∫Calc Practice
Home›Calculus 3›Second-order initial and boundary value problems

Second-order initial and boundary value problems practice problems

Constant-coefficient second-order equations with initial or boundary conditions. 60 problems with worked solutions; in 60 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 6.476

medium
Solve \( \displaystyle 8 y - 4 y' + y'' = 0 \) with \( \displaystyle y(0) = -3,\ y(\frac{\pi}{4}) = 1 \).
  1. \[ m^{2} - 4 m + 8 = \left(m - 2 - 2 i\right) \left(m - 2 + 2 i\right) \]
    The characteristic equation has complex roots 2 ± 2i.✓ Proved
  2. So y = C₁exp(2*x)*cos(2*x) + C₂exp(2*x)*sin(2*x); the two conditions fix C₁ and C₂.
    Reviewed
  3. \[ 8 \left(\sin{\left(2 x \right)} - 3 e^{\frac{\pi}{2}} \cos{\left(2 x \right)}\right) e^{2 x - \frac{\pi}{2}} - 4 \frac{d}{d x} \left(\sin{\left(2 x \right)} - 3 e^{\frac{\pi}{2}} \cos{\left(2 x \right)}\right) e^{2 x - \frac{\pi}{2}} + \frac{d^{2}}{d x^{2}} \left(\sin{\left(2 x \right)} - 3 e^{\frac{\pi}{2}} \cos{\left(2 x \right)}\right) e^{2 x - \frac{\pi}{2}} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \left(\sin{\left(2 x \right)} - 3 e^{\frac{\pi}{2}} \cos{\left(2 x \right)}\right) e^{2 x - \frac{\pi}{2}} \right|_{\substack{ x=0 }} = -3 \]
    ... y(0) = -3 ...✓ Proved
  5. \[ \left. \left(\sin{\left(2 x \right)} - 3 e^{\frac{\pi}{2}} \cos{\left(2 x \right)}\right) e^{2 x - \frac{\pi}{2}} \right|_{\substack{ x=\frac{\pi}{4} }} = 1 \]
    y(pi/4) = 1.✓ Proved
Answer \( y = \left(\sin{\left(2 x \right)} - 3 e^{\frac{\pi}{2}} \cos{\left(2 x \right)}\right) e^{2 x - \frac{\pi}{2}} \)

✓ 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 6.616

medium
Solve \( \displaystyle y - 2 y' + y'' = 0 \) with \( \displaystyle y(0) = 4,\ y(1) = 2 \).
  1. \[ m^{2} - 2 m + 1 = \left(m - 1\right)^{2} \]
    The characteristic equation has a repeated root 1.✓ Proved
  2. So y = C₁exp(x) + C₂x*exp(x); the two conditions fix C₁ and C₂.
    Reviewed
  3. \[ \left(- 4 e x + 2 x + 4 e\right) e^{x - 1} - 2 \frac{d}{d x} \left(- 4 e x + 2 x + 4 e\right) e^{x - 1} + \frac{d^{2}}{d x^{2}} \left(- 4 e x + 2 x + 4 e\right) e^{x - 1} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. 2 \left(- 2 e x + x + 2 e\right) e^{x - 1} \right|_{\substack{ x=0 }} = 4 \]
    ... y(0) = 4 ...✓ Proved
  5. \[ \left. 2 \left(- 2 e x + x + 2 e\right) e^{x - 1} \right|_{\substack{ x=1 }} = 2 \]
    y(1) = 2.✓ Proved
Answer \( y = 2 \left(- 2 e x + x + 2 e\right) e^{x - 1} \)

