Second Order ODEs Practice

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1.1Classify a Second-Order ODE

Exam I | Problem 1.1 |

Classify the differential equation below. State whether it is second-order, linear, homogeneous, and constant-coefficient.

$$ y'' + 3y' - 4y = 0 $$

Solution

The highest derivative is \(y''\), so it is a second-order ODE.

The equation is linear because \(y\), \(y'\), and \(y''\) each appear to the first power and are not multiplied together.

It is homogeneous because the right-hand side is \(0\).

It has constant coefficients because \(1\), \(3\), and \(-4\) are constants.

1.2Write in Normalized Form

Exam I | Problem 1.2 |

Rewrite the equation in normalized form \(y'' + p(x)y' + q(x)y = r(x)\).

$$ 4y'' - 2y' + 7y = 5e^x $$

Solution

Divide every term by \(4\):

y'' - \frac{1}{2}y' + \frac{7}{4}y = \frac{5}{4}e^x

So the normalized form is

y'' - \frac{1}{2}y' + \frac{7}{4}y = \frac{5}{4}e^x.

1.3Check Existence and Uniqueness

Exam I | Problem 1.3 |

Does the existence and uniqueness theorem guarantee a unique local solution near \(x=0\) for the initial value problem below?

y'' = y - y' + \frac{1}{1+x^2}, \qquad y(0)=1, \qquad y'(0)=0

Solution

Write the equation as

y'' = f(x,y,y') = y - y' + \frac{1}{1+x^2}.

The function \(f\) is continuous everywhere, and its partial derivatives with respect to \(y\) and \(y'\) are also continuous everywhere.

Therefore the theorem guarantees a unique local solution near \(x=0\).

1.4Solve a Distinct-Root Homogeneous Equation

Exam I | Problem 1.4 |

Solve the homogeneous equation.

$$ y'' - 5y' + 6y = 0 $$

Solution

Try \(y=e^{rx}\). The characteristic equation is

$$ r^2 - 5r + 6 = 0 $$

Factor:

$$ (r-2)(r-3)=0 $$

So the roots are \(r=2\) and \(r=3\). The general solution is

y = C_1 e^{2x} + C_2 e^{3x}.

1.5Solve a Repeated-Root Equation

Exam I | Problem 1.5 |

Solve the homogeneous equation.

$$ y'' - 4y' + 4y = 0 $$

Solution

Try \(y=e^{rx}\). The characteristic equation is

$$ r^2 - 4r + 4 = 0 = (r-2)^2 $$

This is a repeated root \(r=2\), so

y = (C_1 + C_2 x)e^{2x}.

1.6Solve a Complex-Root Equation

Exam I | Problem 1.6 |

Solve the homogeneous equation.

$$ y'' + 9y = 0 $$

Solution

Try \(y=e^{rx}\). The characteristic equation is

$$ r^2 + 9 = 0 $$

r = \pm 3i.

The general real solution is

y = C_1 \cos(3x) + C_2 \sin(3x).

1.7Choose a Resonant Trial

Exam I | Problem 1.7 |

For the equation below, what trial form should you use for a particular solution in undetermined coefficients?

$$ y'' - 2y' + y = e^x $$

Solution

The homogeneous characteristic equation is

$$ (r-1)^2 = 0, $$

so \(e^x\) and \(xe^x\) already appear in the homogeneous solution.

To avoid duplication, multiply by \(x\) again. A correct trial is

$$ y_p = Ax^2 e^x. $$

1.8Solve an Euler-Cauchy Equation

Exam I | Problem 1.8 |

Solve the equation for \(x>0\).

$$ x^2 y'' + xy' - 4y = 0 $$

Solution

Use the Euler-Cauchy trial \(y=x^m\). Then

y' = mx^{m-1}, \qquad y'' = m(m-1)x^{m-2}.

Substitute:

x^2 m(m-1)x^{m-2} + x(mx^{m-1}) - 4x^m = 0

which simplifies to

$$ m(m-1) + m - 4 = 0. $$

$$ m^2 - 4 = 0, $$

and \(m=\pm 2\). Therefore

y = C_1 x^2 + C_2 x^{-2}.

1.9Compute a Wronskian

Exam I | Problem 1.9 |

Compute the Wronskian of the two functions below.

y_1=x, \qquad y_2=x^2

Solution

Use

W(y_1,y_2)= \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1y_2' - y_1'y_2.

Here,

y_1'=1, \qquad y_2'=2x.

W = x(2x) - 1\cdot x^2 = x^2.

On any interval that does not include \(0\), the functions are linearly independent.

1.10Find the Natural Frequency and Damping Ratio

Exam I | Problem 1.10 |

A mass-spring system has

m=1, \qquad c=4, \qquad k=4.

Find the natural frequency \(\omega_n\), the damping ratio \(\zeta\), and the damping regime.

