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Order and linearity of differential equations practice problems

The order of a differential equation, whether it is linear (and homogeneous), and checking a solution. 60 problems with worked solutions; in 60 of them every equation is proved by a computer algebra system.

14 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 6.411

easy
Classify \( \displaystyle - 4 y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- 2 x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ \frac{d^{2}}{d x^{2}} e^{- 2 x} - 4 e^{- 2 x} = 0 \]
    Substituting y = exp(-2*x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.415

easy
Classify \( \displaystyle 2 y + 3 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ 3 \frac{d}{d x} e^{- x} + \frac{d^{2}}{d x^{2}} e^{- x} + 2 e^{- x} = 0 \]
    Substituting y = exp(-x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.417

easy
Classify \( \displaystyle - 3 y - 2 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ - 2 \frac{d}{d x} e^{- x} + \frac{d^{2}}{d x^{2}} e^{- x} - 3 e^{- x} = 0 \]
    Substituting y = exp(-x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.422

easy
Classify \( \displaystyle 6 y - 5 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{2 x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ 6 e^{2 x} - 5 \frac{d}{d x} e^{2 x} + \frac{d^{2}}{d x^{2}} e^{2 x} = 0 \]
    Substituting y = exp(2*x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.524

easy
Classify \( \displaystyle - 2 y + y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- 2 x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ \frac{d}{d x} e^{- 2 x} + \frac{d^{2}}{d x^{2}} e^{- 2 x} - 2 e^{- 2 x} = 0 \]
    Substituting y = exp(-2*x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.557

easy
Classify \( \displaystyle - 6 y - y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{3 x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ - 6 e^{3 x} - \frac{d}{d x} e^{3 x} + \frac{d^{2}}{d x^{2}} e^{3 x} = 0 \]
    Substituting y = exp(3*x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.559

easy
Classify \( \displaystyle 6 y + 5 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- 2 x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ 5 \frac{d}{d x} e^{- 2 x} + \frac{d^{2}}{d x^{2}} e^{- 2 x} + 6 e^{- 2 x} = 0 \]
    Substituting y = exp(-2*x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.562

easy
Classify \( \displaystyle - 3 y + y' + 3 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{3 x} + 1 \) is a solution.
  1. The highest derivative is y′, so the order is 1.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is nonzero: nonhomogeneous.
    Reviewed
  3. \[ - 3 e^{3 x} + \frac{d}{d x} \left(e^{3 x} + 1\right) = 0 \]
    Substituting y = exp(3*x) + 1 makes the left side 0.✓ Proved
Answer \( \text{order 1, linear and nonhomogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.569

easy
Classify \( \displaystyle - 3 y + 2 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ - 3 e^{x} + 2 \frac{d}{d x} e^{x} + \frac{d^{2}}{d x^{2}} e^{x} = 0 \]
    Substituting y = exp(x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.672

easy
Classify \( \displaystyle 2 y - 3 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ 2 e^{x} - 3 \frac{d}{d x} e^{x} + \frac{d^{2}}{d x^{2}} e^{x} = 0 \]
    Substituting y = exp(x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.674

easy
Classify \( \displaystyle - 2 y + y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ - 2 e^{x} + \frac{d}{d x} e^{x} + \frac{d^{2}}{d x^{2}} e^{x} = 0 \]
    Substituting y = exp(x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.676

easy
Classify \( \displaystyle - 3 y + 2 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- 3 x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ 2 \frac{d}{d x} e^{- 3 x} + \frac{d^{2}}{d x^{2}} e^{- 3 x} - 3 e^{- 3 x} = 0 \]
    Substituting y = exp(-3*x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.733

easy
Classify \( \displaystyle - y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ \frac{d^{2}}{d x^{2}} e^{- x} - e^{- x} = 0 \]
    Substituting y = exp(-x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

Problem 6.738

easy
Classify \( \displaystyle - 3 y - 2 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{3 x} \) is a solution.
  1. The highest derivative is y′′, so the order is 2.
    Reviewed
  2. It is linear: y and its derivatives appear only to the first power, multiplied by functions of x. The term without y is zero: homogeneous.
    Reviewed
  3. \[ - 3 e^{3 x} - 2 \frac{d}{d x} e^{3 x} + \frac{d^{2}}{d x^{2}} e^{3 x} = 0 \]
    Substituting y = exp(3*x) makes the left side 0.✓ Proved
Answer \( \text{order 2, linear and homogeneous} \)

✓ Nihil obstat Lines: 1 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt · How we verify

