Second order ODE initial value problem

In summary, the conversation was about solving the initial value problem y" - y' - 6y = e^-x + 12x, y(0)=1,y'(0)=-2. The general solution was found to be Ae^3x + Be^-2x, and by substituting y=ae^-x + bx + c, the values of a, b, and c were determined to be -1/4, -2, and -1/6 respectively. However, there was some discrepancy in the calculation of the coefficient for b, with one member suggesting it should be -(b+6c). Further clarification was requested to determine the error in the calculation.
  • #1
mkay123321
16
0
So the question is y" - y' - 6y = e^-x + 12x, y(0)=1,y'(0)=-2

First I found the general solution which came out to be, Ae^3x + Be^-2x

I then Substituted y=ae^-x + bx + c
y'=-ae^-x + b
y"=ae^-x
Then I just compared the coefficients to get a=-1/4, B=-2 and C=-1/6

So I am getting y = Ae^3x + Be^-2x -(e^-x)/4 - 2x - 1/6
Im not sure if this is right, I have done the rest but I get some funny answers for A and B so I was wondering if someone could verify if this answer is right. Thanks
 
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  • #2
Your C value isn't correct; try plugging everything back into the equation and see if you get a different C.
 
  • #3
Bohrok said:
Your C value isn't correct; try plugging everything back into the equation and see if you get a different C.

I just tried it again. Got the same answer.
 
  • #4
Could you show what you did?
 
  • #5
Bohrok said:
Could you show what you did?

I did ae^-x - (-ae^-x+b) - 6(ae^-x+bx+c) = e^-x + 12x
Then I got e^-x(-4a-1) - b(1+6c) = x(12+6b)

After that I did -4a = 1 so a = -1/4
6c=-1 so c = -1/6
6b=-12 so b = -2

Could you tell me where I am going wrong?
 
  • #6
mkay123321 said:
I did ae^-x - (-ae^-x+b) - 6(ae^-x+bx+c) = e^-x + 12x
Then I got e^-x(-4a-1) - b(1+6c) = x(12+6b)

the red one. It is -(b+6c)

ehild
 

Related to Second order ODE initial value problem

1. What is a second order ODE initial value problem?

A second order ODE initial value problem is a type of differential equation that involves a second derivative of a function and requires two initial conditions to be solved. The initial conditions specify the values of the function and its first derivative at a given point.

2. How do you solve a second order ODE initial value problem?

To solve a second order ODE initial value problem, you can use various methods such as separation of variables, integrating factors, or the method of undetermined coefficients. These methods involve manipulating the equation to isolate the second derivative and then integrating or solving for the function using the initial conditions.

3. What are some real-world applications of second order ODE initial value problems?

Second order ODE initial value problems have many applications in physics, engineering, and other fields. They can be used to model the motion of objects in a gravitational field, describe the behavior of electrical circuits, and analyze the growth and decay of populations, among other things.

4. Can you provide an example of a second order ODE initial value problem?

One example of a second order ODE initial value problem is the harmonic oscillator equation, which describes the motion of a mass attached to a spring. It can be written as mx'' + kx = 0, where m is the mass, x is the displacement from equilibrium, and k is the spring constant. The initial conditions would specify the initial position and velocity of the mass.

5. What is the difference between a second order ODE initial value problem and a boundary value problem?

A second order ODE initial value problem requires two initial conditions to be solved, while a boundary value problem requires two boundary conditions at different points. In other words, a second order ODE initial value problem specifies the behavior of the function at a single point, while a boundary value problem specifies the behavior at two different points. Additionally, boundary value problems often have multiple solutions, while initial value problems typically have a unique solution.

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