Solution

Compute the natural frequency:

\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{4}{1}} = 2.

Compute the damping ratio:

\zeta = \frac{c}{2\sqrt{mk}} = \frac{4}{2\sqrt{1\cdot 4}} = \frac{4}{4} = 1.

Since \(\zeta=1\), the system is critically damped.

Difficulty

Integrated

2.1Solve an Initial Value Problem

Exam II | Problem 2.1 |

Solve the initial value problem.

y'' - 3y' + 2y = 0, \qquad y(0)=1, \qquad y'(0)=4

Solution

The characteristic equation is

$$ r^2 - 3r + 2 = 0 = (r-1)(r-2). $$

y = C_1 e^x + C_2 e^{2x}.

Use \(y(0)=1\):

$$ C_1 + C_2 = 1. $$

Differentiate:

y' = C_1 e^x + 2C_2 e^{2x}.

Use \(y'(0)=4\):

$$ C_1 + 2C_2 = 4. $$

Subtract the first equation from the second:

$$ C_2 = 3. $$

Then \(C_1 = -2\). So

y = -2e^x + 3e^{2x}.

2.2Solve a Resonant Forced Equation

Exam II | Problem 2.2 |

Solve the initial value problem.

y'' - 2y' + y = e^x, \qquad y(0)=0, \qquad y'(0)=1

Solution

The homogeneous equation has characteristic equation

$$ (r-1)^2=0, $$

$$ y_h = (C_1 + C_2x)e^x. $$

Because the forcing is \(e^x\), use a resonant trial

$$ y_p = Ax^2e^x. $$

Substituting gives \(A=\tfrac12\), so

y = (C_1 + C_2x)e^x + \frac12 x^2 e^x.

Now apply the initial conditions. From \(y(0)=0\),

$$ C_1 = 0. $$

Write

y = e^x\left(C_1 + C_2x + \frac12 x^2\right).

Then

y' = e^x\left(C_1 + C_2x + \frac12 x^2 + C_2 + x\right).

Using \(y'(0)=1\) and \(C_1=0\),

$$ C_2 = 1. $$

So the solution is

y = e^x\left(x + \frac12 x^2\right).

2.3Solve an Euler-Cauchy Initial Value Problem

Exam II | Problem 2.3 |

Solve the initial value problem for \(x>0\).

x^2y'' + xy' - 4y = 0, \qquad y(1)=3, \qquad y'(1)=-1

Solution

Use the Euler-Cauchy trial \(y=x^m\). The auxiliary equation is

$$ m^2 - 4 = 0, $$

so \(m=\pm 2\). Thus

y = C_1x^2 + C_2x^{-2}.

Apply \(y(1)=3\):

$$ C_1 + C_2 = 3. $$

Differentiate:

y' = 2C_1x - 2C_2x^{-3}.

Apply \(y'(1)=-1\):

$$ 2C_1 - 2C_2 = -1. $$

Solve the system to get

C_1 = \frac{5}{4}, \qquad C_2 = \frac{7}{4}.

Therefore

y = \frac{5}{4}x^2 + \frac{7}{4}x^{-2}.

2.4Use a Known Solution to Finish the General Solution

Exam II | Problem 2.4 |

Given that \(y_1=x\) is one nonzero solution of the equation below, find a second linearly independent solution and the general solution.

$$ x^2y'' - 2xy' + 2y = 0 $$

Solution

This is an Euler-Cauchy equation. Using the known solution \(y_1=x\), reduction of order gives a second independent solution \(y_2=x^2\).

So the general solution is

$$ y = C_1x + C_2x^2. $$

2.5Analyze a Boundary Value Problem

Exam II | Problem 2.5 |

Determine whether the boundary value problem has one solution, no solutions, or infinitely many solutions.

y'' + \pi^2 y = 0, \qquad y(0)=0, \qquad y(1)=0

Solution

The characteristic equation is

r^2 + \pi^2 = 0,

so the general solution is

y = C_1\cos(\pi x) + C_2\sin(\pi x).

Use \(y(0)=0\):

$$ C_1=0. $$

Then

y = C_2\sin(\pi x).

Now use \(y(1)=0\):

C_2\sin(\pi)=0,

which is true for every \(C_2\).

So the boundary value problem has infinitely many solutions:

y = C\sin(\pi x).

2.6Solve a Critically Damped Motion Problem

Exam II | Problem 2.6 |

A mass-spring system is modeled by

y'' + 8y' + 16y = 0, \qquad y(0)=2, \qquad y'(0)=0

Solve for \(y(t)\) and classify the damping.

Solution

The characteristic equation is

$$ r^2 + 8r + 16 = 0 = (r+4)^2. $$

This is a repeated root, so

y = (C_1 + C_2t)e^{-4t}.

