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First-order linear equations practice problems

The integrating factor method for y′ + p(x)y = q(x). 70 problems with worked solutions; in 70 of them every equation is proved by a computer algebra system.

11 worked problems

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

Problem 6.17

hard Video
Solve \( \displaystyle y' + 3y = e^{x} \) with \( \displaystyle y(0) = 2 \).
  1. The equation is linear in standard form; the integrating factor is e^(∫3 dx) = e^(3x).
    Reviewed
  2. \[ \frac{d}{d x} Y{\left(x \right)} e^{3 x} = 3 Y{\left(x \right)} e^{3 x} + e^{3 x} \frac{d}{d x} Y{\left(x \right)} \]
    Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
  3. \[ \int e^{4 x}\, dx = \frac{e^{4 x}}{4} \]
    Integrate the right side.✓ Proved
  4. Setting x = 0 and y = 2 fixes the constant of integration: C = 7/4.
    Reviewed
  5. \[ \frac{\left(3 e^{4 x} + 21\right) e^{- 3 x}}{4} + \frac{d}{d x} \frac{\left(e^{4 x} + 7\right) e^{- 3 x}}{4} = e^{x} \]
    The solution satisfies the equation.✓ Proved
  6. \[ 2 \]
    And the initial condition.Reviewed
Answer \( y = \frac{\left(e^{4 x} + 7\right) e^{- 3 x}}{4} \)
The slope field of y' = -3y + e^(x): short segments showing the slope at each point. Several solution curves follow the field; the one through (0, 2) is highlighted, y = (e^(4x) + 7)·e^(-3x)/4.

✓ Nihil obstat Lines: 3 proved, 3 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 6.18

hard Video
Solve \( \displaystyle y' + 1y = 3 x \) with \( \displaystyle y(0) = -2 \).
  1. The equation is linear in standard form; the integrating factor is e^(∫1 dx) = e^(1x).
    Reviewed
  2. \[ \frac{d}{d x} Y{\left(x \right)} e^{x} = Y{\left(x \right)} e^{x} + e^{x} \frac{d}{d x} Y{\left(x \right)} \]
    Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
  3. \[ \int 3 x e^{x}\, dx = \left(3 x - 3\right) e^{x} \]
    Integrate the right side.✓ Proved
  4. Setting x = 0 and y = -2 fixes the constant of integration: C = 1.
    Reviewed
  5. \[ 3 x + \frac{d}{d x} \left(3 x - 3 + e^{- x}\right) - 3 + e^{- x} = 3 x \]
    The solution satisfies the equation.✓ Proved
  6. \[ -2 \]
    And the initial condition.Reviewed
Answer \( y = 3 x - 3 + e^{- x} \)
The slope field of y' = 3x - y: short segments showing the slope at each point. Several solution curves follow the field; the one through (0, -2) is highlighted, y = 3x - 3 + e^(-x).

✓ Nihil obstat Lines: 3 proved, 3 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 6.25

hard Video
Solve \( \displaystyle y' + 4y = 1 \) with \( \displaystyle y(0) = 3 \).
  1. The equation is linear in standard form; the integrating factor is e^(∫4 dx) = e^(4x).
    Reviewed
  2. \[ \frac{d}{d x} Y{\left(x \right)} e^{4 x} = 4 Y{\left(x \right)} e^{4 x} + e^{4 x} \frac{d}{d x} Y{\left(x \right)} \]
    Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
  3. \[ \int e^{4 x}\, dx = \frac{e^{4 x}}{4} \]
    Integrate the right side.✓ Proved
  4. Setting x = 0 and y = 3 fixes the constant of integration: C = 11/4.
    Reviewed
  5. \[ \frac{d}{d x} \left(\frac{1}{4} + \frac{11 e^{- 4 x}}{4}\right) + 1 + 11 e^{- 4 x} = 1 \]
    The solution satisfies the equation.✓ Proved
  6. \[ 3 \]
    And the initial condition.Reviewed
Answer \( y = \frac{1}{4} + \frac{11 e^{- 4 x}}{4} \)
The slope field of y' = 1 - 4y: short segments showing the slope at each point. Several solution curves follow the field; the one through (0, 3) is highlighted, y = 1/4 + 11e^(-4x)/4.

