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Variation of parameters practice problems

A particular solution by variation of parameters, when undetermined coefficients cannot work. 18 problems with worked solutions; in 18 of them every equation is proved by a computer algebra system.

7 worked problems

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

Problem 6.502

hard
Use variation of parameters to find a particular solution of \( \displaystyle y + y'' = \frac{3}{\cos{\left(x \right)}} \), given that \( \displaystyle y_1 = \cos{\left(x \right)} \) and \( \displaystyle y_2 = \sin{\left(x \right)} \) solve the homogeneous equation.
  1. \[ - \sin{\left(x \right)} \frac{d}{d x} \cos{\left(x \right)} + \cos{\left(x \right)} \frac{d}{d x} \sin{\left(x \right)} = 1 \]
    The Wronskian W = y₁y₂′ − y₂y₁′.✓ Proved
  2. \[ \frac{d}{d x} 3 \ln{\left(\cos{\left(x \right)} \right)} = - 3 \tan{\left(x \right)} \]
    u₁′ = −y₂g/W; u₁ = 3*log(cos(x)).✓ Proved
  3. \[ \frac{d}{d x} 3 x = 3 \]
    u₂′ = y₁g/W; u₂ = 3*x.✓ Proved
  4. y_p = u₁y₁ + u₂y₂: 3*x*sin(x) + 3*log(cos(x))*cos(x).
    Reviewed
  5. \[ 3 x \sin{\left(x \right)} + 3 \ln{\left(\cos{\left(x \right)} \right)} \cos{\left(x \right)} + \frac{d^{2}}{d x^{2}} \left(3 x \sin{\left(x \right)} + 3 \ln{\left(\cos{\left(x \right)} \right)} \cos{\left(x \right)}\right) = \frac{3}{\cos{\left(x \right)}} \]
    It satisfies the equation.✓ Proved
Answer \( y_p = 3 x \sin{\left(x \right)} + 3 \ln{\left(\cos{\left(x \right)} \right)} \cos{\left(x \right)} \)

✓ Nihil obstat Lines: 4 proved, 1 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 6.503

hard
Use variation of parameters to find a particular solution of \( \displaystyle y - 2 y' + y'' = \frac{3 e^{x}}{x^{2}} \), given that \( \displaystyle y_1 = e^{x} \) and \( \displaystyle y_2 = x e^{x} \) solve the homogeneous equation.
  1. \[ - x e^{x} \frac{d}{d x} e^{x} + e^{x} \frac{d}{d x} x e^{x} = e^{2 x} \]
    The Wronskian W = y₁y₂′ − y₂y₁′.✓ Proved
  2. \[ \frac{d}{d x} \left(- 3 \ln{\left(x \right)}\right) = - \frac{3}{x} \]
    u₁′ = −y₂g/W; u₁ = -3*log(x).✓ Proved
  3. \[ \frac{d}{d x} \left(- \frac{3}{x}\right) = \frac{3}{x^{2}} \]
    u₂′ = y₁g/W; u₂ = -3/x.✓ Proved
  4. \[ - 3 e^{x} \ln{\left(x \right)} - 3 e^{x} = \left(- 3 \ln{\left(x \right)} - 3\right) e^{x} \]
    y_p = u₁y₁ + u₂y₂.✓ Proved
  5. \[ \left(- 3 \ln{\left(x \right)} - 3\right) e^{x} - 2 \frac{d}{d x} \left(- 3 \ln{\left(x \right)} - 3\right) e^{x} + \frac{d^{2}}{d x^{2}} \left(- 3 \ln{\left(x \right)} - 3\right) e^{x} = \frac{3 e^{x}}{x^{2}} \]
    It satisfies the equation.✓ Proved
Answer \( y_p = 3 \left(- \ln{\left(x \right)} - 1\right) e^{x} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 6.506

