2.7 Exercises
These are homework exercises to accompany Libl's "Differential Equations for Engineering " Textmap. This is a textbook targeted for a one semester first course on differential equations, aimed at engineering students. Prerequisite for the course is the basic calculus sequence.
Your Turn
Prove the superposition principle for nonhomogeneous equations. Suppose that y 1 y_1 L y 1 = f ( x ) Ly_1 = f(x) y 2 y_2 L y 2 = g ( x ) Ly_2 = g(x) L L y = y 1 + y 2 y = y_1 + y_2 L y = f ( x ) + g ( x ) Ly = f(x) + g(x)
Your Turn
For the equation x 2 y ″ − x y ′ = 0 x^2y'' - xy' = 0 y = x ′ y = x'
Equations of the form a x 2 y ″ + b x y ′ + c y = 0 ax^2y'' + bxy' + cy = 0 Euler’s equations or Cauchy-Euler equations . They are solved by trying y = x r y = x^r r r x ⩾ 0 x \geqslant 0
Your Turn
Suppose that ( b − a ) 2 − 4 a c > 0 {(b - a)}^2 - 4ac > 0
Find a formula for the general solution of a x 2 y ″ + b x y ′ + c y = 0 ax^2y'' + bxy' + cy = 0 y = x r y = x^r r r What happens when ( b − a ) 2 − 4 a c = 0 {(b - a)}^2 -4ac = 0 ( b − a ) 2 − 4 a c < 0 {(b - a)}^2 - 4ac < 0 We will revisit the case when ( b − a ) 2 − 4 a c < 0 (b-a)^{2}-4ac<0
Your Turn
Same equation as in Exercise 2.1.5. Suppose ( b − a ) 2 − 4 a c = 0 {(b - a)}^2 - 4ac = 0 a x 2 y ″ + b x y ′ + c y = 0 ax^2y'' + bxy' + cy = 0 y = x r ln x y = x^r \ln x
Your Turn
reduction of order
Suppose y 1 y_1 y ″ + p ( x ) y ′ + q ( x ) y = 0 y'' + p(x)y' + q(x)y = 0
y
2
(
x
)
=
y
1
(
x
)
∫
e
−
∫
p
(
x
)
d
x
(
y
1
(
x
)
)
2
d
x
y_2 (x) = y_1 (x) \int \frac { e^{- \int p(x)dx}}{{(y_1(x))}^2} dx
is also a solution.
Note: If you wish to come up with the formula for reduction of order yourself, start by trying y 2 ( x ) = y 1 ( x ) v ( x ) y_2 (x) = y_1(x)v(x) y 2 y_2 y 1 y_1 w = v ′ w = v' w w w w v v w w
Your Turn
Chebyshev’s equation of order 1
Take ( 1 − x 2 ) y ″ − x y ′ + y = 0 (1 - x^2)y'' - xy' + y = 0
Show that y = x y = x Use reduction of order to find a second linearly independent solution. Write down the general solution. Your Turn
Hermite’s equation of order 2
Take y ″ − 2 x y ′ + 4 y = 0 y'' - 2xy' + 4y = 0
Show that y = 1 − 2 x 2 y = 1 - 2x^2 Use reduction of order to find a second linearly independent solution. Write down the general solution. Your Turn
Are sin ( x ) \sin (x) e x e^x
Answer
Yes. To justify try to find a constant A A sin ( x ) = A e x \sin (x)=Ae^{x} x x
Your Turn
Are e x e^x e x + 2 e^{x+2}
Answer
