Second Order Differential Equations

You can check it by plugging the solution into the differential equation and verifying that it satisfies both the equation and the initial conditions.
  • #1
JamesEllison
13
0

Homework Statement


Solve the ODE with the boundary conditions given

Q''+Q = Sin(2x) where Q(0) = 1 and Q'(0) = 2
So i know i need to solve the general and particular solutions, however, I am a little confused. Any help or advice would be great, Thanks in advance.

Homework Equations


Y = Yg +Yp


The Attempt at a Solution


Assume RHS = 0 for general solution
Q'' + Q = 0

Let Q = e^kx
Q' = k e^kx
Q'' = k^2 e^kx
therefore
k^2e^kx + e^kx = 0
e^kx(k^2) = 0
k=0??
Yg = A e ^kx +Bx e^kx?

Particular Solution
Yp = A Sin 2x + B Cos 2x
Yp' = 2A Cos 2x - 2B Sin 2x
Yp"= -4A Sin 2x - 4B Cos (2x)


-4A Sin 2x - 4B Cos (2x) + A Sin 2x + B Cos 2x = Sin(2x)
-3A Sin 2x = Sin 2x
a = -1/3
 
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  • #2
You've got a mistake. You made the claim that
$$
k^2e^{kx}+e^{kx}=0\\
\Rightarrow k^2e^{kx}=0
$$
That's not true, is it? What's ##k## really?
 
  • #3
JamesEllison said:

Homework Statement


Solve the ODE with the boundary conditions given

Q''+Q = Sin(2x) where Q(0) = 1 and Q'(0) = 2
So i know i need to solve the general and particular solutions, however, I am a little confused. Any help or advice would be great, Thanks in advance.

Homework Equations


Y = Yg +Yp


The Attempt at a Solution


Assume RHS = 0 for general solution
Q'' + Q = 0

Let Q = e^kx
Q' = k e^kx
Q'' = k^2 e^kx
therefore
k^2e^kx + e^kx = 0
e^kx(k^2) = 0
No, factoring out e^{kx} gives e^{kx}(k^2+ 1)= 0.

k=0??
Yg = A e ^kx +Bx e^kx?

Particular Solution
Yp = A Sin 2x + B Cos 2x
Yp' = 2A Cos 2x - 2B Sin 2x
Yp"= -4A Sin 2x - 4B Cos (2x)


-4A Sin 2x - 4B Cos (2x) + A Sin 2x + B Cos 2x = Sin(2x)
-3A Sin 2x = Sin 2x
a = -1/3
 
  • #4
I see. Thanks.
 
  • #5
e^kt(k^2+1) = 0
k = i ---> sqrt(-1)

So then the general assumption would be.

Yg = e^-kt(ASin2x + iBsin2x) ?
 
  • #6
Or just
Yg = A Sin2x + i BCos 2x
?
Sorry about lack of real looking equations I am on my iPhone
 
  • #7
$$
k=\pm i
$$
So the general form of the solution is
$$
Q_g(x)=c_1 e^{ix}+c_2 e^{-ix}
$$
Or equivalently,
$$
Q_g(x)=A\sin(x)+B\cos(x)
$$
Now take a guess at the particular solution. Let's guess that ##Q_p(x)=D\sin(2x)##. Plug this into the differential equation and solve for D. It will give you the particular solution.
 
  • #8
Excellent.
So from there:

Yp = C Sin(2x)
Yp' = 2C Cos(2x)
Yp" = -4C Sin(2x)

-4C Sin(2x) + C Sin(2x) = Sin(2x)
-3C = 1
C = -1/3

Yp = -1/3 Sin (2x)
So then the solution starts as

Y = Yg + Yp
Y = A Sin (x) + B Cos(x) -1/3 Sin(2x)

Sub in initial Conditions Q(0) = 1 or x=0
Y = A Sin (x) + B Cos(x) -1/3 Sin(2x)
1 = 0 + B - 0
B = 1

Sub in next conditions
Q'(0) = 2
Y = A Sin (x) + B Cos(x) -1/3 Sin(2x)
Y' = A Cos(x) - B Sin(x) - 2/3 Cos (2x)
Y'(0) = A - 2/3 = 2
A = 8/3

So the final solution for the equation would be

Y = 8/3 Sin(x) + Cos(x) - 1/3 Sin(2x) ?

:D
 
  • #9
Looks good to me.
 

FAQ: Second Order Differential Equations

1. What is a second order differential equation?

A second order differential equation is a mathematical equation that involves a second derivative, or the rate of change of the rate of change, of a function. It is typically written in the form of y'' = f(x, y, y'), where y'' represents the second derivative, x is the independent variable, y is the dependent variable, and y' is the first derivative.

2. What are some real-life applications of second order differential equations?

Second order differential equations are used to model many physical systems, such as the motion of a pendulum, the growth of a population, and the flow of electricity in a circuit. They are also commonly used in engineering and physics to describe the behavior of systems that involve acceleration or oscillation.

3. How do you solve a second order differential equation?

The method for solving a second order differential equation depends on the specific equation and its initial conditions. Some common techniques include separation of variables, substitution, and using special functions such as Laplace transforms. In some cases, numerical methods may also be used to approximate a solution.

4. What is the difference between a homogeneous and non-homogeneous second order differential equation?

A homogeneous second order differential equation has a right-hand side of 0, meaning that the function f(x, y, y') is equal to 0. This makes the equation easier to solve, as it can be reduced to a first order equation. A non-homogeneous equation has a non-zero right-hand side, requiring additional techniques such as variation of parameters or undetermined coefficients to find a solution.

5. Can all second order differential equations be solved analytically?

No, not all second order differential equations have analytical solutions. In some cases, the equation may be too complex to be solved algebraically, or it may not have a closed-form solution. In these cases, numerical methods may be used to approximate a solution.

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