Równania różniczkowe liniowe rzędu II o stałych współczynnikach:

1. y’’-2y’+y=0

2. y’’-y’-2y=0

3. y’’+6y’+25y=0

4. y’’-2y=0

5. y’’+3y’+2y=0

6. y’’+2y’+2y=0

a) y’’-2y’+y=0

τ² − 2τ + 1 = 0 

=(−2)² − 4 * 1 * 1 = 0

 τ₁ = τ₂ 

$$\ \tau_{1} = \frac{- b}{2a} = \frac{2}{2} = 1\ $$

y(t) = c₁ * e^t +  c₂ * te^t

b)y’’-y’-2y=0

τ² − τ − 2 = 0

  = 1² − 4 * (−2) * 1 = 9

$$\text{\ \ }\sqrt{} = 3\ $$

$$\text{\ \ }\tau_{1} = \frac{- b - \ \sqrt{}}{2a} = \frac{1 - 3}{2} = - 1 = > y_{1} = e^{- 1t}\ $$

$$\text{\ \ \ }\tau_{2} = \frac{- b + \ \sqrt{}}{2a} = \frac{1 + 3}{2} = 2 = > y_{2} = e^{2t}\ $$

y(t) = c₁ * e^−t +  c₂ * e^2t

c)y’’+6y’+25y=0

τ² + 6τ + 25 = 0

  = 6² − 4 * 25 * 1 = 36 − 100 = 64j²

$$\text{\ \ }\sqrt{} = 8j\ $$

$$\text{\ \ }\tau_{1} = \frac{- b - \ \sqrt{}}{2a} = \frac{- 6 - 8j}{2} = - 3 - 4j = > y_{1} = e^{- 3t}cos4t\ $$

$$\text{\ \ \ }\tau_{2} = \frac{- b + \ \sqrt{}}{2a} = \frac{- 6 + 8j}{2} = - 3 + 4 = > y_{2} = e^{- 3t}sin4t\ $$

y(t) = e^−3t(c₁cos4t+c₂sin4t) c_1,c₂ ∈ R

e) y’’+3y’+2y=0

τ² + 3τ + 2 = 0

  = 3² − 4 * 2 * 1 = 1

$$\sqrt{} = 1$$

$$\text{\ \ }\tau_{1} = \frac{- b - \ \sqrt{}}{2a} = \frac{- 3 - 1}{2} = - 2 = > y_{1} = e^{- 2t}\ $$

$$\text{\ \ \ }\tau_{2} = \frac{- b + \ \sqrt{}}{2a} = \frac{- 3 + 1}{2} = - 1 = > y_{2} = e^{- t}$$

y(t) = c₁ * e^−2t +  c₂ * te^−t

F)y’’+2y’+2y=0

τ² + 2τ + 2 = 0

  = 2² − 4 * 2 * 1 = 0  

τ₁ = τ₂

$$\ \ \tau = \frac{- b}{2a} = \frac{- 2}{2} = - 1 = > y_{1} = e^{- t}\ $$

y(t) = c₁ * e^−t +  c₂ * te^−t