No. e x + 2 = e 2 e x e^{x+2}=e^{2}e^{x}
Your Turn
Find the general solution to x y ″ + y ′ = 0 xy'' + y' = 0 y ′ y'
Answer
y
=
C
1
ln
(
x
)
+
C
2
y=C_{1}\ln (x)+C_{2}
Your Turn
Write down an equation (guess) for which we have the solutions e x e^{x} e 2 x e^{2x} y ″ + A y ′ + B y = 0 y''+Ay'+By=0 A A B B e x e^{x} e 2 x e^{2x} A A B B
Answer
y
″
−
3
y
′
+
2
y
=
0
y''-3y'+2y=0
Your Turn
Solve y ″ − 8 y ′ + 16 y = 0 y'' - 8y' + 16y = 0 y ( 0 ) = 2 , y ′ ( 0 ) = 0 y(0) = 2, y'(0) = 0
Your Turn
Solve y ″ + 9 y ′ = 0 y'' + 9y' = 0 y ( 0 ) = 1 , y ′ ( 0 ) = 1 y(0) = 1, y'(0) = 1
Your Turn
The method of this section applies to equations of other orders than two. We will see higher orders later. Try to solve the first order equation 2 y ′ + 3 y = 0 2y' + 3y = 0
Your Turn
Let us revisit Euler’s equations of Exercise 1. Suppose now that ( b − a ) 2 − 4 a c < 0 {(b - a)}^2 - 4ac < 0 a x 2 y ″ + b x y ′ + c y = 0 ax^2y'' + bxy' + cy = 0 x r = e r ln x x^r = e^{r \ln x}
Your Turn
Find the solution to y ″ − ( 2 α ) y ′ + α 2 y = 0 y''-(2\alpha )y'+\alpha^{2}y=0 y ( 0 ) = a y(0)=a y ′ ( 0 ) = b y'(0)=b α \alpha a a b b
Your Turn
Construct an equation such that y = C 1 e − 2 x cos ( 3 x ) + C 2 e − 2 x sin ( 3 x ) y=C_{1}e^{-2x}\cos (3x)+C_{2}e^{-2x}\sin (3x)
Your Turn
Find the general solution to y ″ + 4 y ′ + 2 y = 0 y'' + 4y' + 2y = 0
Answer
y
=
C
1
e
(
−
2
+
2
)
x
+
C
2
e
(
−
2
−
2
)
x
y=C_{1}e^{(-2+\sqrt{2})x}+C_{2}e^{(-2-\sqrt{2})x}
Your Turn
Find the general solution to y ″ − 6 y ′ + 9 y = 0 y'' - 6y' + 9y = 0
Answer
y
=
C
1
e
3
x
+
C
2
x
e
3
x
y=C_{1}e^{3x}+C_{2}xe^{3x}
Your Turn
Find the solution to 2 y ″ + y ′ + y = 0 , y ( 0 ) = 1 , y ′ ( 0 ) = − 2 2y'' + y' + y = 0, y(0) = 1, y'(0) = -2
Answer
y
=
e
−
x
/
4
cos
(
(
7
4
)
x
)
−
7
e
−
x
/
4
sin
(
(
7
4
)
x
)
y=e^{-x/4}\cos\left((\frac{\sqrt{7}}{4})x\right)-\sqrt{7}e^{-x/4}\sin\left((\frac{\sqrt{7}}{4})x\right)
Your Turn
Find the solution to 2 y ″ + y ′ − 3 y = 0 , y ( 0 ) = a , y ′ ( 0 ) = b 2y'' + y' - 3y = 0, y(0) = a, y'(0) = b
Answer
y
=
2
(
a
−
b
)
5
e
−
3
x
/
2
+
3
a
+
2
b
5
e
x
y=\frac{2(a-b)}{5}e^{-3x/2}+\frac{3a+2b}{5}e^{x}
Your Turn
Find the solution to z ″ ( t ) = − 2 z ′ ( t ) − 2 z ( t ) , z ( 0 ) = 2 , z ′ ( 0 ) = − 2 z''(t) = -2z'(t) - 2z (t), z(0) = 2, z'(0) = -2
Answer
z
(
t
)
=
2
e
−
t
cos
(
t
)
z(t)=2e^{-t}\cos (t)
Your Turn
Find the solution to y ″ − ( α + β ) y ′ + α β y = 0 y''-(\alpha +\beta )y'+\alpha\beta y=0 y ( 0 ) = a y(0)=a y ′ ( 0 ) = b y'(0)=b α \alpha β \beta a a b b α ≠ β \alpha\neq\beta
