How Does the General Solution of a Second-Order Differential Equation Look?

[radwa]

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)

 

[HallsofIvy]

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= e^kx here, v'= k e^kx and v"= k²e^kx so the equation becomes (k²e+ ë²)e^kx= 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 e^ëix and e^-ëix but it is simpler to use e^ëix= 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.

 

[M98Ranger]

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