✓ Nihil obstat Lines: 3 proved, 3 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 6.27

hard Video
Solve \( \displaystyle y' + 2y = 2 e^{- 2 x} \) with \( \displaystyle y(0) = 3 \).
  1. The equation is linear in standard form; the integrating factor is e^(∫2 dx) = e^(2x).
    Reviewed
  2. \[ \frac{d}{d x} Y{\left(x \right)} e^{2 x} = 2 Y{\left(x \right)} e^{2 x} + e^{2 x} \frac{d}{d x} Y{\left(x \right)} \]
    Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
  3. \[ \int 2\, dx = 2 x \]
    Integrate the right side.✓ Proved
  4. Setting x = 0 and y = 3 fixes the constant of integration: C = 3.
    Reviewed
  5. \[ \left(4 x + 6\right) e^{- 2 x} + \frac{d}{d x} \left(2 x + 3\right) e^{- 2 x} = 2 e^{- 2 x} \]
    The solution satisfies the equation.✓ Proved
  6. \[ 3 \]
    And the initial condition.Reviewed
Answer \( y = \left(2 x + 3\right) e^{- 2 x} \)
The slope field of y' = -2y + 2e^(-2x): short segments showing the slope at each point. Several solution curves follow the field; the one through (0, 3) is highlighted, y = (2x + 3)·e^(-2x).

✓ Nihil obstat Lines: 3 proved, 3 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 6.61

hard Video
Solve \( \displaystyle y' + 2y = x \) with \( \displaystyle y(0) = 2 \).
  1. The equation is linear in standard form; the integrating factor is e^(∫2 dx) = e^(2x).
    Reviewed
  2. \[ \frac{d}{d x} Y{\left(x \right)} e^{2 x} = 2 Y{\left(x \right)} e^{2 x} + e^{2 x} \frac{d}{d x} Y{\left(x \right)} \]
    Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
  3. \[ \int x e^{2 x}\, dx = \frac{\left(2 x - 1\right) e^{2 x}}{4} \]
    Integrate the right side.✓ Proved
  4. Setting x = 0 and y = 2 fixes the constant of integration: C = 9/4.
    Reviewed
  5. \[ x + \frac{d}{d x} \left(\frac{x}{2} - \frac{1}{4} + \frac{9 e^{- 2 x}}{4}\right) - \frac{1}{2} + \frac{9 e^{- 2 x}}{2} = x \]
    The solution satisfies the equation.✓ Proved
  6. \[ 2 \]
    And the initial condition.Reviewed
Answer \( y = \frac{x}{2} - \frac{1}{4} + \frac{9 e^{- 2 x}}{4} \)
The slope field of y' = x - 2y: short segments showing the slope at each point. Several solution curves follow the field; the one through (0, 2) is highlighted, y = x/2 - 1/4 + 9e^(-2x)/4.

✓ Nihil obstat Lines: 3 proved, 3 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 6.63

hard Video
Solve \( \displaystyle y' + 1y = 2 x \) with \( \displaystyle y(0) = 1 \).
  1. The equation is linear in standard form; the integrating factor is e^(∫1 dx) = e^(1x).
    Reviewed
  2. \[ \frac{d}{d x} Y{\left(x \right)} e^{x} = Y{\left(x \right)} e^{x} + e^{x} \frac{d}{d x} Y{\left(x \right)} \]
    Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
  3. \[ \int 2 x e^{x}\, dx = \left(2 x - 2\right) e^{x} \]
    Integrate the right side.✓ Proved
  4. Setting x = 0 and y = 1 fixes the constant of integration: C = 3.
    Reviewed
  5. \[ 2 x + \frac{d}{d x} \left(2 x - 2 + 3 e^{- x}\right) - 2 + 3 e^{- x} = 2 x \]
    The solution satisfies the equation.✓ Proved
  6. \[ 1 \]
    And the initial condition.Reviewed
Answer \( y = 2 x - 2 + 3 e^{- x} \)
The slope field of y' = 2x - y: short segments showing the slope at each point. Several solution curves follow the field; the one through (0, 1) is highlighted, y = 2x - 2 + 3e^(-x).