Answer
y
=
α
β
−
b
β
−
α
e
α
x
+
b
−
a
α
β
−
α
e
β
x
y=\frac{\alpha\beta -b}{\beta -\alpha}e^{\alpha x}+\frac{b-a\alpha}{\beta -\alpha}e^{\beta x}
Your Turn
Construct an equation such that y = C 1 e 3 x + C 2 e − 2 x y=C_{1}e^{3x}+C_{2}e^{-2x}
Answer
y
″
−
y
′
−
6
y
=
0
y''-y'-6y=0
Your Turn
Suppose the characteristic equation for a differential equation is ( r − 1 ) 2 ( r − 2 ) 2 = 0 {(r - 1)}^2{(r -2)}^2 = 0
Find such a differential equation. Find its general solution. Your Turn
Let f ( x ) = e x − cos x , g ( x ) = e x + cos x f(x) = e^x - \cos x, g(x) = e^x + \cos x h ( x ) = cos x h(x) = \cos x f ( x ) , g ( x ) , f(x), g(x), h ( x ) h(x)
Your Turn
Let f ( x ) = 0 , g ( x ) = cos x f(x) = 0, g(x) = \cos x h ( x ) = sin x h(x) = \sin x f ( x ) , g ( x ) , f(x), g(x), h ( x ) h(x)
Your Turn
Are x , x 2 , a n d x 4 x, x^2, and x^4
Your Turn
Are e x , x e x , a n d x 2 e x e^x, xe^x, and x^2e^x
Your Turn
Find the general solution of y ( 5 ) − y ( 4 ) = 0 y^{(5)} - y^{(4)} = 0
Answer
y
=
C
1
e
x
+
C
2
x
3
+
C
3
x
2
+
C
4
x
+
C
5
y=C_{1}e^{x}+C_{2}x^{3}+C_{3}x^{2}+C_{4}x+C_{5}
Your Turn
Suppose that the characteristic equation of a third order differential equation has roots 3 ± 2 i 3 \pm 2i
What is the characteristic equation? Find the corresponding differential equation. Find the general solution. Answer
r
3
−
3
r
2
+
4
r
−
12
=
0
r^{3}-3r^{2}+4r-12=0
y
‴
−
3
y
″
+
4
y
′
−
12
y
=
0
y'''-3y''+4y'-12y=0
y
=
C
1
e
3
x
+
C
2
sin
(
2
x
)
+
C
3
cos
(
2
x
)
y=C_{1}e^{3x}+C_{2}\sin (2x)+C_{3}\cos (2x)
Your Turn
Solve 1001 y ‴ + 3.2 y ″ + π y ′ − 4 y = 0 , y ( 0 ) = 0 , y ′ ( 0 ) = 0 , y ″ ( 0 ) = 0 1001y''' + 3.2y'' + \pi y' - \sqrt {4} y = 0, y(0) = 0, y' (0) = 0, y'' (0) = 0
Answer
y
=
0
y=0
Your Turn
Are e x , e x + 1 , e 2 x , sin ( x ) e^x, e^{x+1}, e^{2x}, \sin (x)
Answer
No. e 1 e x − e x + 1 = 0 e^{1}e^{x}-e^{x+1}=0
Your Turn
Are sin ( x ) , x , x sin ( x ) \sin (x), x, x \sin (x)
Answer
Yes. (Hint: First note that sin ( x ) \sin (x) x x x sin ( x ) x\sin (x)
Your Turn
Find an equation such that y = cos ( x ) y=\cos (x) y = sin ( x ) y=\sin(x) y = e x y=e^{x}
Answer
y
‴
−
y
″
+
y
′
−
y
=
0
y'''-y''+y'-y=0
Your Turn
Consider a mass and spring system with a mass m = 2 m = 2 k = 3 k = 3 c = 1 c = 1
Set up and find the general solution of the system. Is the system underdamped, overdamped or critically damped? If the system is not critically damped, find a c c Your Turn
