2nd Order Homogeneous, Real Roots, Initial Value

Destroxia
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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.

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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...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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