Use \(y(0)=2\):

$$ C_1 = 2. $$

Differentiate:

y' = e^{-4t}\bigl(C_2 - 4C_1 - 4C_2t\bigr).

Use \(y'(0)=0\):

$$ C_2 - 4(2)=0, $$

$$ C_2 = 8. $$

Thus

y = (2 + 8t)e^{-4t}.

Because the root is repeated, the motion is critically damped.

2.7Find the Steady-State Response

Exam II | Problem 2.7 |

Find a particular solution for the forced vibration equation.

y'' + 9y = 6\sin(2t)

Solution

Since the forcing is \(\sin(2t)\) and there is no resonance, try

y_p = A\cos(2t) + B\sin(2t).

Then

y_p'' = -4A\cos(2t) - 4B\sin(2t).

Substitute into the equation:

y_p'' + 9y_p = 5A\cos(2t) + 5B\sin(2t).

Match coefficients with \(6\sin(2t)\):

5A = 0, \qquad 5B = 6.

So \(A=0\) and \(B=\frac{6}{5}\). A steady-state response is

y_p = \frac{6}{5}\sin(2t).

2.8Find the Interval of Guaranteed Uniqueness

Exam II | Problem 2.8 |

For the initial value problem below, on what largest open interval containing \(x=2\) is a unique local solution guaranteed?

y'' + \frac{1}{x-1}y' + (\ln x)y = e^x, \qquad y(2)=0, \qquad y'(2)=1

Solution

In normalized form, the coefficients are

p(x)=\frac{1}{x-1}, \qquad q(x)=\ln x, \qquad r(x)=e^x.

The function \(p(x)\) is undefined at \(x=1\), and \(q(x)=\ln x\) requires \(x>0\).

So the coefficients are continuous on

(1,\infty).

That is the largest open interval containing \(x=2\) on which a unique local solution is guaranteed.

Difficulty

Applied

3.1Model a Simple Mass-Spring Motion

Final | Problem 3.1 |

A \(2\)-kg mass is attached to a spring with spring constant \(18\) N/m. The mass is displaced \(1/2\) meter from equilibrium and released from rest.

Find the equation of motion and the first time it returns to equilibrium.

Solution

For an undamped spring, the equation is

$$ m y'' + ky = 0. $$

Here,

$$ 2y'' + 18y = 0, $$

$$ y'' + 9y = 0. $$

The general solution is

y = C_1\cos(3t) + C_2\sin(3t).

Use \(y(0)=1/2\):

C_1 = \frac12.

Released from rest means \(y'(0)=0\). Since

y' = -3C_1\sin(3t) + 3C_2\cos(3t),

we get \(C_2=0\).

y(t)=\frac12\cos(3t).

The first return to equilibrium happens when \(y(t)=0\), so

\cos(3t)=0 \quad \Rightarrow \quad 3t=\frac{\pi}{2}.

Thus the first return time is

t=\frac{\pi}{6}.

3.2Solve a Critically Damped IVP

Final | Problem 3.2 |

Solve the initial value problem.

y'' + 8y' + 16y = 0, \qquad y(0)=2, \qquad y'(0)=0

Solution

The characteristic equation is

$$ r^2 + 8r + 16 = 0 = (r+4)^2. $$

y = (C_1 + C_2t)e^{-4t}.

Apply the initial conditions:

y(0)=2 \Rightarrow C_1=2.

Differentiate:

y' = e^{-4t}\bigl(C_2 - 4C_1 - 4C_2t\bigr).

Then

y'(0)=0 \Rightarrow C_2 - 8 = 0,

so \(C_2=8\).

Therefore

y = (2+8t)e^{-4t}.

3.3Find a Particular Solution for Forced Oscillation

Final | Problem 3.3 |

Find a particular solution for the forced vibration equation below.

y'' + 9y = 6\sin(2t)

Solution

Try

y_p = A\cos(2t) + B\sin(2t).

Then

y_p'' + 9y_p = 5A\cos(2t) + 5B\sin(2t).

Match coefficients with \(6\sin(2t)\):

A=0, \qquad B=\frac65.

So a particular solution is

y_p = \frac65\sin(2t).

3.4A Boundary Value Problem with a Unique Solution

Final | Problem 3.4 |

Solve the boundary value problem.

y'' + \pi^2 y = 0, \qquad y(0)=0, \qquad y\left(\frac12\right)=0

Solution

The general solution is

y = C_1\cos(\pi x) + C_2\sin(\pi x).

Use \(y(0)=0\):

$$ C_1=0. $$

y = C_2\sin(\pi x).

Now apply \(y\left(\frac12\right)=0\):

C_2\sin\left(\frac{\pi}{2}\right)=C_2=0.

