Finding a Particular Solution for a Non-Homogeneous Differential Equation

One of the key challenges in solving differential equations is to find the particular solution that satisfies the given non-homogeneous equation. In this article, we will guide you through the process to find a particular solution for the differential equation: [ y''' - 3y'' - 5y cos(x)e^x ]

Let's break down the process into manageable steps, ensuring each step is clear and easy to follow.

Simplifying the Equation

The given differential equation can be simplified as:

[ y''' - 3y'' - 5y cos(x)e^x. ]

Writing this in a more standard form, we have:

[ y''' - 3y'' - 5y cos(x)e^x. ]

This is a non-homogeneous linear differential equation. The next steps will focus on finding the particular solution.

Step 1: Solve the Homogeneous Equation

First, we solve the associated homogeneous equation:

[ y''' - 3y'' - 5y 0. ]

To solve this, we find the characteristic equation:

[ r^3 - 3r^2 - 5 0. ]

Using the quadratic formula for the coefficients of (r^2), (r^1), and (r^0), we get:

[ r frac{-b pm sqrt{b^2 - 4ac}}{2a} ]

Here, (a 1), (b -3), and (c -5). Plugging in these values, we solve for (r):

[ r frac{3 pm sqrt{(-3)^2 - 4 cdot 1 cdot (-5)}}{2 cdot 1} frac{3 pm sqrt{9 20}}{2} frac{3 pm sqrt{29}}{2} ]

Let:

[ r_1 frac{-3 sqrt{29}}{2} quad text{and} quad r_2 frac{-3 - sqrt{29}}{2}. ]

The general solution to the homogeneous equation is:

[ y_h C_1 e^{r_1 x} C_2 e^{r_2 x}, ]

where (C_1) and (C_2) are constants.

Step 2: Find a Particular Solution

Next, we find a particular solution (y_p) to the non-homogeneous equation. The right-hand side is (cos(x)e^x), which suggests using the method of undetermined coefficients.

We assume a particular solution of the form:

[ y_p e^x(A cos(x) B sin(x)). ]

Where (A) and (B) are constants to be determined.

Step 3: Compute Derivatives

Now, we compute the first and second derivatives of (y_p):

First derivative:

[ y_p' e^x(A cos(x) B sin(x))' e^x(-A sin(x) B cos(x)) ]

Second derivative:

[ y_p'' e^x(-A sin(x) B cos(x))' e^x(-A cos(x) - B sin(x)) ]

Third derivative:

[ y_p''' e^x(-A cos(x) - B sin(x))' e^x(A sin(x) - B cos(x)) ]

Substituting these back into the original differential equation:

[ y_p''' - 3y_p'' - 5y_p e^x(A sin(x) - B cos(x)) - 3e^x(-A cos(x) - B sin(x)) - 5e^x(A cos(x) B sin(x)) cos(x)e^x ]

Combining like terms, we get:

[ e^x(A sin(x) - B cos(x) 3A cos(x) 3B sin(x) - 5A cos(x) - 5B sin(x)) cos(x)e^x ]

Simplifying further:

[ e^x((A - B 3B - 5B) sin(x) (A 3A - 5A) cos(x)) cos(x)e^x ]

Setting the coefficients of (cos(x)) and (sin(x)) equal gives you a system of equations to solve for (A) and (B).

Setting the coefficients of (cos(x)) and (sin(x)):

[ -2A 1 quad text{and} quad 2B 0 ]

Solving these, we get:

[ A -frac{1}{2} quad text{and} quad B 0. ]

The particular solution is:

[ y_p -frac{1}{2} e^x cos(x). ]

Step 4: Combine the Solutions

Finally, the general solution is the sum of the homogeneous solution and the particular solution:

[ y y_h y_p C_1 e^{frac{-3 sqrt{29}}{2} x} C_2 e^{frac{-3 - sqrt{29}}{2} x} - frac{1}{2} e^x cos(x). ]

This is the complete solution to the differential equation.

Conclusion

Apart from following the steps above, if you need further assistance with any specific calculations or steps, feel free to ask. Understanding these methods will help you solve a wide range of differential equations involving non-homogeneous terms.