✓ Nihil obstat Lines: 3 proved, 3 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 6.71

hard Video
Solve \( \displaystyle y' + 3y = 3 e^{x} \) with \( \displaystyle y(0) = -2 \).
  1. The equation is linear in standard form; the integrating factor is e^(∫3 dx) = e^(3x).
    Reviewed
  2. \[ \frac{d}{d x} Y{\left(x \right)} e^{3 x} = 3 Y{\left(x \right)} e^{3 x} + e^{3 x} \frac{d}{d x} Y{\left(x \right)} \]
    Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
  3. \[ \int 3 e^{4 x}\, dx = \frac{3 e^{4 x}}{4} \]
    Integrate the right side.✓ Proved
  4. Setting x = 0 and y = -2 fixes the constant of integration: C = -11/4.
    Reviewed
  5. \[ \frac{\left(9 e^{4 x} - 33\right) e^{- 3 x}}{4} + \frac{d}{d x} \frac{\left(3 e^{4 x} - 11\right) e^{- 3 x}}{4} = 3 e^{x} \]
    The solution satisfies the equation.✓ Proved
  6. \[ -2 \]
    And the initial condition.Reviewed
Answer \( y = \frac{\left(3 e^{4 x} - 11\right) e^{- 3 x}}{4} \)
The slope field of y' = -3y + 3e^(x): short segments showing the slope at each point. Several solution curves follow the field; the one through (0, -2) is highlighted, y = (3e^(4x) - 11)·e^(-3x)/4.

✓ Nihil obstat Lines: 3 proved, 3 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 6.74

hard Video
Solve \( \displaystyle y' + 4y = 2 e^{x} \) with \( \displaystyle y(0) = 1 \).
  1. The equation is linear in standard form; the integrating factor is e^(∫4 dx) = e^(4x).
    Reviewed
  2. \[ \frac{d}{d x} Y{\left(x \right)} e^{4 x} = 4 Y{\left(x \right)} e^{4 x} + e^{4 x} \frac{d}{d x} Y{\left(x \right)} \]
    Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
  3. \[ \int 2 e^{5 x}\, dx = \frac{2 e^{5 x}}{5} \]
    Integrate the right side.✓ Proved
  4. Setting x = 0 and y = 1 fixes the constant of integration: C = 3/5.
    Reviewed
  5. \[ \frac{\left(8 e^{5 x} + 12\right) e^{- 4 x}}{5} + \frac{d}{d x} \frac{\left(2 e^{5 x} + 3\right) e^{- 4 x}}{5} = 2 e^{x} \]
    The solution satisfies the equation.✓ Proved
  6. \[ 1 \]
    And the initial condition.Reviewed
Answer \( y = \frac{\left(2 e^{5 x} + 3\right) e^{- 4 x}}{5} \)
The slope field of y' = -4y + 2e^(x): short segments showing the slope at each point. Several solution curves follow the field; the one through (0, 1) is highlighted, y = (2e^(5x) + 3)·e^(-4x)/5.

✓ Nihil obstat Lines: 3 proved, 3 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 6.172

hard Video
Solve \( \displaystyle y' + 4y = 5 \) with \( \displaystyle y(0) = -1 \).
  1. The equation is linear in standard form; the integrating factor is e^(∫4 dx) = e^(4x).
    Reviewed
  2. \[ \frac{d}{d x} Y{\left(x \right)} e^{4 x} = 4 Y{\left(x \right)} e^{4 x} + e^{4 x} \frac{d}{d x} Y{\left(x \right)} \]
    Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
  3. \[ \int 5 e^{4 x}\, dx = \frac{5 e^{4 x}}{4} \]
    Integrate the right side.✓ Proved
  4. Setting x = 0 and y = -1 fixes the constant of integration: C = -9/4.
    Reviewed
  5. \[ \frac{d}{d x} \left(\frac{5}{4} - \frac{9 e^{- 4 x}}{4}\right) + 5 - 9 e^{- 4 x} = 5 \]
    The solution satisfies the equation.✓ Proved
  6. \[ -1 \]
    And the initial condition.Reviewed
Answer \( y = \frac{5}{4} - \frac{9 e^{- 4 x}}{4} \)
The slope field of y' = 5 - 4y: short segments showing the slope at each point. Several solution curves follow the field; the one through (0, -1) is highlighted, y = 5/4 - 9e^(-4x)/4.