Do Exercise 2.4.1 for m = 3 , k = 12 , a n d c = 12 m = 3, k = 12, and c = 12
Your Turn
Using the mks units (meters-kilograms-seconds), suppose you have a spring with spring constant 4 N m 4 \dfrac {N}{m}
You count and find that the frequency is 0.8 Hz 0.8\text{ Hz} Find a formula for the mass m m w w Hz \text{Hz} Your Turn
Suppose we add possible friction to Exercise 2.4.3. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg 1\text{ kg} 2 kg 2\text{ kg} 1 kg 1\text{ kg} 1.1 Hz 1.1\text{ Hz} 2 kg 2\text{ kg} 0.8 Hz 0.8\text{ Hz}
Find k k c c Find a formula for the mass in terms of the frequency in Hz \text{Hz} For an unknown object you measured 0.2 Hz 0.2\text{ Hz} Your Turn
Suppose you wish to measure the friction a mass of 0.1 kg 0.1\text{ kg} c c k = 5 N m k = 5 \dfrac {N}{m} 1 Hz 1\text{ Hz}
Your Turn
A mass of 2 2 k k t = 0 t = 0 2 2 k k 3 3
Answer
k = 8 9 k=\frac{8}{9}
Your Turn
Suppose we have an RLC circuit with a resistor of 100 100 0.1 0.1 50 50 0.05 0.05 5 5
Set up the ODE equation for the current I I Find the general solution. Solve for I ( 0 ) = 10 I (0) = 10 I ′ ( 0 ) = 0 I' (0) = 0 Answer
0.05
I
″
+
0.1
I
′
+
(
1
5
)
I
=
0
0.05I''+0.1I'+\left(\frac{1}{5}\right)I=0
I
=
C
e
−
t
cos
(
3
t
−
γ
)
I=Ce^{-t}\cos (\sqrt{3}t-\gamma )
I
=
10
e
−
t
cos
(
3
t
)
+
10
3
e
−
t
sin
(
3
t
)
I=10e^{-t}\cos (\sqrt{3} t)+\frac{10}{\sqrt{3}}e^{-t}\sin (\sqrt{3}t)
Your Turn
A 5000 kg 5000\text{ kg} 1 m s 1 \dfrac {m}{s} 0.1 m 0.1\text{ m}
Find k k Find out how far does the spring compress when a 10000 kg 10000\text{ kg} If the spring would break if it compresses further than 0.3 m 0.3\text{ m} 1 m s 1 \dfrac {m}{s} What is the maximum mass of a railcar that can hit the spring without breaking at 2 m s 2 \dfrac {m}{s} Answer
k
=
500000
k=500000
1
5
2
≈
0.141
\frac{1}{5\sqrt{2}}\approx 0.141
45000
kg
45000\text{ kg}
11250
kg
11250\text{ kg}
Your Turn
A mass of m kg m\text{ kg} k = 3 N m k=3\frac{\text{N}}{\text{m}} c = 2 Ns m c=2\frac{\text{Ns}}{\text{m}} m 0 m_{0} m < m 0 m<m_{0}
Answer
m 0 = 1 2 m_{0}=\frac{1}{2} m < m 0 m<m_{0}
Your Turn
Solve the initial value problem y ″ + 9 y = cos ( 3 x ) + sin ( 3 x ) y'' + 9y = \cos (3x) + \sin (3x) y ( 0 ) = 2 , y ′ ( 0 ) = 1 y(0) = 2, y'(0) = 1
Your Turn
Setup the form of the particular solution but do not solve for the coefficients for y ( 4 ) − 2 y ‴ + y ″ = e x y^{(4)} - 2y''' + y'' = e^x