✓ 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 6.621

medium
Solve \( \displaystyle 10 y + 2 y' + y'' = 0 \) with \( \displaystyle y(0) = 3,\ y(1) = 3 \).
  1. \[ m^{2} + 2 m + 10 = \left(m + 1 - 3 i\right) \left(m + 1 + 3 i\right) \]
    The characteristic equation has complex roots -1 ± 3i.✓ Proved
  2. So y = C₁exp(-x)*cos(3*x) + C₂exp(-x)*sin(3*x); the two conditions fix C₁ and C₂.
    Reviewed
  3. \[ \frac{10 \left(3 e \sin{\left(3 x \right)} - 3 \sin{\left(3 x - 3 \right)}\right) e^{- x}}{\sin{\left(3 \right)}} + 2 \frac{d}{d x} \frac{\left(3 e \sin{\left(3 x \right)} - 3 \sin{\left(3 x - 3 \right)}\right) e^{- x}}{\sin{\left(3 \right)}} + \frac{d^{2}}{d x^{2}} \frac{\left(3 e \sin{\left(3 x \right)} - 3 \sin{\left(3 x - 3 \right)}\right) e^{- x}}{\sin{\left(3 \right)}} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \frac{3 \left(e \sin{\left(3 x \right)} - \sin{\left(3 x - 3 \right)}\right) e^{- x}}{\sin{\left(3 \right)}} \right|_{\substack{ x=0 }} = 3 \]
    ... y(0) = 3 ...✓ Proved
  5. \[ \left. \frac{3 \left(e \sin{\left(3 x \right)} - \sin{\left(3 x - 3 \right)}\right) e^{- x}}{\sin{\left(3 \right)}} \right|_{\substack{ x=1 }} = 3 \]
    y(1) = 3.✓ Proved
Answer \( y = \frac{3 \left(e \sin{\left(3 x \right)} - \sin{\left(3 x - 3 \right)}\right) e^{- x}}{\sin{\left(3 \right)}} \)

✓ 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 6.624

medium
Solve \( \displaystyle 6 y - 5 y' + y'' = 0 \) with \( \displaystyle y(0) = 2,\ y(1) = -1 \).
  1. \[ m^{2} - 5 m + 6 = \left(m - 3\right) \left(m - 2\right) \]
    The characteristic equation has distinct real roots 3 and 2.✓ Proved
  2. So y = C₁exp(3*x) + C₂exp(2*x); the two conditions fix C₁ and C₂.
    Reviewed
  3. \[ \frac{6 \left(\left(1 + 2 e^{2}\right) e^{x} - 2 e^{3} - 1\right) e^{2 x - 2}}{1 - e} - 5 \frac{d}{d x} \frac{\left(\left(1 + 2 e^{2}\right) e^{x} - 2 e^{3} - 1\right) e^{2 x - 2}}{1 - e} + \frac{d^{2}}{d x^{2}} \frac{\left(\left(1 + 2 e^{2}\right) e^{x} - 2 e^{3} - 1\right) e^{2 x - 2}}{1 - e} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \frac{\left(\left(1 + 2 e^{2}\right) e^{x} - 2 e^{3} - 1\right) e^{2 x - 2}}{1 - e} \right|_{\substack{ x=0 }} = 2 \]
    ... y(0) = 2 ...✓ Proved
  5. \[ \left. \frac{\left(\left(1 + 2 e^{2}\right) e^{x} - 2 e^{3} - 1\right) e^{2 x - 2}}{1 - e} \right|_{\substack{ x=1 }} = -1 \]
    y(1) = -1.✓ Proved
Answer \( y = \frac{\left(\left(1 + 2 e^{2}\right) e^{x} - 2 e^{3} - 1\right) e^{2 x - 2}}{1 - e} \)

