1M2 Solution Sheet 1: Second Order ODEs
1.1. Solve the following initial value problems
2x
− 5 + 3 ex
(i) 5y′′ − 3y′ − 2y = 0, y(0) = −1, y′ (0) = 1, y(x) = − 10
7e
7
(ii) y′′ − 6y′ + 10y = 0, y(0) = 2, y′ (0) = 0, y(x) = 2e3x [cos x − 3 sin x]
(iii) y′′ − 4y′ + 3y = 0, y(0) = 0, y′ (0) = 1, y(x) = − 12 ex + 12 e3x
1.2. Find the general solutions of the following ODEs:
h
i
√
√
x
(i) 4y′′ − 2y′ + 7y = 0, y(x) = e 4 A cos( 3 43x ) + B sin( 3 43x )
√
√
(ii) y′′ + 6y′ − 4y = 0, y(x) = Ae(−3+ 13)x + Be(−3− 13)x
h
i
√
√
x
3x
3x
−
′′
′
2
(iii) 3y + 3y + 3y = 0, y(x) = e
A cos( 2 ) + B sin( 2 )
1.3. Consider the ODE
y′′ + 5πy′ + γy = 0
with a free parameter γ. What value does γ have to take such that the solution y oscillates
with wavelength 2?
The characteristic equation of the ODE is λ2 + 5πλ + γ = 0 leading to the solutions
r
5
25 2
λ1,2 = − π ±
π − γ.
2
4
2
In order for y(x) to oscillate we need to have 25
4 π − γ < 0. In that case the values λ1,2 can
be represented as λ1,2 = α ± iβ for some real numbers α and β and y(x) has the form
y(x) = eαx (A cos βx + B sin βx).
This is a damped or growing oscillatory motion with wavelength (or period in terms of
time dependent problems) 2π
β . We want the wavelength to be 2. This leads to the condition
β = π giving
r
25 2
π − γ = iπ.
4
2
The solution of this equation is γ = 29
4 π . The general solution of the ODE with this value
is
5πx
y(x) = e− 2 [A cos πx + B sin πx] .
This is an exponentially decaying oscillation with wavelength 2.
1.4. Find the general solutions of the following ODEs
h
√
√ i
3
1
(i) y′′ − 3y′ + 4y = cos 4x − sin 4x, y(x) = − 12
cos 4x + e 2 x A cos 27x + B cos 27x
(ii) 9y′′ − 12y′ + 4y = e−3x , y(x) =
(iii)
2x
1 −3x
+ Ae 3
121 e
2y′′ +4y′ −7y = 7 cos 2x, y(x) = − 105
289
2x
+ Bxe 3
−1− √3 x
−1+ √3 x
56
2
2
cos 2x+ 289 sin 2x+Ae
+Be
(iv)
2y′′+4y′ −7y
= 7 cos 2x+e−x , y(x) = − 19 e−x − 105
289
−1+ √3 x
−1− √3 x
56
2
cos 2x+ 289 sin 2x+Ae
+
2
Be
(Hint: Just use the previous result plus the principle of superposition to
obtain the solution of this ODE)
1.5. Consider the forced harmonic oscillator
y′′ + 2y′ + 2y = sin 2x
with given initial derivative y′ (0) = 0. What is the value y(0) if we know that y(π/2) = 1?
The general solution of this ODE is
1
1
y(x) = − cos 2x − sin 2x + e−x [A cos x + B sin x] .
5
10
The two conditions are
1
y′ (0) = − − A + B = 0
5
π
1
y(π/2) = + Be− 2 = 1
5
π
π
This leads to B = 54 e 2 and A = − 15 + 45 e 2 .
The solution of the ODE is therefore
1
1
1 4 π
4 π
−x
2
2
y(x) = − cos 2x − sin 2x + e
(− + e ) cos x + e sin x ,
5
10
5 5
5
π
from which it immediately follows that y(0) = − 52 + 45 e 2 .