2nd Order Homogeneous, Real Roots, Initial Value

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Destroxia
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Homework Statement



Solve the initial value problem

Handwritten fragment showing the characters "0," followed by a partial letter y, likely part of an initial condition such as y(0) = ...


Homework Equations



Quadratic Formula

The Attempt at a Solution



Handwritten solution to 2y'' + y' - 4y = 0 with discriminant Δ = 1 + 32 = 33, roots r1 = (-1 + √33)/4 and r2 = (-1 - √33)/4, and general solution yH = c1 e^(r1 t) + c2 e^(r2 t).


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

Fragment of a printed initial condition reading "0," followed by a partially visible "y" at the right edge, cropped from a larger expression.
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



Handwritten solution of the second-order ODE 2y'' + y' - 4y = 0: discriminant Δ = 1 + 32 = 33, roots r1 and r2 given with square root of 33, then y1 and y2 expressions with exponentials and c1, c2 terms, and a general solution yg.


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...