Linear 2nd order diff. eqn.

In summary, the problem involves solving a non-homogeneous differential equation with given boundary conditions. The solution consists of a general solution to the homogenous equation and a particular solution to the non-homogeneous equation. The particular solution can be determined by plugging in the given function for F(t) and solving for u(t). The general solution can be found by solving the homogenous equation and using the boundary conditions to determine the constants. The solution provided in the textbook is consistent with the general solution and satisfies the given conditions.
  • #1
musemonkey
25
0
1. This is exercise 3.19.15 from Boyce & DiPrima's Differential Equations.

[tex] u'' + u = F(t), u(0) = 0, u'(0) = 0 [/tex]

[tex] 0 \leq t < \pi , F(t) = F_0t [/tex]
[tex] \pi \leq t \leq 2\pi , F(t) = F_0(2\pi-t) [/tex]
[tex] t > 2\pi , F(t) = 0 [/tex]

Solve for [tex] u(t) [/tex].



2. The solution to the homogenous equation is

[tex] u'' + u = 0 [/tex]

is [tex] u(t) = c_1 cos(t) + c_2 sin(t) [/tex].



3. Plugging in the boundary conditions yields

[tex] u(0) = c_1 + 0 = 0 [/tex]
[tex] u'(t) = - c_1 sin(t) + c_2 cos(t) [/tex]
[tex] u'(0) = c_2 = 0 [/tex].

Then a particular solution of the form [tex] u_p = A t [/tex] works

[tex] u'' + u = At = F_0t [/tex],

and so the solution for [tex] 0 \leq t < \pi [/tex] I find to be [tex] u_1(t) = F_0 t [/tex]. But the solution in the back of the textbook states [tex] u_1(t) = F_0t - F_0

sin(t) [/tex], which clearly also satisfies the differential equation and the initial value conditions, but it's inconsistent with what I get. Where did I go wrong?
 
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  • #2
You went wrong in not finding the second solution. The general solution of the non-homogeneous equation is given byu(t) = c_1 cos(t) + c_2 sin(t) + F_0 t - F_0 sin(t), where c_1 and c_2 are constants determined by the initial conditions. In this case, c_1 = 0 and c_2 = 0 so the solution is indeed u_1(t) = F_0t - F_0 sin(t).
 

1. What is a 2nd order linear differential equation?

A 2nd order linear differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It contains terms with the function, its first derivative, and its second derivative, and the coefficients of these terms are constants.

2. What is the general form of a 2nd order linear differential equation?

The general form of a 2nd order linear differential equation is y'' + p(x)y' + q(x)y = r(x), where p(x), q(x), and r(x) are functions of x and y' and y'' represent the first and second derivatives of the unknown function y.

3. How do you solve a 2nd order linear differential equation?

To solve a 2nd order linear differential equation, you can use the method of undetermined coefficients or variation of parameters. These methods involve finding a particular solution and the general solution using initial conditions or boundary conditions.

4. What are some real-life applications of 2nd order linear differential equations?

2nd order linear differential equations are commonly used in physics and engineering to model systems that involve motion, such as a swinging pendulum or a spring-mass system. They are also used in population growth models and in finance to calculate interest rates.

5. How do you verify a solution to a 2nd order linear differential equation?

To verify a solution to a 2nd order linear differential equation, you can substitute the solution into the original equation and its initial conditions or boundary conditions. If the equation holds true, then the solution is valid.

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