Second Order Differential Equations

In summary, the conversation discusses finding a particular solution to the second order differential equation y'' - 4y' + 4y = f(x), where f(x) = 25cos(x). The suggested approach is to let y = Asin(x) + Bcos(x) and substitute it back into the original formula to find values for A and B. The conversation notes that the correct answer is y(p) = 3cos(x) - 4sin(x) and suggests gathering the sinx terms and factoring them, then doing the same for the cosx terms. The final step is to set the coefficient for sinx to zero and solve for A in terms of B, then set the coefficient for cosx
  • #1
Illusionist
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Homework Statement


Consider the second order differential equation y'' - 4y' + 4y = f(x)
Find a particular solution if f(x) = 25cos(x)

Homework Equations


I believe for this type of question I should let y = Asin(x) + B cos(x)
Hence y' = Acos(x) - Bsin(x) and
y'' = -Asin(x) - Bsin(x)

The Attempt at a Solution


I think everything above is right, and for the rest of the question I should just be able to substitute back into the original formula and find values for A and B.
The problem is when I do substitute back in all I get is a mess and a headache!

Here is what I got by substitution:
-Asin(x) - Bcos(x) - 4[Acos(x) - Bsin(x)] + 4[Asin(x) + Bcos(x)] = 25cos(x)
Hence
-Asin(x) + 4[Asin(x) + Bsin(x)] - Bcos(x) - 4[Acos(x) - Bcos(x)] = 25cos(x)

That's about all I think I can do and I have no idea how to obtain values for A and B. I know the answer is y(p) = 3cos(x) - 4sin(x).
How to get this answer from the above I have no idea, I'm obviously missing something.
Thanks in advance for any advice or help.
 
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  • #2
You are just about there. Gather your sinx terms then factor the sin, same for the cosx terms, now, since sin x does not appear in the particular solution set its coefficient to zero, solve for A in terms of B. Set the coefficient for the cos x term = 25 and find B.
 

FAQ: Second Order Differential Equations

What is a second order differential equation?

A second order differential equation is a mathematical equation that involves a function, its derivatives, and independent variables. It is represented in the form of y'' + p(x)y' + q(x)y = r(x), where y is the function, x is the independent variable, p(x) and q(x) are known functions, and r(x) is the source term.

What is the difference between a first and second order differential equation?

The main difference between first and second order differential equations is the highest derivative present in the equation. In a first order differential equation, the highest derivative is y', while in a second order differential equation, the highest derivative is y''. This means that a second order differential equation involves two levels of change, whereas a first order differential equation involves only one.

What are some real-life applications of second order differential equations?

Second order differential equations have many applications in physics, engineering, and other sciences. They are used to model systems with acceleration, such as a mass-spring system or an RLC circuit. They can also be used to model oscillatory motion, such as the motion of a pendulum or a vibrating string. Additionally, they are used in the study of fluid mechanics, heat transfer, and population dynamics.

How do you solve a second order differential equation?

The process of solving a second order differential equation involves finding a function that satisfies the equation. This can be done analytically, using techniques such as separation of variables, substitution, or the method of undetermined coefficients. It can also be solved numerically using computer software. However, not all second order differential equations have closed-form solutions, and in these cases, numerical methods are often used.

What is the significance of the initial conditions in solving a second order differential equation?

The initial conditions, also known as boundary conditions, are necessary in solving a second order differential equation. They are values given for the function and its derivatives at a particular point, which help to determine the unique solution to the equation. Without these conditions, the solution to the equation may have multiple possible forms, making it impossible to find a single solution.

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