Differential Equations – Section 5.2 – Homework: 1, 3, 5, 9, 13, 17
First we will solve the boundary-value problem
eigenfunctions.

y   y  0,
y (0)  0,
y ( L)  0 , using eigenvalues and
is called a parameter.
Case 1:
  0.
The general form of the solution is
Case 2:
  0.
The general form of the solution is y  c1 cosh
  x  c 2 sinh
Case 3:
  0.
The general form of the solution is y  c1 cos
 x  c 2 sin  x .
y  c1 x  c2 . What happens?
  x . What happens?
What happens?
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We will now consider structures that are constructed using beams and find the deflection of the beams. The deflection
curve approximates the shape of the beam.
In the following formula E = Young’s modulus of elasticity of the material of the beam and
I = the moment of inertia of a cross-section of the beam.
The product EI is called the flexural rigidity of the beam.
The deflection y(x) satisfies the first-order differential equation
EI
d4y
 w( x) .
dx 4
Boundary conditions depend on how the ends of the beam are supported.
Ends of the Beam
Embedded
Free
Simply Supported
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Boundary Conditions
y  0, y   0
y   0, y   0
y  0, y   0
Consider a long slender vertical column of uniform cross-section and length L. Let y(x) denote the deflection of the
column when a constant vertical compressive force, or load, P is applied to its top. Then
d2y
EI 2  Py  0 .
dx
 nx 
y ( L)  0 . As we saw earlier, y n ( x)  c 2 sin 
,
 L 
2 2
2
corresponding to the eigenvalues  n  Pn / EI  n  / L , n  1,2,3,... . Physically this means that the column will
If we let
  P / EI
we have y   y  0,
y (0)  0,
buckle or deflect only when the compressive force is one of the values Pn  n
2
 2 EI / L2 .
called critical loads. The deflection curve corresponding to the smallest critical load
load, is
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y1 ( x)  c2 sin x / L and is known as the first buckling mode.
These different forces are
P1   2 EI / L2 , called the Euler