Second differential equations

In summary, the conversation discusses the general solution to a linear, non-homogeneous, differential equation with constant coefficients. The standard method for solving such equations involves finding the solutions to the corresponding homogeneous equation and then adding a particular solution to the equation. The conversation also mentions the use of exponential solutions and provides a step-by-step example of solving the given equation. The conversation concludes by mentioning that this topic is covered in basic texts on differential equations.
  • #1
radwa
5
0
will you please help
v" + ë^2v = [-WL/(2EI)]x + [W/(2EI)]x^2
where ë^2= PE/I
P, E, I, W are constants;
how will the general solution of this equation be
v(x)=Acosëx + Bsinëx + (W/2P)*((x^2)-Lx-2/ë^2)
 
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  • #2
This is a linear, non-homogeneous, differential equation with constant coefficients. A standard way of solving these is to first look at the corresponding homogeneous equation. In this case that is
v" + ë2v = 0. Such equations typically have exponential (or related)solutions. If you try v= ekx here, v'= k ekx and v"= k2ekx so the equation becomes (k2e+ ë2)ekx= 0. Since an exponential is never 0, we must have k2+ ë2= 0. The solutions to that equation are k= ëi and -ëi. Two independent solutions to the differential equation are eëix and e-ëix but it is simpler to use eëix= cos(ëx)+ i sin(ëx).
The general solution to the homogeneous equation is
v(x)= C1cos(ëx)+ C2 sin(ëx).

Now "look for" a solution to the entire equation- Since the "right hand side", [-WL/(2EI)]x + [W/(2EI)]x2, is a quadratic, try a solution of the form v(x)= Ax2+ Bx+ C. v'= 2Ax+ B and v"= 2A. The equation becomes 2A+ ë2(Ax2+ Bx+ C)= ë2Ax2+ ë2Bx+ 2A+ ë2C= [-WL/(2EI)]x + [W/(2EI)]x2. Equating coefficients of like powers, ë2A= W/(2EI), ë2B= -WL/(2EI), and 2A+ ë2C= 0.

After solving for A, B, C, add that to the solution to the homogeneous equation: v(x)= C1cos(ëx)+ C2 sin(ëx)+ Ax2+ Bx+ C, the solution you give.

Solving linear differential equations with constant coefficients is covered in any basic text on differential equations.
 
  • #3


Sure, I would be happy to help you with this second-order differential equation. To solve this equation, we can use the method of undetermined coefficients. This involves finding a particular solution and then adding it to the general solution of the homogeneous equation.

First, let's find the particular solution. Since the right-hand side of the equation is a polynomial, we can assume a particular solution of the form v(x) = Cx^2 + Dx + E. Plugging this into the equation, we get:

v" + ë^2v = [-WL/(2EI)]x + [W/(2EI)]x^2
2C + ë^2(Cx^2 + Dx + E) = [-WL/(2EI)]x + [W/(2EI)]x^2
(2C + ë^2Cx^2) + (ë^2Dx) + (2C + ë^2E) = [-WL/(2EI)]x + [W/(2EI)]x^2

Comparing coefficients, we get:
2C + ë^2C = [W/(2EI)]
ë^2D = [-WL/(2EI)]
2C + ë^2E = 0

Solving for C, D, and E, we get:
C = W/(2ë^2EI)
D = -WL/(2ë^4EI)
E = -W/(4ë^2EI)

Therefore, the particular solution is v(x) = (W/2ë^2EI)x^2 - (WL/2ë^4EI)x - (W/4ë^2EI).

Next, we need to find the general solution of the homogeneous equation, which is the solution when the right-hand side of the equation is equal to 0. This can be done by solving the characteristic equation:

r^2 + ë^2 = 0
r = ±ië

Therefore, the general solution of the homogeneous equation is v(x) = Acos(ëx) + Bsin(ëx), where A and B are constants.

Finally, the general solution of the original equation is the sum of the particular solution and the general solution of the homogeneous equation:

v(x) = Acos(ëx) + Bsin(ëx) + (W/2ë^2EI)x^2 - (WL/2
 

1. What is a second differential equation?

A second differential equation is a mathematical expression that relates the second derivative of a function to the function itself. It is commonly used in mathematical modeling and can describe a wide range of physical phenomena.

2. How is a second differential equation different from a first differential equation?

A first differential equation involves the first derivative of a function, while a second differential equation involves the second derivative. This means that a second differential equation is more complex and usually requires more information to solve.

3. What is the general form of a second differential equation?

The general form of a second differential equation is y'' = f(x,y,y'), where y' represents the first derivative of y and y'' represents the second derivative of y. The function f(x,y,y') can be any algebraic, trigonometric, or exponential function.

4. How are second differential equations used in real-life applications?

Second differential equations are used in many areas of science and engineering, such as physics, chemistry, biology, and economics. They can be used to model the motion of objects, the growth of populations, the spread of diseases, and the behavior of electrical circuits, among other things.

5. What techniques are used to solve second differential equations?

There are several techniques for solving second differential equations, including separation of variables, variation of parameters, undetermined coefficients, and Laplace transforms. The appropriate technique to use depends on the specific form of the differential equation and the initial conditions given.

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