Math 432. Differential Equations Fourth Exam – Solutions Spring 2011

Math 432. Differential Equations
Fourth Exam – Solutions
1. Find the general solutions to the homogeneous equations
(a)
(b)
(c)
These are all Cauchy-Euler equations.
(a) The characteristic equation is r(r – 1) + r – 4 = 0.
r2 – 4 = 0
r = ±2
y1 = t 2 and y2 = t –2
The general solution is
y = c1t 2 + c2t –2.
(b) The differential equation in standard form: 9t 2y'' + 15t y' + y = 0.
The characteristic equation: 9r(r – 1) + 15r + 1 = 0.
9r 2 – 9r + 15r + 1 = 0
9r 2 + 6r + 1 = 0
(3r + 1)2 = 0
Repeated root: r = –1/3.
The general solution is
(c) The differential in standard form: t 2y'' – t y' + 5y = 0.
The characteristic equation is r(r – 1) – r + 5 = 0.
r 2 – 2r + 5 = 0
The general solution is
Spring 2011
2. Given that y1 = x + 1 is one solution to the differential equation (x2 + 2x – 1) y'' – 2(x + 1) y' + 2y = 0, find
the general solution to this differential equation.
You need to use reduction of order to do this one:
To get started, you have to have the differential equation in standard form.
Dividing through by (x2 + 2x – 1) gives
So we have
.
So the general solution is
3. Use the definition of the Laplace transform to find
for
4. Use the method of Laplace transforms to solve the initial-value problem y'' + 5y' – y = et – 1; y(0) = 1,
y'(0) = 1. {I changed this in class to just finding Y(s).}
5. Solve the initial-value problem t y'' – t y' + y = 2; y(0) = 2, y'(0) = –1.
This has to be done using Laplace transforms.
This is linear. In preferred form we have . . .
Taking the inverse Laplace transform:
Using the initial condition to evaluate C:
y' = C
y'(0) = C = –1.
So the solution is
.
y = 2 – t.
6. Solve the initial-value problem t 2 y'' + t y' = t 2; y(1) = 1, y'(1) = –1/2.
This is a Cauchy-Euler equation.
Solve the homogeneous equation first.
r(r – 1) + r + 0 = 0
r2 = 0
Repeated root: r = 0. So y1 = t 0 = 1 and y2 = ln t.
Then use variation of parameters to find yp.
First you need the standard form of the equation:
So g(t) = 1.
.
Parts: Let u = ln t and dv = t dt. Then du = 1/t and v = t2/2. Then
So
So far, we know the general solution:
Using the initial conditions to evaluate c1 and c2:
So the solution to this initial-value problem is