More order and linearity of differential equations practice

Classify \( \displaystyle - y^{2} + y' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = \frac{1}{2 - x} \) is a solution.
Problem 6.412easy✓ Every equation proved
Classify \( \displaystyle x^{2} y'' - 2 y = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = x^{2} \) is a solution.
Problem 6.413easy✓ Every equation proved
Classify \( \displaystyle - 3 y + y' - 3 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{3 x} - 1 \) is a solution.
Problem 6.414easy✓ Every equation proved
Classify \( \displaystyle 3 y + y' - 3 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 1 + e^{- 3 x} \) is a solution.
Problem 6.416easy✓ Every equation proved
Classify \( \displaystyle x + y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = - x + \sin{\left(x \right)} \) is a solution.
Problem 6.418easy✓ Every equation proved
Classify \( \displaystyle - x + y y' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = \sqrt{x^{2} + 1} \) is a solution.
Problem 6.419easy✓ Every equation proved
Classify \( \displaystyle - y^{2} + y' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = \frac{1}{3 - x} \) is a solution.
Problem 6.420easy✓ Every equation proved
Classify \( \displaystyle 2 y + y' - 6 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 3 + e^{- 2 x} \) is a solution.
Problem 6.421easy✓ Every equation proved
Classify \( \displaystyle - 3 x + y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 3 x + \sin{\left(x \right)} \) is a solution.
Problem 6.423easy✓ Every equation proved
Classify \( \displaystyle - y + y' - 2 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{x} - 2 \) is a solution.
Problem 6.424easy✓ Every equation proved
Classify \( \displaystyle - x + y y' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = \sqrt{x^{2} + 2} \) is a solution.
Problem 6.425easy✓ Every equation proved
Classify \( \displaystyle 3 x + y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = - 3 x + \sin{\left(x \right)} \) is a solution.
Problem 6.516easy✓ Every equation proved
Classify \( \displaystyle - x + y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = x + \sin{\left(x \right)} \) is a solution.
Problem 6.517easy✓ Every equation proved
Classify \( \displaystyle - 2 x + y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 2 x + \sin{\left(x \right)} \) is a solution.
Problem 6.518easy✓ Every equation proved
Classify \( \displaystyle - x + y y' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = \sqrt{x^{2} + 3} \) is a solution.
Problem 6.519easy✓ Every equation proved
Classify \( \displaystyle - y + y' + 3 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{x} + 3 \) is a solution.
Problem 6.520easy✓ Every equation proved
Classify \( \displaystyle - 2 y - y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- x} \) is a solution.
Problem 6.521easy✓ Every equation proved
Classify \( \displaystyle - 3 y + y' - 6 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{3 x} - 2 \) is a solution.
Problem 6.522easy✓ Every equation proved
Classify \( \displaystyle - y + y' - 1 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{x} - 1 \) is a solution.
Problem 6.523easy✓ Every equation proved
Classify \( \displaystyle 3 y + y' + 9 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = -3 + e^{- 3 x} \) is a solution.
Problem 6.525easy✓ Every equation proved
Classify \( \displaystyle - 4 y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{2 x} \) is a solution.
Problem 6.556easy✓ Every equation proved
Classify \( \displaystyle - y + y' - 3 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{x} - 3 \) is a solution.
Problem 6.558easy✓ Every equation proved
Classify \( \displaystyle - 2 y + y' - 6 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{2 x} - 3 \) is a solution.
Problem 6.560easy✓ Every equation proved
Classify \( \displaystyle 3 y + y' - 6 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 2 + e^{- 3 x} \) is a solution.
Problem 6.561easy✓ Every equation proved
Classify \( \displaystyle 2 x + y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = - 2 x + \sin{\left(x \right)} \) is a solution.
Problem 6.563easy✓ Every equation proved
Classify \( \displaystyle - 3 y + y' + 6 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{3 x} + 2 \) is a solution.
Problem 6.564easy✓ Every equation proved
Classify \( \displaystyle 3 y - 4 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{3 x} \) is a solution.
Problem 6.565easy✓ Every equation proved
Classify \( \displaystyle - y^{2} + y' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = \frac{1}{4 - x} \) is a solution.
Problem 6.566easy✓ Every equation proved
Classify \( \displaystyle 6 y + 5 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- 3 x} \) is a solution.
Problem 6.567easy✓ Nihil obstat
Classify \( \displaystyle 2 y + 3 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- 2 x} \) is a solution.
Problem 6.568easy✓ Nihil obstat
Classify \( \displaystyle y + y' - 1 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 1 + e^{- x} \) is a solution.
Problem 6.570easy✓ Every equation proved
Classify \( \displaystyle - 6 y + y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- 3 x} \) is a solution.
Problem 6.669easy✓ Nihil obstat
Classify \( \displaystyle - 2 y + y' + 4 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{2 x} + 2 \) is a solution.
Problem 6.670easy✓ Every equation proved
Classify \( \displaystyle - 9 y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- 3 x} \) is a solution.
Problem 6.671easy✓ Every equation proved
Classify \( \displaystyle 2 y + y' - 2 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = 1 + e^{- 2 x} \) is a solution.
Problem 6.673easy✓ Every equation proved
Classify \( \displaystyle - 2 y + y' - 4 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{2 x} - 2 \) is a solution.
Problem 6.675easy✓ Every equation proved
Classify \( \displaystyle 3 y + 4 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- x} \) is a solution.
Problem 6.677easy✓ Nihil obstat
Classify \( \displaystyle - 2 y - y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{2 x} \) is a solution.
Problem 6.678easy✓ Nihil obstat
Classify \( \displaystyle 3 y + y' + 3 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = -1 + e^{- 3 x} \) is a solution.
Problem 6.729easy✓ Every equation proved
Classify \( \displaystyle - y + y' + 2 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{x} + 2 \) is a solution.
Problem 6.730easy✓ Every equation proved
Classify \( \displaystyle 2 y - 3 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{2 x} \) is a solution.
Problem 6.731easy✓ Nihil obstat
Classify \( \displaystyle - 2 y + y' + 6 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{2 x} + 3 \) is a solution.
Problem 6.732easy✓ Every equation proved
Classify \( \displaystyle 2 y + y' + 4 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = -2 + e^{- 2 x} \) is a solution.
Problem 6.734easy✓ Every equation proved
Classify \( \displaystyle - 9 y + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{3 x} \) is a solution.
Problem 6.735easy✓ Every equation proved
Classify \( \displaystyle - 2 y + y' - 2 = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{2 x} - 1 \) is a solution.
Problem 6.736easy✓ Every equation proved
Classify \( \displaystyle 3 y + 4 y' + y'' = 0 \): its order, whether it is linear and, if so, whether it is homogeneous. Then verify that \( \displaystyle y = e^{- 3 x} \) is a solution.
Problem 6.737easy✓ Every equation proved