Title 

Non-homogeneous Equation; Method of Undetermined Coefficients

                                                            Overview

           The topic is about Non-homogeneous equation, with method of undetermined Coefficients. For our better understanding we all should know what homogeneous equation is. Homogeneous equation is a differential equation, which is equal to zero. However, before we proceed to solve the Non-homogeneous equation, with method of undetermined Coefficients, we must look for some key factors into our differential equation. For example, the differential equations must be linear and should not be more than second order. The method of undetermined coefficients is a use full technique determining a particular solution to a differential equation with linear constant-Coefficient.

    Theorem

The form of the nonhomogeneous second-order differential equation, looks like this y”+p(t)y’+q(t)y=g(t) Where p, q and g are given continuous function on an open interval I. On the other hand, the homogenous part will look like, y’’+p(t)y’+q(t)y=0

Sample problem

A Sample Problem of Non-homogeneous Equation; Method of Undetermined Coefficients below:

1. y” – 4y’+ 3y= – 168e^7t , particular solution satisfying y(0)= 9 and y’(0)= -15.

Step 1:

Make this differential as a homogeneous differential equation
y” – 4y’+ 3y= 0	Note: The differential equation is now homogenous

Step 2:

Now we can write the homogenous differential to a characteristic equation.
r2­ – 4r + 3 = 0	Note: The differential equation is now characteristic polynomial.

Step 3:

Solve the characteristic equation
r2­ – 4r + 3 = 0r2 – 3r – r +3 = 0r ( r -3 ) – 1 (r – 3)=0 (r – 3) ( r -1)=0		Note: 3 and 1 are the roots of the characteristic polynomial. After we find the roots of the polynomial, the Y1 and Y2 will take form of eat. However, there are some exceptions and we will address it later ( at the end of the page )
(r – 3)=0 r= 3	( r -1)=0 Or  r = 1
Y1 = e3t	Y2 = et
y(t)= C1 e3t + C2 et		Note: This is homogeneous part of solution. Where C1 and C2 are the arbitrary constants.

Step 4:

Guessing Table
If the right side g(t)	We will guess
Ae7t	Ce7t
Example: – 168e7t	Ce7t       Note: C is for arbitrary constants and e7t is from the differential equation.
A sin(t)	C sin (t) + D Cos (t)
A cos (t)	C sin (t) + D Cos (t)
t2+ 7	Ct2 + Dt + E

Step 5:

After appropriate guessing, now, we can process to solve the Non-homogeneous part of differential equation. Now we have to take the first and second derivative guess.

Guess:

Y(t) = Ce^7t

^y’ = C (7e^7t )

y’’= C (49e^7t)

Then plug in the y(t), y’(t) and y’’(t) into the original differential equation. In this case, we will plug it into y” – 4y’+ 3y= – 168e^7t.

Non-homogeneous part of differential equation
y” – 4y’+ 3y= – 168e7t  C (49e7t) – C (4*7e7t ) + C (3e7t)= – 168e7t C =(- 168e7t)/(24e7t) C = -7e7t	Note:   Non-homogeneous solution to   differential equation is -7e7t
Y(t)= C1 e3t + C2 et -7e7t  Note: General solution is ,Y(t)= C1 e3t +  C2et -7e7t

 

 

Step 6:

Particular solution

Condition I, y(0) = 9
9 = C1 + C2 – 7
9 + 7 = C1 + C2
C1 = 9 + 7 – C2
C1 = 9 + 7 – 7
C1 = 9

Note: Set 9 equal to our General solution, but make sure do not take the whole mathematical expression. Instated, take only arbitrary constants as a
C1 and C2. Then write the differential equation expression in tram of C1.

Condition II, y’(0) = – 15
Y (t) = C1e3t + C2 et -7e7t
y’ (t)= 3C1e3t + C2et – 49 e7t
-15 = 3C1 + C2 – 49
-15 = 3 ( 9 + 7 – C2 ) + C2 – 49
– 15 = 3 (16 – C2 ) + C2 – 49
– 15 = 48 – 3 C2 + C2 – 49
– 15 – 48 + 49 = – 3 C2 + C2
– 14 = – 2 C2
C2 = (- 14 )/(– 2 )
C2 = 7
Note: In condition II, we have to take our general solution y(t) then take the first derivative . Then Set y’(0)= -15 equal to Coefficients of the differential equation with arbitrary constants as a  C1 and C2. Then plug in the value of C1 into our new differential equation expression. Finally, solve for C2.
The particular solution of differential equation is : Y(t)= 9e3t + 7 et – 7e7t

 

CAUTION: (very important) In step 3, before writing homogeneous part of the solution, we have to consider some of the important facts. If homogeneous part of the solution , contain same roots or double roots, such as, if we would find r = 7 0r r= 7 and r = 7, then we must introduce another t into our homogeneous part of the solution . How to identify homogeneous part of the solution has single root or double roots? Good question. If value of r is same as g(t) or in another word , value of r and the exponent of Nonhomogeneous equation are same then it has single root . If two values of r are same as an exponent of Nonhomogeneous,g(t), equation, then it has double roots. {An example, if r= 7 then Y₁ = (e^7t)*t}. In another case, we might find complex roots such as,   r = -4 ± 3i then y(t) = C₁ e^{– 4t}cos(3t) + C₂ e^{– 4t} sin(3t), because of imaginary part 3i, we have Cos (3t) and Sin (3t).

 

If you still have any questions and concerns then please contact me at Rana.Das@mail.citytech.cuny.edu

Thank you,

Rana das