✓ 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 6.625

medium
Solve \( \displaystyle 2 y - 2 y' + y'' = 0 \) with \( \displaystyle y(0) = 4,\ y'(0) = 1 \).
  1. \[ m^{2} - 2 m + 2 = \left(m - 1 - i\right) \left(m - 1 + i\right) \]
    The characteristic equation has complex roots 1 ± 1i.✓ Proved
  2. So y = C₁exp(x)*cos(x) + C₂exp(x)*sin(x); the two conditions fix C₁ and C₂.
    Reviewed
  3. \[ 2 \left(- 3 \sin{\left(x \right)} + 4 \cos{\left(x \right)}\right) e^{x} - 2 \frac{d}{d x} \left(- 3 \sin{\left(x \right)} + 4 \cos{\left(x \right)}\right) e^{x} + \frac{d^{2}}{d x^{2}} \left(- 3 \sin{\left(x \right)} + 4 \cos{\left(x \right)}\right) e^{x} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \left(- 3 \sin{\left(x \right)} + 4 \cos{\left(x \right)}\right) e^{x} \right|_{\substack{ x=0 }} = 4 \]
    ... y(0) = 4 ...✓ Proved
  5. \[ \left. \frac{d}{d x} \left(- 3 \sin{\left(x \right)} + 4 \cos{\left(x \right)}\right) e^{x} \right|_{\substack{ x=0 }} = 1 \]
    y′(0) = 1.✓ Proved
Answer \( y = \left(- 3 \sin{\left(x \right)} + 4 \cos{\left(x \right)}\right) e^{x} \)

✓ 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 6.648

medium
Solve \( \displaystyle 20 y - 4 y' + y'' = 0 \) with \( \displaystyle y(0) = -2,\ y(1) = -3 \).
  1. \[ m^{2} - 4 m + 20 = \left(m - 2 - 4 i\right) \left(m - 2 + 4 i\right) \]
    The characteristic equation has complex roots 2 ± 4i.✓ Proved
  2. So y = C₁exp(2*x)*cos(4*x) + C₂exp(2*x)*sin(4*x); the two conditions fix C₁ and C₂.
    Reviewed
  3. \[ \frac{20 \left(- 3 \sin{\left(4 x \right)} + 2 e^{2} \sin{\left(4 x - 4 \right)}\right) e^{2 x - 2}}{\sin{\left(4 \right)}} - 4 \frac{d}{d x} \frac{\left(- 3 \sin{\left(4 x \right)} + 2 e^{2} \sin{\left(4 x - 4 \right)}\right) e^{2 x - 2}}{\sin{\left(4 \right)}} + \frac{d^{2}}{d x^{2}} \frac{\left(- 3 \sin{\left(4 x \right)} + 2 e^{2} \sin{\left(4 x - 4 \right)}\right) e^{2 x - 2}}{\sin{\left(4 \right)}} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \frac{\left(- 3 \sin{\left(4 x \right)} + 2 e^{2} \sin{\left(4 x - 4 \right)}\right) e^{2 x - 2}}{\sin{\left(4 \right)}} \right|_{\substack{ x=0 }} = -2 \]
    ... y(0) = -2 ...✓ Proved
  5. \[ \left. \frac{\left(- 3 \sin{\left(4 x \right)} + 2 e^{2} \sin{\left(4 x - 4 \right)}\right) e^{2 x - 2}}{\sin{\left(4 \right)}} \right|_{\substack{ x=1 }} = -3 \]
    y(1) = -3.✓ Proved
Answer \( y = \frac{\left(- 3 \sin{\left(4 x \right)} + 2 e^{2} \sin{\left(4 x - 4 \right)}\right) e^{2 x - 2}}{\sin{\left(4 \right)}} \)

✓ 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 6.651

medium
Solve \( \displaystyle 12 y + 7 y' + y'' = 0 \) with \( \displaystyle y(0) = -1,\ y'(0) = -2 \).
  1. \[ m^{2} + 7 m + 12 = \left(m + 3\right) \left(m + 4\right) \]
    The characteristic equation has distinct real roots -4 and -3.✓ Proved
  2. So y = C₁exp(-4*x) + C₂exp(-3*x); the two conditions fix C₁ and C₂.
    Reviewed
  3. \[ 12 \left(5 - 6 e^{x}\right) e^{- 4 x} + 7 \frac{d}{d x} \left(5 - 6 e^{x}\right) e^{- 4 x} + \frac{d^{2}}{d x^{2}} \left(5 - 6 e^{x}\right) e^{- 4 x} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \left(5 - 6 e^{x}\right) e^{- 4 x} \right|_{\substack{ x=0 }} = -1 \]
    ... y(0) = -1 ...✓ Proved
  5. \[ \left. \frac{d}{d x} \left(5 - 6 e^{x}\right) e^{- 4 x} \right|_{\substack{ x=0 }} = -2 \]
    y′(0) = -2.✓ Proved
Answer \( y = \left(5 - 6 e^{x}\right) e^{- 4 x} \)

