)dx = 1 6 e^−3xsin(6x) + 3 6 ∫e^−3xsin(6x)dx = 1 6 e^−3xsin(6x) + 12 I ₁
⇒ 2I ₃ − I ₁ = 13e^−3xsin(6x)    ...  (5)
I ₄ = ∫e^−3xcos(2x)dx = 1 2 e^−3xsin(2x) + 32 ∫e^−3xsin(2x)dx = 12e^−3xsin(2x) + 32 I ₂

⇒ 2I ₄ − 3I ₂ = e^−3xsin(2x)    ...  (6)

Solve equations (3) and (5) simultaneously:

2I ₁ + I ₃ = − 13e^−3xcos(6x)    ...  (3)

2I ₃ − I ₁ = 13e^−3xsin(6x)    ...  (5)

Multiply equation (5) by 2 and add them together (term I ₁ will neutralize):

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

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

⇒ I ₃ = 115e^−3x[2sin(6x) − cos(6x)]

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

⇒ 5I ₁ = − 23e^−3xcos(6x) − 13e^−3xsin(6x)

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

⇒ I ₁ = − 115e^−3x[2cos(6x) + sin(6x)]

Solve equations (4) and (6) simultaneously:

2I ₂ + 3I ₄ = − e^−3xcos(2x)    ...  (4)

2I ₄ − 3I ₂ = e^−3xsin(2x)    ...  (6)

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

⇒ 13I ₄ = − 3e^−3xcos(2x) + 2e^−3xsin(2x)

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

⇒ I ₄ = 113e^−3x[2sin(2x) − 3cos(2x)]

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

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

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

⇒ I ₂ = − 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^−3x 4[−13(2cos(6x)+sin(6x))+15(2 cos⁡(2x)+3sin(2x))]

∫ y₁(x)f(x) W(y₁, y₂) dx

= 195 4 ∫e^−3x[cos(6x) + cos(2x)]dx    ...  (2)

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

= e^−3x 4[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^−3x 4[−13(2cos(6x)+sin(6x)) + 15(2 cos⁡(2x)+3sin(2x))] + e^3xsin(2x) e^−3x 4[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 d²y dx² − 6 dy dx + 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

Find a Particular Solution of Y 3y 2y Sin T

Source: https://www.mathsisfun.com/calculus/differential-equations-variation-parameters.html

Share :