hard
Use variation of parameters to find a particular solution of \( \displaystyle y + 2 y' + y'' = \frac{3 e^{- x}}{x} \), given that \( \displaystyle y_1 = e^{- x} \) and \( \displaystyle y_2 = x e^{- x} \) solve the homogeneous equation.
  1. \[ - x e^{- x} \frac{d}{d x} e^{- x} + e^{- x} \frac{d}{d x} x e^{- x} = e^{- 2 x} \]
    The Wronskian W = y₁y₂′ − y₂y₁′.✓ Proved
  2. \[ \frac{d}{d x} \left(- 3 x\right) = -3 \]
    u₁′ = −y₂g/W; u₁ = -3*x.✓ Proved
  3. \[ \frac{d}{d x} 3 \ln{\left(x \right)} = \frac{3}{x} \]
    u₂′ = y₁g/W; u₂ = 3*log(x).✓ Proved
  4. \[ 3 x e^{- x} \ln{\left(x \right)} - 3 x e^{- x} = 3 x \left(\ln{\left(x \right)} - 1\right) e^{- x} \]
    y_p = u₁y₁ + u₂y₂.✓ Proved
  5. \[ 3 x \left(\ln{\left(x \right)} - 1\right) e^{- x} + 2 \frac{d}{d x} 3 x \left(\ln{\left(x \right)} - 1\right) e^{- x} + \frac{d^{2}}{d x^{2}} 3 x \left(\ln{\left(x \right)} - 1\right) e^{- x} = \frac{3 e^{- x}}{x} \]
    It satisfies the equation.✓ Proved
Answer \( y_p = 3 x \left(\ln{\left(x \right)} - 1\right) e^{- x} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 6.507

hard
Use variation of parameters to find a particular solution of \( \displaystyle y - 2 y' + y'' = \frac{e^{x}}{x} \), given that \( \displaystyle y_1 = e^{x} \) and \( \displaystyle y_2 = x e^{x} \) solve the homogeneous equation.
  1. \[ - x e^{x} \frac{d}{d x} e^{x} + e^{x} \frac{d}{d x} x e^{x} = e^{2 x} \]
    The Wronskian W = y₁y₂′ − y₂y₁′.✓ Proved
  2. \[ \frac{d}{d x} \left(- x\right) = -1 \]
    u₁′ = −y₂g/W; u₁ = -x.✓ Proved
  3. \[ \frac{d}{d x} \ln{\left(x \right)} = \frac{1}{x} \]
    u₂′ = y₁g/W; u₂ = log(x).✓ Proved
  4. \[ x e^{x} \ln{\left(x \right)} - x e^{x} = x \left(\ln{\left(x \right)} - 1\right) e^{x} \]
    y_p = u₁y₁ + u₂y₂.✓ Proved
  5. \[ x \left(\ln{\left(x \right)} - 1\right) e^{x} - 2 \frac{d}{d x} x \left(\ln{\left(x \right)} - 1\right) e^{x} + \frac{d^{2}}{d x^{2}} x \left(\ln{\left(x \right)} - 1\right) e^{x} = \frac{e^{x}}{x} \]
    It satisfies the equation.✓ Proved
Answer \( y_p = x \left(\ln{\left(x \right)} - 1\right) e^{x} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 6.509

hard
Use variation of parameters to find a particular solution of \( \displaystyle y + y'' = \frac{2}{\sin{\left(x \right)}} \), given that \( \displaystyle y_1 = \cos{\left(x \right)} \) and \( \displaystyle y_2 = \sin{\left(x \right)} \) solve the homogeneous equation.
  1. \[ - \sin{\left(x \right)} \frac{d}{d x} \cos{\left(x \right)} + \cos{\left(x \right)} \frac{d}{d x} \sin{\left(x \right)} = 1 \]
    The Wronskian W = y₁y₂′ − y₂y₁′.✓ Proved
  2. \[ \frac{d}{d x} \left(- 2 x\right) = -2 \]
    u₁′ = −y₂g/W; u₁ = -2*x.✓ Proved
  3. \[ \frac{d}{d x} 2 \ln{\left(\sin{\left(x \right)} \right)} = \frac{2}{\tan{\left(x \right)}} \]
    u₂′ = y₁g/W; u₂ = 2*log(sin(x)).✓ Proved
  4. y_p = u₁y₁ + u₂y₂: -2*x*cos(x) + 2*log(sin(x))*sin(x).
    Reviewed
  5. \[ - 2 x \cos{\left(x \right)} + 2 \ln{\left(\sin{\left(x \right)} \right)} \sin{\left(x \right)} + \frac{d^{2}}{d x^{2}} \left(- 2 x \cos{\left(x \right)} + 2 \ln{\left(\sin{\left(x \right)} \right)} \sin{\left(x \right)}\right) = \frac{2}{\sin{\left(x \right)}} \]
    It satisfies the equation.✓ Proved
Answer \( y_p = - 2 x \cos{\left(x \right)} + 2 \ln{\left(\sin{\left(x \right)} \right)} \sin{\left(x \right)} \)

