Method of Undetermined Coefficient

Method of Undetermined Coefficient
Written by: Robert Pinterson

Earned my Ph.D. in mathematics from University of North Carolina at Chapel Hill.
I am a lecturer with over 5 years of teaching experience and an active researcher in the field of quantum information. Passionate about everything connected with maths in particular and science in general.

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To make things simple and easy, we only look at the case as given below:

$$\frac{d^2y}{dx^2}+P\frac{dy}{dx}+qy=f(x)$$

Fact: Where p and q are constant

The complete solution to such an equation can be found by combining two type of solution which is names as particular solution and Complementary Solution:

SOLUTION:

  • The general solution of the homogeneous equation

    $$\frac{d^2y}{dx^2}+P\frac{dy}{dx}+qy=0$$

  • Particular solution of the non homogeneous equation

    $$\frac{d^2y}{dx^2}+P\frac{dy}{dx}+qy=f(x)$$

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

  • At the end we will combine both general and particular solution

The homo generous equation are as follows:

$$\frac{d^2y}{dx^2}-y=0$$

Complementary Solution:

$$y=Ae^x+Be^{-x}$$

The non-homo generous equation

$$\frac{d^2y}{dx^2}-y=2x^2-x-3$$

has a particular solution

$$y=-2x^2+x-1$$

So, the complete solution of differential equation

$$\frac{d^2y}{dx^2}-y=2x^2-x-3$$

is

$$y=Ae^x+Be^{-x}-2x^2+x+1$$

Fact: we have combine these two solutions

Checking:

$$y=Ae^x+Be^{-x}-2x^2+x+1$$

1st and 2nd order derivatives are as follows:

  • $$y=Ae^x-Be^{-x}-4x+1$$

  • $$y=Ae^x+Be^{-x}-1$$

So,

$$\frac{d^2y}{dx^2}-y=Ae^x+Be^{-x}-4-(Ae^x+Be^{-x}-2x^2+x-1)$$

$$\frac{d^2y}{dx^2}-y=Ae^x+Be^{-x}-4-Ae^x-Be^{-x}+2x^2-x+1$$

$$\frac{d^2y}{dx^2}-y=2x^2-x-3$$

How do we find a particular solution?

Introduction to limits

Example

Find particular Solution of Differential Equation:

$$Y’’+-4y’-12y=3e^{4t}$$

Solution: First we make it homogeneous equation

$$Y’’+-4y’-12y=3e^{4t}$$

The factor equation is:

$$r^2-4r-12=0$$

$$(r-6)(r+2)=0$$

$$r^1=6$$

$$r^2=-2$$

Complementary Solution:

$$y_c=Ae^{6t}+Be^{-2t}$$

Particular Solution:

  • Putting into differential equation after taking derivatives:

    After solving we have,

    $$3e^{4t}=-7Ae^{4t}$$

    $$A=-\frac{3}{7}$$

  • Write right side of equation as

    $$y_p=3e^{4t}$$

Particular Solution:

$$y_p=\frac{-3}{7}e^{4t}$$