✓ 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 6.654

medium
Solve \( \displaystyle - 12 y + y' + y'' = 0 \) with \( \displaystyle y(0) = 2,\ y(1) = -2 \).
  1. \[ m^{2} + m - 12 = \left(m - 3\right) \left(m + 4\right) \]
    The characteristic equation has distinct real roots -4 and 3.✓ Proved
  2. So y = C₁exp(-4*x) + C₂exp(3*x); the two conditions fix C₁ and C₂.
    Reviewed
  3. \[ \frac{\left(- 24 \left(1 + e^{4}\right) e^{7 x} + 24 \left(1 + e^{3}\right) e^{4}\right) e^{- 4 x}}{1 - e^{7}} + \frac{d}{d x} \frac{\left(2 \left(1 + e^{4}\right) e^{7 x} - 2 \left(1 + e^{3}\right) e^{4}\right) e^{- 4 x}}{1 - e^{7}} + \frac{d^{2}}{d x^{2}} \frac{\left(2 \left(1 + e^{4}\right) e^{7 x} - 2 \left(1 + e^{3}\right) e^{4}\right) e^{- 4 x}}{1 - e^{7}} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \frac{2 \left(\left(1 + e^{4}\right) e^{7 x} - \left(1 + e^{3}\right) e^{4}\right) e^{- 4 x}}{1 - e^{7}} \right|_{\substack{ x=0 }} = 2 \]
    ... y(0) = 2 ...✓ Proved
  5. \[ \left. \frac{2 \left(\left(1 + e^{4}\right) e^{7 x} - \left(1 + e^{3}\right) e^{4}\right) e^{- 4 x}}{1 - e^{7}} \right|_{\substack{ x=1 }} = -2 \]
    y(1) = -2.✓ Proved
Answer \( y = \frac{2 \left(\left(1 + e^{4}\right) e^{7 x} - \left(1 + e^{3}\right) e^{4}\right) e^{- 4 x}}{1 - e^{7}} \)

✓ 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 6.712

medium
Solve \( \displaystyle 6 y - 5 y' + y'' = 0 \) with \( \displaystyle y(0) = -2,\ y(1) = 4 \).
  1. \[ m^{2} - 5 m + 6 = \left(m - 3\right) \left(m - 2\right) \]
    The characteristic equation has distinct real roots 3 and 2.✓ Proved
  2. So y = C₁exp(3*x) + C₂exp(2*x); the two conditions fix C₁ and C₂.
    Reviewed
  3. \[ \frac{6 \left(2 \left(- e^{2} - 2\right) e^{x} + 4 + 2 e^{3}\right) e^{2 x - 2}}{1 - e} - 5 \frac{d}{d x} \frac{\left(2 \left(- e^{2} - 2\right) e^{x} + 4 + 2 e^{3}\right) e^{2 x - 2}}{1 - e} + \frac{d^{2}}{d x^{2}} \frac{\left(2 \left(- e^{2} - 2\right) e^{x} + 4 + 2 e^{3}\right) e^{2 x - 2}}{1 - e} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \frac{2 \left(\left(- e^{2} - 2\right) e^{x} + 2 + e^{3}\right) e^{2 x - 2}}{1 - e} \right|_{\substack{ x=0 }} = -2 \]
    ... y(0) = -2 ...✓ Proved
  5. \[ \left. \frac{2 \left(\left(- e^{2} - 2\right) e^{x} + 2 + e^{3}\right) e^{2 x - 2}}{1 - e} \right|_{\substack{ x=1 }} = 4 \]
    y(1) = 4.✓ Proved
Answer \( y = \frac{2 \left(- \left(2 + e^{2}\right) e^{x} + 2 + e^{3}\right) e^{2 x - 2}}{1 - e} \)

