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Solutions 2.3-Page 131
Problem 1
Find the general solutions of the differential equations.
y ′′ − 4 y = 0
The first step is to find the roots of the characteristic equation.
r2 − 4 = 0
(r + 2)(r − 2) = 0
r = −2,2
Since the roots are real and distinct, the general solution is
y ( x) = c1e −2 x + c 2 e 2 x
Problem 9
Find the general solutions of the differential equations.
y ′′ + 8 y ′ + 25 y = 0
The characteristic equation is r 2 + 8r + 25 = 0 . The roots can be found using the
quadratic equation formula or a calculator.
The roots are r = −4 ± 3i .
The general solution is based on Theorem 3 on page 128.
y ( x) = e −4 x (c1 cos 3x + c 2 sin 3 x)
Problem 21
Solve the initial value problems.
y′′ − 4 y′ + 3 y = 0;
y (0) = 7, y′(0) = 11
The characteristic equation is r 2 − 4r + 3 = 0 . The roots can be found using the quadratic
equation formula or a calculator.
The roots are r = 3,1 .
The general solution is y = c1e 3 x + c 2 e x . The initial conditions are used to find the
constants.
y ′ = 3c1e 3 x + c 2 e x . Substituting the initial conditions gives:
y (0) = 7 = c1 + c 2
y ′(0) = 11 = 3c1 + c 2
Solving the system of equations gives c1 = 2, c 2 = 5
y ( x) = 2e 3 x + 5e x
Problem 32
Find the general solutions of the equations. First find a small integral root of the
characteristic equation by inspection; then factor by division.
y ( 4 ) + y ( 3) − 3 y ′′ − 5 y ′ − 2 y = 0
The characteristic equation is r 4 + r 3 − 3r 2 − 5r − 2 = 0 . The root obtained from
inspection is r = 2. Long division of the characteristic equation by (r-2) yields:
(r + 1) 3 = 0
With the root r=-1 repeated three times, the general solution is
y ( x) = c1e 2 x + (c 2 + c3 x + c 4 x 2 )e − x