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:

Step 2:

Step 3:

Step 4:

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.

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