✓ 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 6.715

medium
Solve \( \displaystyle - 4 y + y'' = 0 \) with \( \displaystyle y(0) = 2,\ y(1) = -3 \).
  1. \[ m^{2} - 4 = \left(m - 2\right) \left(m + 2\right) \]
    The characteristic equation has distinct real roots 2 and -2.✓ Proved
  2. So y = C₁exp(2*x) + C₂exp(-2*x); the two conditions fix C₁ and C₂.
    Reviewed
  3. \[ \frac{\left(- 4 \left(2 + 3 e^{2}\right) e^{4 x} + 4 \left(3 + 2 e^{2}\right) e^{2}\right) e^{- 2 x}}{1 - e^{4}} + \frac{d^{2}}{d x^{2}} \frac{\left(\left(2 + 3 e^{2}\right) e^{4 x} - \left(3 + 2 e^{2}\right) e^{2}\right) e^{- 2 x}}{1 - e^{4}} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \frac{\left(\left(2 + 3 e^{2}\right) e^{4 x} - \left(3 + 2 e^{2}\right) e^{2}\right) e^{- 2 x}}{1 - e^{4}} \right|_{\substack{ x=0 }} = 2 \]
    ... y(0) = 2 ...✓ Proved
  5. \[ \left. \frac{\left(\left(2 + 3 e^{2}\right) e^{4 x} - \left(3 + 2 e^{2}\right) e^{2}\right) e^{- 2 x}}{1 - e^{4}} \right|_{\substack{ x=1 }} = -3 \]
    y(1) = -3.✓ Proved
Answer \( y = \frac{\left(\left(2 + 3 e^{2}\right) e^{4 x} - \left(3 + 2 e^{2}\right) e^{2}\right) e^{- 2 x}}{1 - e^{4}} \)

✓ 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 6.799

medium
Solve \( \displaystyle 4 y + 5 y' + y'' = 0 \) with \( \displaystyle y(0) = 4,\ y(2) = -4 \).
  1. \[ m^{2} + 5 m + 4 = \left(m + 1\right) \left(m + 4\right) \]
    The characteristic equation has distinct real roots -4 and -1.✓ Proved
  2. So y = C₁exp(-4*x) + C₂exp(-x); the two conditions fix C₁ and C₂.
    Reviewed
  3. \[ \frac{4 \left(4 \left(1 + e^{8}\right) e^{3 x} - 4 \left(1 + e^{2}\right) e^{6}\right) e^{- 4 x}}{1 - e^{6}} + 5 \frac{d}{d x} \frac{\left(4 \left(1 + e^{8}\right) e^{3 x} - 4 \left(1 + e^{2}\right) e^{6}\right) e^{- 4 x}}{1 - e^{6}} + \frac{d^{2}}{d x^{2}} \frac{\left(4 \left(1 + e^{8}\right) e^{3 x} - 4 \left(1 + e^{2}\right) e^{6}\right) e^{- 4 x}}{1 - e^{6}} = 0 \]
    It satisfies the equation ...✓ Proved
  4. \[ \left. \frac{4 \left(\left(1 + e^{8}\right) e^{3 x} - \left(1 + e^{2}\right) e^{6}\right) e^{- 4 x}}{1 - e^{6}} \right|_{\substack{ x=0 }} = 4 \]
    ... y(0) = 4 ...✓ Proved
  5. \[ \left. \frac{4 \left(\left(1 + e^{8}\right) e^{3 x} - \left(1 + e^{2}\right) e^{6}\right) e^{- 4 x}}{1 - e^{6}} \right|_{\substack{ x=2 }} = -4 \]
    y(2) = -4.✓ Proved
Answer \( y = \frac{4 \left(\left(1 + e^{8}\right) e^{3 x} - \left(1 + e^{2}\right) e^{6}\right) e^{- 4 x}}{1 - e^{6}} \)