Your Turn
Setup the form of the particular solution but do not solve for the coefficients for y ( 4 ) − 2 y ‴ + y ″ = e x + x + sin x y^{(4)} - 2y''' + y'' = e^x + x + \sin x
Your Turn
Using variation of parameters find a particular solution of y ″ − 2 y ′ + y = e x y'' - 2y' + y = e^x Find a particular solution using undetermined coefficients. Are the two solutions you found the same? What is going on? See also Exercise 2.5.9 Your Turn
Find a particular solution of y ″ − 2 y ′ + y = sin ( x 2 ) y'' - 2y' + y = \sin (x^2)
Your Turn
For an arbitrary constant c c y ″ − y = e c x y'' - y = e^{cx} c c
Your Turn
Using variation of parameters find a particular solution of y ″ − y = e x y''-y=e^{x} Find a particular solution using undetermined coefficients. Are the two solutions you found the same? What is going on? Your Turn
Find a polynomial P ( x ) P(x) y = 2 x 2 + 3 x + 4 y=2x^{2}+3x+4 y ″ + 5 y ′ + y = P ( x ) y''+5y'+y=P(x)
Your Turn
Find a particular solution to y ″ − y ′ + y = 2 sin ( 3 x ) y'' - y' + y = 2 \sin (3x)
Answer
y
=
−
16
sin
(
3
x
)
+
6
cos
(
3
x
)
73
y=\frac{-16\sin (3x)+6\cos (3x)}{73}
Your Turn
Find a particular solution to y ″ + 2 y = e x + x 3 y'' + 2y = e^x + x^3 Find the general solution. Answer
y
=
2
e
x
+
3
x
3
−
9
x
6
y=\frac{2e^{x}+3x^{3}-9x}{6}
y
=
C
1
cos
(
2
x
)
+
C
2
sin
(
2
x
)
+
2
e
x
+
3
x
3
−
9
x
6
y=C_{1}\cos (\sqrt{2}x)+C_{2}\sin (\sqrt{2}x)+\frac{2e^{x}+3x^{3}-9x}{6}
Your Turn
Solve y ″ + 2 y ′ + y = x 2 , y ( 0 ) = 1 , y ′ ( 0 ) = 2 y'' + 2y' + y = x^2, y(0) = 1, y'(0) = 2
Answer
y
(
x
)
=
x
2
−
4
x
+
6
+
e
−
x
(
x
−
5
)
y(x)=x^{2}-4x+6+e^{-x}(x-5)
Your Turn
Use variation of parameters to find a particular solution of y ″ − y = 1 e x + e − x y'' - y = \dfrac {1}{e^x + e^{-x}}
Answer
y
=
2
x
e
x
−
(
e
x
+
e
−
x
)
log
(
e
2
x
+
1
)
4
y=\frac{2xe^{x}-(e^{x}+e^{-x})\log (e^{2x}+1)}{4}
Your Turn
For an arbitrary constant c c y ″ − 2 y = sin ( x + c ) y'' - 2y = \sin (x + c)
Answer
y
=
−
sin
(
x
+
c
)
3
+
C
1
e
2
x
+
C
2
e
−
2
x
y=\frac{-\sin (x+c)}{3}+C_{1}e^{\sqrt{2}x}+C_{2}e^{-\sqrt{2}x}
Your Turn
Undetermined coefficients can sometimes be used to guess a particular solution to other equations than constant coefficients. Find a polynomial y ( x ) y(x) y ′ + x y = x 3 + 2 x 2 + 5 x + 2 y'+xy=x^{3}+2x^{2}+5x+2
Note: Not every right hand side will allow a polynomial solution, for example, y ′ + x y = 1 y'+xy=1 Chapter 7 .