✓ Nihil obstat Lines: 3 proved, 3 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 6.62

medium Video
Solve \( \displaystyle y' + 2y = e^{- 2 x} \) with \( \displaystyle y(0) = 4 \).
  1. The equation is linear in standard form; the integrating factor is e^(∫2 dx) = e^(2x).
    Reviewed
  2. \[ \frac{d}{d x} Y{\left(x \right)} e^{2 x} = 2 Y{\left(x \right)} e^{2 x} + e^{2 x} \frac{d}{d x} Y{\left(x \right)} \]
    Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
  3. \[ \int 1\, dx = x \]
    Integrate the right side.✓ Proved
  4. Setting x = 0 and y = 4 fixes the constant of integration: C = 4.
    Reviewed
  5. \[ \left(2 x + 8\right) e^{- 2 x} + \frac{d}{d x} \left(x + 4\right) e^{- 2 x} = e^{- 2 x} \]
    The solution satisfies the equation.✓ Proved
  6. \[ 4 \]
    And the initial condition.Reviewed
Answer \( y = \left(x + 4\right) e^{- 2 x} \)
The slope field of y' = -2y + e^(-2x): short segments showing the slope at each point. Several solution curves follow the field; the one through (0, 4) is highlighted, y = (x + 4)·e^(-2x).

✓ Nihil obstat Lines: 3 proved, 3 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 6.101

medium Video
Solve \( \displaystyle y' + 3y = e^{- 3 x} \) with \( \displaystyle y(0) = -2 \).
  1. The equation is linear in standard form; the integrating factor is e^(∫3 dx) = e^(3x).
    Reviewed
  2. \[ \frac{d}{d x} Y{\left(x \right)} e^{3 x} = 3 Y{\left(x \right)} e^{3 x} + e^{3 x} \frac{d}{d x} Y{\left(x \right)} \]
    Multiplying by e^(ax) turns the left side into (e^(ax) y)'.✓ Proved
  3. \[ \int 1\, dx = x \]
    Integrate the right side.✓ Proved
  4. Setting x = 0 and y = -2 fixes the constant of integration: C = -2.
    Reviewed
  5. \[ \left(3 x - 6\right) e^{- 3 x} + \frac{d}{d x} \left(x - 2\right) e^{- 3 x} = e^{- 3 x} \]
    The solution satisfies the equation.✓ Proved
  6. \[ -2 \]
    And the initial condition.Reviewed
Answer \( y = \left(x - 2\right) e^{- 3 x} \)
The slope field of y' = -3y + e^(-3x): short segments showing the slope at each point. Several solution curves follow the field; the one through (0, -2) is highlighted, y = (x - 2)·e^(-3x).

✓ Nihil obstat Lines: 3 proved, 3 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 first-order linear equations practice

