Brushing up on the basics of diff eq

In summary: C_1 . In summary, the goal is to solve for C1 and C2 so that the equation y(x) = C1sin(2x) + C2cos(2x) satisfies the initial conditions y(∏/8) = 0 and y'(∏/8) = √2. To do this, you can multiply the equation C1(1/2)(√2) + C2(1/2)(√2) + 1 = 0 by √2 and then use the resulting equation and the initial condition y(∏/8) = 0 to solve for C1. Once
  • #1
cowmoo32
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Homework Statement


y(x) = C1sin(2x) + C2cos(2x)

Homework Equations


y(∏/8) = 0


The Attempt at a Solution


C1(1/2)(√2) + C2(1/2)(√2) + 1 = 0


The book jumps to the equation below and I'm having trouble figuring out how they got there.
C1 + C2 = -√2
 
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  • #2
cowmoo32 said:

Homework Statement


y(x) = C1sin(2x) + C2cos(2x)

Homework Equations


y(∏/8) = 0


The Attempt at a Solution


C1(1/2)(√2) + C2(1/2)(√2) + 1 = 0


The book jumps to the equation below and I'm having trouble figuring out how they got there.
C1 + C2 = -√2
What is the dif. eq. you're trying to solve?

Multiply your "attempt at a solution" by √2 .
 
  • #3
Determine C1 and C2 so that y(x) = C1sin(2x) + C2cos(2x) will satisfy the initial conditions y(∏/8) = 0 and y'(∏/8) = √2
 
  • #4
Multiply your "attempt at a solution" in your original post by √2 .
 
  • #5
I guess it's safe to assume that C1/2 = C1 and C2/2 = C2 considering they're both unknowns at this point?
 
  • #6
cowmoo32 said:
I guess it's safe to assume that C1/2 = C1 and C2/2 = C2 considering they're both unknowns at this point?
That doesn't make sense, unless you are asking if you can replace C1/2 by C1 and C2/2 by C2 .

If [itex]C_1+C_2=-\sqrt{2}\,,[/itex] then you can replace [itex]C_2[/itex] with [itex]-C_1-\sqrt{2}[/itex]
 

FAQ: Brushing up on the basics of diff eq

What is the purpose of learning differential equations?

Differential equations are used to describe and model real-world phenomena in various fields such as physics, engineering, economics, and biology. They are essential for understanding the behavior of systems and making predictions, making them a fundamental tool for scientists and engineers.

What are the basic concepts of differential equations?

The basic concepts of differential equations include the order of the equation, types of equations (ordinary vs partial), initial and boundary conditions, and methods of solving (analytical vs numerical). Additionally, understanding terms such as derivatives, integrals, and variables is crucial for working with differential equations.

How do I solve a differential equation?

There are various methods for solving differential equations, including separation of variables, integrating factors, and series solutions. The specific method used depends on the type of equation and its order. It is also important to check the solutions for consistency and to verify that they satisfy the initial or boundary conditions.

Can differential equations be solved analytically or numerically?

Yes, differential equations can be solved both analytically and numerically. Analytical solutions involve finding a formula or expression that represents the solution to the equation, while numerical solutions use algorithms and computer programs to approximate the solution. Both methods have their advantages and are used depending on the complexity of the equation and the desired level of accuracy.

What are some applications of differential equations?

Differential equations have numerous applications in various fields, including physics (mechanics, electromagnetism), engineering (control systems, fluid mechanics), economics (supply and demand models), and biology (population growth, chemical reactions). They are also used in everyday life, such as predicting weather patterns or modeling the spread of diseases.

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