2nd Order Homogeneous, Real Roots, Initial Value

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SUMMARY

The discussion centers on solving the initial value problem for the second-order homogeneous differential equation 2y'' + y' - 4y = 0 with initial conditions y(0) = 0 and y'(0) = 1. The main issue raised was the inability to solve for constants due to apparent cancellations when substituting the initial conditions. A key insight provided was that evaluating e^0 = 1 leads to the equation c_1 + c_2 = 0, indicating that the constants do not cancel out as initially thought. This clarification allows for the proper determination of the constants involved in the solution.

PREREQUISITES
  • Understanding of second-order homogeneous differential equations
  • Familiarity with the Quadratic Formula
  • Knowledge of initial value problems in differential equations
  • Basic algebra skills for solving equations
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  • Study the method of characteristic equations for solving differential equations
  • Learn about the application of initial conditions in differential equations
  • Explore the concept of homogeneous vs. non-homogeneous differential equations
  • Practice solving various initial value problems to reinforce understanding
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Homework Statement



Solve the initial value problem

question.jpg


Homework Equations



Quadratic Formula

The Attempt at a Solution



work.jpg


My problem is that I don't understand how to solve the constants now, I understand, 2 equations, 2 unknowns, but when I plug the y(0) = 0 into the YsubH equation, everything cancels out, I can't solve for anything. I don't see anyway to cancel anything in the equations either to make it not equivalent to 0, or to be able to solve it.

[/B]
 
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RyanTAsher said:

Homework Statement



Solve the initial value problem

question.jpg
For simple differential equations such as this, it's better to type the equation and initial condition than to post an image. What I see is a thumbnail that shows "0, y". I have to click the image to see the full image.

Your equation is
2y'' + y' - 4y = 0, y(0) = 0, y'(0) = 1
RyanTAsher said:

Homework Equations



Quadratic Formula

The Attempt at a Solution



work.jpg


My problem is that I don't understand how to solve the constants now, I understand, 2 equations, 2 unknowns, but when I plug the y(0) = 0 into the YsubH equation, everything cancels out, I can't solve for anything. I don't see anyway to cancel anything in the equations either to make it not equivalent to 0, or to be able to solve it.
[/B]
Your solution looks fine to me. Your error is in how you evaluated it. Note that e0 = 1, so you get ##c_1 + c_2 = 0## for your first equation, so everything does not cancel out.
 
Mark44 said:
For simple differential equations such as this, it's better to type the equation and initial condition than to post an image. What I see is a thumbnail that shows "0, y". I have to click the image to see the full image.

Your equation is
2y'' + y' - 4y = 0, y(0) = 0, y'(0) = 1

Your solution looks fine to me. Your error is in how you evaluated it. Note that e0 = 1, so you get ##c_1 + c_2 = 0## for your first equation, so everything does not cancel out.

Okay, I'll remember that, the next time that I post... and thank you so much for catching my error, I feel terrible that I made such an algebra mistake...
 

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