Solve \( \displaystyle y' + 3y = e^{- 3 x} \) with \( \displaystyle y(0) = 3 \).
Problem 6.24medium Video✓ Every equation proved
Solve \( \displaystyle y' + 1y = 1 \) with \( \displaystyle y(0) = 0 \).
Problem 6.29medium Video✓ Every equation proved
Solve \( \displaystyle y' + 2y = 2 \) with \( \displaystyle y(0) = -3 \).
Problem 6.30medium Video✓ Nihil obstat
Solve \( \displaystyle y' + 4y = e^{- 4 x} \) with \( \displaystyle y(0) = 0 \).
Problem 6.64medium Video✓ Every equation proved
Solve \( \displaystyle y' + 1y = 2 e^{x} \) with \( \displaystyle y(0) = -1 \).
Problem 6.69medium Video✓ Nihil obstat
Solve \( \displaystyle y' + 1y = 2 e^{x} \) with \( \displaystyle y(0) = 0 \).
Problem 6.99medium Video✓ Every equation proved
Solve \( \displaystyle y' + 1y = e^{- x} \) with \( \displaystyle y(0) = 0 \).
Problem 6.102medium Video✓ Every equation proved
Solve \( \displaystyle y' + 1y = 3 \) with \( \displaystyle y(0) = -1 \).
Problem 6.103medium Video✓ Nihil obstat
Solve \( \displaystyle y' + 1y = e^{- x} \) with \( \displaystyle y(0) = -3 \).
Problem 6.105medium Video✓ Every equation proved
Solve \( \displaystyle y' + 3y = e^{- 3 x} \) with \( \displaystyle y(0) = 1 \).
Problem 6.106medium Video✓ Every equation proved
Solve \( \displaystyle y' + 1y = 3 e^{- x} \) with \( \displaystyle y(0) = -1 \).
Problem 6.125medium Video✓ Every equation proved
Solve \( \displaystyle y' + 1y = e^{- x} \) with \( \displaystyle y(0) = 2 \).
Problem 6.128medium Video✓ Every equation proved
Solve \( \displaystyle y' + 1y = 6 \) with \( \displaystyle y(0) = 3 \).
Problem 6.129medium Video✓ Nihil obstat
Solve \( \displaystyle y' + 2y = e^{- 2 x} \) with \( \displaystyle y(0) = -2 \).
Problem 6.148medium Video✓ Every equation proved
Solve \( \displaystyle y' + 1y = 1 \) with \( \displaystyle y(0) = 4 \).
Problem 6.150medium Video✓ Nihil obstat
Solve \( \displaystyle y' + 2y = 6 \) with \( \displaystyle y(0) = -3 \).
Problem 6.174medium Video✓ Nihil obstat
Solve \( \displaystyle y' + 3y = 3 \) with \( \displaystyle y(0) = -2 \).
Problem 6.176medium Video✓ Nihil obstat
Solve \( \displaystyle y' + 1y = 2 e^{x} \) with \( \displaystyle y(0) = 1 \).
Problem 6.196medium Video✓ Every equation proved
Solve \( \displaystyle y' + 1y = 2 e^{x} \) with \( \displaystyle y(0) = 3 \).
Problem 6.200medium Video✓ Every equation proved
Solve \( \displaystyle y' + 1y = 4 \) with \( \displaystyle y(0) = 0 \).
Problem 6.201medium Video✓ Nihil obstat
Solve \( \displaystyle y' + 1y = 3 e^{- x} \) with \( \displaystyle y(0) = -3 \).
Problem 6.202medium Video✓ Every equation proved
Solve \( \displaystyle y' + 3y = 2 e^{- 3 x} \) with \( \displaystyle y(0) = 1 \).
Problem 6.16hard Video✓ Every equation proved
Solve \( \displaystyle y' + 4y = 3 e^{- 4 x} \) with \( \displaystyle y(0) = -1 \).
Problem 6.19hard Video✓ Every equation proved
Solve \( \displaystyle y' + 4y = 2 x \) with \( \displaystyle y(0) = -2 \).
Problem 6.20hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 4y = 6 \) with \( \displaystyle y(0) = 0 \).
Problem 6.21hard Video✓ Every equation proved
Solve \( \displaystyle y' + 2y = 2 e^{- 2 x} \) with \( \displaystyle y(0) = 4 \).
Problem 6.22hard Video✓ Every equation proved
Solve \( \displaystyle y' + 4y = 3 x \) with \( \displaystyle y(0) = -2 \).
Problem 6.23hard Video✓ Every equation proved
Solve \( \displaystyle y' + 2y = 3 e^{- 2 x} \) with \( \displaystyle y(0) = -2 \).
Problem 6.26hard Video✓ Every equation proved
Solve \( \displaystyle y' + 2y = 4 x \) with \( \displaystyle y(0) = 4 \).
Problem 6.28hard Video✓ Every equation proved