Answer
y
=
x
2
+
2
x
+
3
y=x^{2}+2x+3
Your Turn
Derive a formula for x s p x_{sp} m x ″ + c x ′ + k x = F 0 sin ( ω t ) mx'' + cx' + kx = F_0 \sin ( \omega t) c > 0 c > 0
Your Turn
Derive a formula for x s p x_{sp} m x ″ + c x ′ + k x = F 0 cos ( ω t ) + F 1 cos ( 3 ω t ) mx'' + cx' + kx = F_0 \cos (\omega t) + F_1 \cos (3 \omega t) c > 0 c > 0
Your Turn
Take m x ″ + c x ′ + k x = F 0 cos ( ω t ) mx'' + cx' + kx = F_0 \cos ( \omega t) m > 0 m > 0 k > 0 k > 0 C ( ω ) C(\omega ) c c m , k , m, k, F 0 F_0 c c C ( ω ) C(\omega ) ω > 0 \omega > 0
Your Turn
Take m x ″ + c x ′ + k x = F 0 cos ( ω t ) mx'' + cx' + kx = F_0 \cos (\omega t) c > 0 c > 0 k > 0 k > 0 C ( ω ) C(\omega ) m m c , k , c, k, F 0 F_0 m m C ( ω ) C(\omega ) ω > 0 \omega > 0
Your Turn
Suppose a water tower in an earthquake acts as a mass-spring system. Assume that the container on top is full and the water does not move around. The container then acts as a mass and the support acts as the spring, where the induced vibrations are horizontal. Suppose that the container with water has a mass of m = 10,000 k g m = 10, 000 kg 1000 1000 1 1
Suppose that an earthquake induces an external force F ( t ) = m A ω 2 cos ( ω t ) . F(t) = mA \omega^2 \cos (\omega t).
What is the natural frequency of the water tower? If ω \omega Suppose A = 1 A = 1 0.5 0.5 1.5 1.5 Your Turn
A mass of 4 kg 4\text{ kg} k = 4 k = 4 c = 1 c = 1 F 0 = 2 F_0 = 2 F 0 cos ( ω t ) F_0 \cos (\omega t) ω \omega
Answer
ω
=
31
4
2
≈
0.984
C
(
ω
)
=
16
3
7
≈
2.016
\omega=\frac{\sqrt{31}}{4\sqrt{2}}\approx 0.984\quad C(\omega )=\frac{16}{3\sqrt{7}}\approx 2.016
Your Turn
Derive a formula for x s p x_{sp} m x ″ + c x ′ + k x = F 0 cos ( ω t ) + A mx'' + cx' + kx = F_0 \cos (\omega t) + A A A c > 0 c > 0
Answer
x s p = ( ω 0 2 − ω 2 ) F 0 m ( 2 ω p ) 2 + m ( ω 0 2 − ω 2 ) 2 cos ( ω t ) + 2 ω p F 0 m ( 2 ω p ) 2 + m ( ω 0 2 − ω 2 ) 2 sin ( ω t ) + A k x_{sp}=\frac{(\omega_{0}^{2}-\omega^{2})F_{0}}{m(2\omega p)^{2}+m(\omega_{0}^{2}-\omega^{2})^{2}}\cos (\omega t)+\frac{2\omega pF_{0}}{m(2\omega p)^{2}+m(\omega_{0}^{2}-\omega^{2})^{2}}\sin (\omega t)+\frac{A}{k} p = c 2 m p=\frac{c}{2m} ω 0 = k m \omega_{0}=\sqrt{\frac{k}{m}}
Your Turn
Suppose there is no damping in a mass and spring system with m = 5 , k = 20 , m = 5, k = 20, F 0 = 5 F_0 = 5 ω \omega
Find ω \omega Find the amplitude of the oscillations at time t = 100 t = 100 Answer
ω
=
2
\omega =2
25
25
Adapted from Differential Equations for Engineers by Jiří Lebl (https://www.jirka.org/diffyqs/), © Jiří Lebl, licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0 .