✓ Nihil obstat Lines: 4 proved, 1 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 6.512

hard
Use variation of parameters to find a particular solution of \( \displaystyle y - 2 y' + y'' = \frac{e^{x}}{x^{2}} \), given that \( \displaystyle y_1 = e^{x} \) and \( \displaystyle y_2 = x e^{x} \) solve the homogeneous equation.
  1. \[ - x e^{x} \frac{d}{d x} e^{x} + e^{x} \frac{d}{d x} x e^{x} = e^{2 x} \]
    The Wronskian W = y₁y₂′ − y₂y₁′.✓ Proved
  2. \[ \frac{d}{d x} \left(- \ln{\left(x \right)}\right) = - \frac{1}{x} \]
    u₁′ = −y₂g/W; u₁ = -log(x).✓ Proved
  3. \[ \frac{d}{d x} \left(- \frac{1}{x}\right) = \frac{1}{x^{2}} \]
    u₂′ = y₁g/W; u₂ = -1/x.✓ Proved
  4. \[ - e^{x} \ln{\left(x \right)} - e^{x} = \left(- \ln{\left(x \right)} - 1\right) e^{x} \]
    y_p = u₁y₁ + u₂y₂.✓ Proved
  5. \[ \left(- \ln{\left(x \right)} - 1\right) e^{x} - 2 \frac{d}{d x} \left(- \ln{\left(x \right)} - 1\right) e^{x} + \frac{d^{2}}{d x^{2}} \left(- \ln{\left(x \right)} - 1\right) e^{x} = \frac{e^{x}}{x^{2}} \]
    It satisfies the equation.✓ Proved
Answer \( y_p = \left(- \ln{\left(x \right)} - 1\right) e^{x} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 6.515

hard
Use variation of parameters to find a particular solution of \( \displaystyle y + 2 y' + y'' = \frac{e^{- x}}{x} \), given that \( \displaystyle y_1 = e^{- x} \) and \( \displaystyle y_2 = x e^{- x} \) solve the homogeneous equation.
  1. \[ - x e^{- x} \frac{d}{d x} e^{- x} + e^{- x} \frac{d}{d x} x e^{- x} = e^{- 2 x} \]
    The Wronskian W = y₁y₂′ − y₂y₁′.✓ Proved
  2. \[ \frac{d}{d x} \left(- x\right) = -1 \]
    u₁′ = −y₂g/W; u₁ = -x.✓ Proved
  3. \[ \frac{d}{d x} \ln{\left(x \right)} = \frac{1}{x} \]
    u₂′ = y₁g/W; u₂ = log(x).✓ Proved
  4. \[ x e^{- x} \ln{\left(x \right)} - x e^{- x} = x \left(\ln{\left(x \right)} - 1\right) e^{- x} \]
    y_p = u₁y₁ + u₂y₂.✓ Proved
  5. \[ x \left(\ln{\left(x \right)} - 1\right) e^{- x} + 2 \frac{d}{d x} x \left(\ln{\left(x \right)} - 1\right) e^{- x} + \frac{d^{2}}{d x^{2}} x \left(\ln{\left(x \right)} - 1\right) e^{- x} = \frac{e^{- x}}{x} \]
    It satisfies the equation.✓ Proved
Answer \( y_p = x \left(\ln{\left(x \right)} - 1\right) e^{- x} \)

✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. 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 variation of parameters practice

Use variation of parameters to find a particular solution of \( \displaystyle y - 2 y' + y'' = \frac{2 e^{x}}{x^{2}} \), given that \( \displaystyle y_1 = e^{x} \) and \( \displaystyle y_2 = x e^{x} \) solve the homogeneous equation.
Problem 6.501hard✓ Every equation proved
Use variation of parameters to find a particular solution of \( \displaystyle y - 2 y' + y'' = \frac{3 e^{x}}{x} \), given that \( \displaystyle y_1 = e^{x} \) and \( \displaystyle y_2 = x e^{x} \) solve the homogeneous equation.
Problem 6.504hard✓ Every equation proved
Use variation of parameters to find a particular solution of \( \displaystyle 4 y + y'' = \frac{3}{\cos{\left(2 x \right)}} \), given that \( \displaystyle y_1 = \cos{\left(2 x \right)} \) and \( \displaystyle y_2 = \sin{\left(2 x \right)} \) solve the homogeneous equation.
Problem 6.505hard✓ Every equation proved
Use variation of parameters to find a particular solution of \( \displaystyle y + y'' = \frac{2}{\cos{\left(x \right)}} \), given that \( \displaystyle y_1 = \cos{\left(x \right)} \) and \( \displaystyle y_2 = \sin{\left(x \right)} \) solve the homogeneous equation.
Problem 6.508hard✓ Every equation proved
Use variation of parameters to find a particular solution of \( \displaystyle 4 y + y'' = \frac{2}{\cos{\left(2 x \right)}} \), given that \( \displaystyle y_1 = \cos{\left(2 x \right)} \) and \( \displaystyle y_2 = \sin{\left(2 x \right)} \) solve the homogeneous equation.
Problem 6.510hard✓ Every equation proved
Use variation of parameters to find a particular solution of \( \displaystyle y - 2 y' + y'' = \frac{2 e^{x}}{x} \), given that \( \displaystyle y_1 = e^{x} \) and \( \displaystyle y_2 = x e^{x} \) solve the homogeneous equation.
Problem 6.511hard✓ Every equation proved
Use variation of parameters to find a particular solution of \( \displaystyle y + 2 y' + y'' = \frac{2 e^{- x}}{x} \), given that \( \displaystyle y_1 = e^{- x} \) and \( \displaystyle y_2 = x e^{- x} \) solve the homogeneous equation.
Problem 6.513hard✓ Nihil obstat
Use variation of parameters to find a particular solution of \( \displaystyle y + y'' = \frac{1}{\sin{\left(x \right)}} \), given that \( \displaystyle y_1 = \cos{\left(x \right)} \) and \( \displaystyle y_2 = \sin{\left(x \right)} \) solve the homogeneous equation.
Problem 6.514hard✓ Nihil obstat
Use variation of parameters to find a particular solution of \( \displaystyle y + y'' = \frac{3}{\sin{\left(x \right)}} \), given that \( \displaystyle y_1 = \cos{\left(x \right)} \) and \( \displaystyle y_2 = \sin{\left(x \right)} \) solve the homogeneous equation.
Problem 6.646hard✓ Every equation proved
Use variation of parameters to find a particular solution of \( \displaystyle y + y'' = \frac{1}{\cos{\left(x \right)}} \), given that \( \displaystyle y_1 = \cos{\left(x \right)} \) and \( \displaystyle y_2 = \sin{\left(x \right)} \) solve the homogeneous equation.
Problem 6.647hard✓ Nihil obstat
Use variation of parameters to find a particular solution of \( \displaystyle 4 y + y'' = \frac{1}{\cos{\left(2 x \right)}} \), given that \( \displaystyle y_1 = \cos{\left(2 x \right)} \) and \( \displaystyle y_2 = \sin{\left(2 x \right)} \) solve the homogeneous equation.
Problem 6.668hard✓ Every equation proved