Thus the only solution is

$$ y=0. $$

3.5Predict Long-Term Behavior from Roots

Final | Problem 3.5 |

For the equation below, describe the long-term behavior of the general solution as \(x\to\infty\).

$$ y'' - y' - 2y = 0 $$

Solution

The characteristic equation is

$$ r^2 - r - 2 = 0 = (r-2)(r+1). $$

y = C_1e^{2x} + C_2e^{-x}.

The \(e^{2x}\) term grows as \(x\to\infty\), while the \(e^{-x}\) term decays.

So a generic solution grows without bound, and the equilibrium is unstable unless \(C_1=0\).

Difficulty

Challenge

4.1Solve a Resonant Forced Problem

Final | Problem 4.1 |

Solve the initial value problem.

y'' + 4y = 8\cos(2x), \qquad y(0)=0, \qquad y'(0)=0

Solution

The homogeneous equation has characteristic roots \(\pm 2i\), so

y_h = C_1\cos(2x) + C_2\sin(2x).

Because the forcing is \(\cos(2x)\), which resonates with the homogeneous solution, try

y_p = x\bigl(A\sin(2x) + B\cos(2x)\bigr).

For this choice,

y_p'' + 4y_p = 4A\cos(2x) - 4B\sin(2x).

Match coefficients with \(8\cos(2x)\):

A=2, \qquad B=0.

y = C_1\cos(2x) + C_2\sin(2x) + 2x\sin(2x).

Apply the initial conditions:

y(0)=0 \Rightarrow C_1=0,

and

y'(0)=0 \Rightarrow C_2=0.

Therefore

y = 2x\sin(2x).

4.2Solve a Damped Forced Oscillator

Final | Problem 4.2 |

Solve the initial value problem.

y'' + 2y' + 5y = 10e^{-x}, \qquad y(0)=1, \qquad y'(0)=0

Solution

The characteristic equation is

$$ r^2 + 2r + 5 = 0, $$

r = -1 \pm 2i.

Thus

y_h = e^{-x}\bigl(C_1\cos(2x) + C_2\sin(2x)\bigr).

For the forcing term \(10e^{-x}\), try

y_p = Ae^{-x}.

Substitute:

y_p'' + 2y_p' + 5y_p = (1 - 2 + 5)Ae^{-x} = 4Ae^{-x}.

Set this equal to \(10e^{-x}\):

A = \frac52.

y = e^{-x}\bigl(C_1\cos(2x) + C_2\sin(2x)\bigr) + \frac52 e^{-x}.

Use \(y(0)=1\):

C_1 + \frac52 = 1 \Rightarrow C_1 = -\frac32.

Differentiate in the form \(y=e^{-x}u(x)\), where

u(x)=C_1\cos(2x) + C_2\sin(2x) + \frac52.

Then

y' = e^{-x}(u' - u).

At \(x=0\),

u(0)=C_1+\frac52 = 1, \qquad u'(0)=2C_2.

Using \(y'(0)=0\),

$$ 2C_2 - 1 = 0, $$

C_2 = \frac12.

Therefore

y = e^{-x}\left(-\frac32\cos(2x) + \frac12\sin(2x) + \frac52\right).

4.3Solve an Euler-Cauchy Boundary Value Problem

Final | Problem 4.3 |

Solve the boundary value problem for \(x>0\).

x^2y'' + xy' - y = 0, \qquad y(1)=2, \qquad y(2)=3

Solution

Use the Euler-Cauchy trial \(y=x^m\). The auxiliary equation is

$$ m(m-1) + m - 1 = 0, $$

which simplifies to

$$ m^2 - 1 = 0. $$

So \(m=\pm 1\), and

y = C_1x + C_2x^{-1}.

Apply \(y(1)=2\):

$$ C_1 + C_2 = 2. $$

Apply \(y(2)=3\):

2C_1 + \frac{C_2}{2} = 3.

Solve the system:

C_1 = \frac43, \qquad C_2 = \frac23.

So the solution is

y = \frac43x + \frac23x^{-1}.

4.4Use Variation of Parameters

Final | Problem 4.4 |

Solve the differential equation on any interval where the forcing term is defined.

y'' + y = \sec x

Solution

The homogeneous equation \(y''+y=0\) has fundamental solutions

y_1=\cos x, \qquad y_2=\sin x.

Their Wronskian is

W = y_1y_2' - y_1'y_2 = \cos x(\cos x) - (-\sin x)(\sin x) = 1.

For variation of parameters,

u_1' = -\frac{y_2\,\sec x}{W} = -\sin x\,\sec x = -\tan x,

u_1 = \ln|\cos x|.

Also,

u_2' = \frac{y_1\,\sec x}{W} = \cos x\,\sec x = 1,

$$ u_2 = x. $$

Thus a particular solution is

y_p = u_1y_1 + u_2y_2 = \cos x\ln|\cos x| + x\sin x.

Therefore the general solution is

y = C_1\cos x + C_2\sin x + \cos x\ln|\cos x| + x\sin x.