2nd Order Homogeneous, Real Roots, Initial Value

In summary, the student attempted to solve a simple differential equation but was having trouble because of an error in how they evaluated the solution.
  • #1
Destroxia
204
7

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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  • #2
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.
 
  • #3
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...
 

Related to 2nd Order Homogeneous, Real Roots, Initial Value

1. What is a 2nd order homogeneous equation?

A 2nd order homogeneous equation is a type of differential equation in which all the terms contain the dependent variable and its derivatives, with no other independent variables. This means that the equation can be written in the form of y'' + p(x)y' + q(x)y = 0, where p(x) and q(x) are functions of x.

2. What are real roots in the context of 2nd order homogeneous equations?

In the context of 2nd order homogeneous equations, real roots refer to the solutions of the equation that are real numbers. This means that when solving for the dependent variable, y, the solution will not involve any imaginary numbers.

3. How do you find the initial value for a 2nd order homogeneous equation?

The initial value for a 2nd order homogeneous equation is typically given in the form of the value of the dependent variable, y, at a specific value of the independent variable, x. This means that when solving the equation, you will use this initial value to find the specific solution for y at that given value of x.

4. What is the general solution for a 2nd order homogeneous equation with real roots?

The general solution for a 2nd order homogeneous equation with real roots is y = c1e^(r1x) + c2e^(r2x), where c1 and c2 are constants and r1 and r2 are the real roots of the equation. This solution includes all possible solutions for y that satisfy the equation.

5. How do you solve a 2nd order homogeneous equation with real roots using initial values?

To solve a 2nd order homogeneous equation with real roots using initial values, you will first find the general solution using the equation y = c1e^(r1x) + c2e^(r2x). Then, you will use the given initial values to set up a system of equations to solve for the constants c1 and c2. Once the values of c1 and c2 are determined, you can plug them back into the general solution to find the specific solution for y.

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