Solving Linear Differential Equations with Undetermined Coefficients and Characteristic Equations

Introduction

In this article, we will explore a step-by-step method to solve three specific linear differential equations using the method of undetermined coefficients and the characteristic equation. These techniques are fundamental in solving differential equations which often arise in various scientific and engineering applications.

Solving Problem 1: D2 - 2D y ex

Let's start with the first problem:

D2 - 2D y ex

Step 1: Find the Complementary Solution

The complementary solution is found by solving the homogeneous part of the equation:

D2 - 2D y 0

The characteristic equation is:

r2 - 2r 0

Solving this, we get:

(r - 1)2 0

This gives a double root:

r 1 (with multiplicity 2)

Therefore, the complementary solution is:

yc C1e^x C2xe^x

Step 2: Find the Particular Solution

Since the right-hand side is ex, we try:

yp A e^x

Computing the derivatives:

y'p A e^x

y''p A e^x

Substituting into the differential equation:

A e^x - 2A e^x A e^x 0

Since e^x is part of the complementary solution, we multiply by x:

yp A x^2 e^x

Substituting back to find A:

A x^2 e^x - 2A x e^x A e^x e^x

A(x^2 - 2x 1) 1

A 1/2

Thus, the particular solution is:

yp (1/2) x^2 e^x

Step 3: General Solution

The general solution is the sum of the complementary and particular solutions:

y yc yp

y C1e^x C2xe^x (1/2) x^2 e^x

Solving Problem 2: D3 - 6D2 - 11D - 6 y e-3x

Now, let's solve the second problem:

D3 - 6D2 - 11D - 6 y e-3x

Step 1: Find the Complementary Solution

The complementary solution is found by solving the homogeneous part of the equation:

D3 - 6D2 - 11D - 6 0

The characteristic equation is:

r3 - 6r2 - 11r - 6 0

Factoring gives:

(r - 1)(r - 2)(r - 3) 0

Therefore, the roots are:

r 1, 2, 3

Thus, the complementary solution is:

yc C1e^x C2e2x C3e3x

Step 2: Find the Particular Solution

Since the right-hand side is e-3x, we try:

yp A e-3x

Computing the derivatives:

y'p -3A e-3x

y''p 9A e-3x

y'''p -27A e-3x

Substituting into the differential equation:

-27A 54A - 33A - 6A e-3x

-12A 1

A -1/12

Thus, the particular solution is:

yp (-1/12) e-3x

Step 3: General Solution

The general solution is the sum of the complementary and particular solutions:

y yc yp

y C1e^x C2e2x C3e3x - (1/12) e-3x

Solving Problem 3: D2 - 5D - 6 y e-2x cos x

Finally, let's solve the third problem:

D2 - 5D - 6 y e-2x cos x

Step 1: Find the Complementary Solution

The complementary solution is found by solving the homogeneous part of the equation:

D2 - 5D - 6 0

The characteristic equation is:

r2 - 5r - 6 0

Factoring gives:

(r - 2)(r 3) 0

Therefore, the roots are:

r -2, -3

Thus, the complementary solution is:

yc C1e-2x C2e-3x

Step 2: Find the Particular Solution

For e-2x as a part of the complementary solution, we try:

y1p A x2 e-2x

For cos x, we try:

y2p B cos x C sin x

Substituting and solving for coefficients yields:

A -1/6

B 0

C 1/5

Thus, the particular solution is:

yp -1/6 x2 e-2x 1/5 sin x

Step 3: General Solution

The general solution is the sum of the complementary and particular solutions:

y yc yp

y C1e-2x C2e-3x - 1/6 x2 e-2x 1/5 sin x

Conclusion

By following the method of undetermined coefficients and the characteristic equation, we have effectively solved three different linear differential equations. Each step involves finding the complementary solution first and then the particular solution, and finally combining them to obtain the general solution. Understanding these techniques is crucial for solving a wide range of differential equations that appear in various fields such as physics, engineering, and applied mathematics.

Further Reading and Resources

For further elaboration and additional resources on solving differential equations, I recommend exploring tutorials and textbooks on differential equations. These can be found in academic libraries, online resources, and course materials from educational institutions.