Solve \( \displaystyle y' + 4y = 2 e^{- 4 x} \) with \( \displaystyle y(0) = -2 \).
Problem 6.65hard Video✓ Every equation proved
Solve \( \displaystyle y' + 4y = e^{x} \) with \( \displaystyle y(0) = 4 \).
Problem 6.66hard Video✓ Every equation proved
Solve \( \displaystyle y' + 2y = 3 e^{x} \) with \( \displaystyle y(0) = -2 \).
Problem 6.67hard Video✓ Every equation proved
Solve \( \displaystyle y' + 3y = 1 \) with \( \displaystyle y(0) = 1 \).
Problem 6.68hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 4y = 5 \) with \( \displaystyle y(0) = 3 \).
Problem 6.70hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 3y = e^{x} \) with \( \displaystyle y(0) = 0 \).
Problem 6.72hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 3y = e^{x} \) with \( \displaystyle y(0) = 1 \).
Problem 6.73hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 2y = 2 e^{- 2 x} \) with \( \displaystyle y(0) = 1 \).
Problem 6.75hard Video✓ Every equation proved
Solve \( \displaystyle y' + 4y = 3 e^{x} \) with \( \displaystyle y(0) = 2 \).
Problem 6.100hard Video✓ Every equation proved
Solve \( \displaystyle y' + 4y = 3 x \) with \( \displaystyle y(0) = -1 \).
Problem 6.104hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 2y = 2 e^{x} \) with \( \displaystyle y(0) = 3 \).
Problem 6.123hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 4y = x \) with \( \displaystyle y(0) = 2 \).
Problem 6.124hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 2y = x \) with \( \displaystyle y(0) = 4 \).
Problem 6.126hard Video✓ Every equation proved
Solve \( \displaystyle y' + 2y = 2 e^{x} \) with \( \displaystyle y(0) = -3 \).
Problem 6.127hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 2y = 2 e^{- 2 x} \) with \( \displaystyle y(0) = -3 \).
Problem 6.130hard Video✓ Every equation proved
Solve \( \displaystyle y' + 1y = 4 x \) with \( \displaystyle y(0) = -2 \).
Problem 6.147hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 1y = e^{x} \) with \( \displaystyle y(0) = 4 \).
Problem 6.149hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 4y = 3 e^{- 4 x} \) with \( \displaystyle y(0) = 2 \).
Problem 6.151hard Video✓ Every equation proved
Solve \( \displaystyle y' + 4y = 4 x \) with \( \displaystyle y(0) = -3 \).
Problem 6.152hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 1y = 4 x \) with \( \displaystyle y(0) = 3 \).
Problem 6.153hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 4y = 4 x \) with \( \displaystyle y(0) = 0 \).
Problem 6.154hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 2y = 3 e^{x} \) with \( \displaystyle y(0) = 3 \).
Problem 6.171hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 4y = 2 x \) with \( \displaystyle y(0) = 4 \).
Problem 6.173hard Video✓ Every equation proved
Solve \( \displaystyle y' + 4y = 2 x \) with \( \displaystyle y(0) = -1 \).
Problem 6.175hard Video✓ Every equation proved
Solve \( \displaystyle y' + 1y = e^{x} \) with \( \displaystyle y(0) = 2 \).
Problem 6.177hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 3y = 3 x \) with \( \displaystyle y(0) = -1 \).
Problem 6.178hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 2y = 2 x \) with \( \displaystyle y(0) = 3 \).
Problem 6.195hard Video✓ Every equation proved
Solve \( \displaystyle y' + 3y = x \) with \( \displaystyle y(0) = 3 \).
Problem 6.197hard Video✓ Nihil obstat
Solve \( \displaystyle y' + 2y = 2 e^{x} \) with \( \displaystyle y(0) = 2 \).
Problem 6.198hard Video✓ Every equation proved
Solve \( \displaystyle y' + 2y = 4 x \) with \( \displaystyle y(0) = 3 \).
Problem 6.199hard Video✓ Every equation proved