✓ 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

More second-order initial and boundary value problems practice

Solve \( \displaystyle 8 y + 6 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y(1) = 1 \).
Problem 6.471medium✓ Every equation proved
Solve \( \displaystyle 8 y + 6 y' + y'' = 0 \) with \( \displaystyle y(0) = 3,\ y'(0) = 0 \).
Problem 6.472medium✓ Every equation proved
Solve \( \displaystyle 5 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 2,\ y'(0) = -4 \).
Problem 6.473medium✓ Every equation proved
Solve \( \displaystyle 8 y + 6 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y'(0) = 2 \).
Problem 6.474medium✓ Every equation proved
Solve \( \displaystyle 2 y - 2 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y'(0) = -3 \).
Problem 6.475medium✓ Every equation proved
Solve \( \displaystyle - 4 y + 3 y' + y'' = 0 \) with \( \displaystyle y(0) = 4,\ y'(0) = 4 \).
Problem 6.477medium✓ Every equation proved
Solve \( \displaystyle - 9 y + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y'(0) = 4 \).
Problem 6.478medium✓ Every equation proved
Solve \( \displaystyle 5 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 2,\ y'(0) = 3 \).
Problem 6.479medium✓ Every equation proved
Solve \( \displaystyle 20 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 1,\ y'(0) = 2 \).
Problem 6.480medium✓ Every equation proved
Solve \( \displaystyle 13 y - 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 3,\ y'(0) = 2 \).
Problem 6.481medium✓ Every equation proved
Solve \( \displaystyle y - 2 y' + y'' = 0 \) with \( \displaystyle y(0) = 1,\ y'(0) = -2 \).
Problem 6.482medium✓ Every equation proved
Solve \( \displaystyle 6 y - 5 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y(1) = -3 \).
Problem 6.483medium✓ Every equation proved
Solve \( \displaystyle - 2 y + y' + y'' = 0 \) with \( \displaystyle y(0) = -2,\ y'(0) = 1 \).
Problem 6.484medium✓ Every equation proved
Solve \( \displaystyle 20 y - 4 y' + y'' = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = -2 \).
Problem 6.485medium✓ Every equation proved
Solve \( \displaystyle 4 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 2,\ y(1) = -2 \).
Problem 6.617medium✓ Every equation proved
Solve \( \displaystyle y + y'' = 0 \) with \( \displaystyle y(0) = 4,\ y(\frac{\pi}{4}) = -1 \).
Problem 6.618medium✓ Every equation proved
Solve \( \displaystyle - 3 y + 2 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y'(0) = 1 \).
Problem 6.619medium✓ Every equation proved
Solve \( \displaystyle 5 y - 2 y' + y'' = 0 \) with \( \displaystyle y(0) = 4,\ y'(0) = 0 \).
Problem 6.620medium✓ Every equation proved
Solve \( \displaystyle 9 y - 6 y' + y'' = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = 3 \).
Problem 6.622medium✓ Every equation proved
Solve \( \displaystyle 3 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = -2 \).
Problem 6.623medium✓ Every equation proved
Solve \( \displaystyle 5 y + 2 y' + y'' = 0 \) with \( \displaystyle y(0) = -2,\ y(1) = -3 \).
Problem 6.626medium✓ Every equation proved
Solve \( \displaystyle - 4 y + 3 y' + y'' = 0 \) with \( \displaystyle y(0) = -1,\ y'(0) = 2 \).
Problem 6.627medium✓ Every equation proved
Solve \( \displaystyle 4 y - 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y'(0) = 4 \).
Problem 6.628medium✓ Every equation proved
Solve \( \displaystyle y + 2 y' + y'' = 0 \) with \( \displaystyle y(0) = -1,\ y(1) = -3 \).
Problem 6.629medium✓ Every equation proved
Solve \( \displaystyle - 2 y - y' + y'' = 0 \) with \( \displaystyle y(0) = 4,\ y'(0) = -3 \).
Problem 6.630medium✓ Every equation proved
Solve \( \displaystyle 4 y + y'' = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = -4 \).
Problem 6.649medium✓ Every equation proved
Solve \( \displaystyle 3 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = -4 \).
Problem 6.650medium✓ Every equation proved
Solve \( \displaystyle - 6 y - y' + y'' = 0 \) with \( \displaystyle y(0) = -2,\ y'(0) = 1 \).
Problem 6.652medium✓ Every equation proved
Solve \( \displaystyle 8 y - 4 y' + y'' = 0 \) with \( \displaystyle y(0) = -3,\ y(\frac{\pi}{4}) = 2 \).
Problem 6.653medium✓ Nihil obstat
Solve \( \displaystyle 4 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y(2) = 0 \).
Problem 6.655medium✓ Every equation proved
Solve \( \displaystyle 3 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 2,\ y(1) = 2 \).
Problem 6.656medium✓ Nihil obstat
Solve \( \displaystyle 9 y + 6 y' + y'' = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = -4 \).
Problem 6.657medium✓ Nihil obstat
Solve \( \displaystyle 4 y - 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 2,\ y'(0) = -3 \).
Problem 6.709medium✓ Every equation proved
Solve \( \displaystyle 3 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = -2,\ y'(0) = -2 \).
Problem 6.710medium✓ Every equation proved
Solve \( \displaystyle 4 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = 2 \).
Problem 6.711medium✓ Every equation proved
Solve \( \displaystyle - 12 y + y' + y'' = 0 \) with \( \displaystyle y(0) = -3,\ y'(0) = -4 \).
Problem 6.713medium✓ Every equation proved
Solve \( \displaystyle 3 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 3,\ y'(0) = -4 \).
Problem 6.714medium✓ Every equation proved
Solve \( \displaystyle - 3 y + 2 y' + y'' = 0 \) with \( \displaystyle y(0) = 3,\ y'(0) = 0 \).
Problem 6.716medium✓ Every equation proved
Solve \( \displaystyle 6 y - 5 y' + y'' = 0 \) with \( \displaystyle y(0) = -2,\ y'(0) = -1 \).
Problem 6.717medium✓ Every equation proved
Solve \( \displaystyle 6 y + 5 y' + y'' = 0 \) with \( \displaystyle y(0) = -1,\ y'(0) = -4 \).
Problem 6.718medium✓ Every equation proved
Solve \( \displaystyle 3 y + 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 4,\ y(2) = 2 \).
Problem 6.800medium✓ Nihil obstat
Solve \( \displaystyle 5 y - 4 y' + y'' = 0 \) with \( \displaystyle y(0) = 1,\ y'(0) = 4 \).
Problem 6.801medium✓ Nihil obstat
Solve \( \displaystyle 9 y + 6 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y'(0) = 1 \).
Problem 6.802medium✓ Every equation proved
Solve \( \displaystyle - 2 y + y' + y'' = 0 \) with \( \displaystyle y(0) = 2,\ y'(0) = -4 \).
Problem 6.803medium✓ Every equation proved
Solve \( \displaystyle 9 y + 6 y' + y'' = 0 \) with \( \displaystyle y(0) = 3,\ y'(0) = -1 \).
Problem 6.804medium✓ Every equation proved
Solve \( \displaystyle - 2 y - y' + y'' = 0 \) with \( \displaystyle y(0) = -2,\ y'(0) = 3 \).
Problem 6.805medium✓ Every equation proved
Solve \( \displaystyle - 6 y + y' + y'' = 0 \) with \( \displaystyle y(0) = -1,\ y'(0) = 0 \).
Problem 6.806medium✓ Every equation proved
Solve \( \displaystyle 6 y - 5 y' + y'' = 0 \) with \( \displaystyle y(0) = 0,\ y(2) = 1 \).
Problem 6.807medium✓ Every equation proved
Solve \( \displaystyle 9 y - 6 y' + y'' = 0 \) with \( \displaystyle y(0) = 4,\ y(1) = -4 \).
Problem 6.808medium✓ Every equation proved