Math Problem Statement

Löse mit Rechnung

Lösen Sie die folgenden Differentialgleichungen: a) y'' + y' - 6y = 0 b) y'' + 6y' + 9y = 0 c) y'' - 4y' + 5y = 0 d) y'' + 2y' + 2y = 0

Solution

The provided image contains a set of linear homogeneous second-order differential equations. Let's solve each of them step-by-step:

General Approach:

For second-order linear homogeneous differential equations of the form:

y+ay+by=0y'' + a y' + b y = 0

we can find solutions by solving the characteristic equation:

r2+ar+b=0r^2 + a r + b = 0

The nature of the roots r1r_1 and r2r_2 will determine the general solution of the differential equation:

  1. If r1r_1 and r2r_2 are real and distinct: y(t)=C1er1t+C2er2ty(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}
  2. If r1=r2r_1 = r_2 (real and equal): y(t)=(C1+C2t)er1ty(t) = (C_1 + C_2 t) e^{r_1 t}
  3. If r1,r2r_1, r_2 are complex conjugates: y(t)=eαt(C1cos(βt)+C2sin(βt))y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) where r1,r2=α±iβr_1, r_2 = \alpha \pm i \beta

Now let's solve each part.

a) y+y6y=0y'' + y' - 6y = 0

  1. Characteristic equation: r2+r6=0r^2 + r - 6 = 0

  2. Solve using the quadratic formula: r=1±124(1)(6)2(1)=1±1+242=1±52r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-6)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2} Roots are r1=2r_1 = 2 and r2=3r_2 = -3.

  3. General solution: y(t)=C1e2t+C2e3ty(t) = C_1 e^{2t} + C_2 e^{-3t}

b) y+6y+9y=0y'' + 6y' + 9y = 0

  1. Characteristic equation: r2+6r+9=0r^2 + 6r + 9 = 0

  2. Solve using the quadratic formula: r=6±624(1)(9)2(1)=6±36362=62=3r = \frac{-6 \pm \sqrt{6^2 - 4(1)(9)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 36}}{2} = \frac{-6}{2} = -3 Root r1=r2=3r_1 = r_2 = -3 (repeated root).

  3. General solution: y(t)=(C1+C2t)e3ty(t) = (C_1 + C_2 t) e^{-3t}

c) y4y+5y=0y'' - 4y' + 5y = 0

  1. Characteristic equation: r24r+5=0r^2 - 4r + 5 = 0

  2. Solve using the quadratic formula: r=4±(4)24(1)(5)2(1)=4±16202=4±42r = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)} = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2} r=2±ir = 2 \pm i, which are complex roots.

  3. General solution: y(t)=e2t(C1cos(t)+C2sin(t))y(t) = e^{2t} (C_1 \cos(t) + C_2 \sin(t))

d) y+2y+2y=0y'' + 2y' + 2y = 0

  1. Characteristic equation: r2+2r+2=0r^2 + 2r + 2 = 0

  2. Solve using the quadratic formula: r=2±224(1)(2)2(1)=2±482=2±42r = \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} r=1±ir = -1 \pm i, which are complex roots.

  3. General solution: y(t)=et(C1cos(t)+C2sin(t))y(t) = e^{-t} (C_1 \cos(t) + C_2 \sin(t))

Summary of Solutions:

  • a) y(t)=C1e2t+C2e3ty(t) = C_1 e^{2t} + C_2 e^{-3t}
  • b) y(t)=(C1+C2t)e3ty(t) = (C_1 + C_2 t) e^{-3t}
  • c) y(t)=e2t(C1cos(t)+C2sin(t))y(t) = e^{2t} (C_1 \cos(t) + C_2 \sin(t))
  • d) y(t)=et(C1cos(t)+C2sin(t))y(t) = e^{-t} (C_1 \cos(t) + C_2 \sin(t))

Would you like further explanation on any of these? Let me know!

Here are 5 related questions:

  1. How do we handle non-homogeneous differential equations?
  2. Can we use other methods, like Laplace transforms, to solve these equations?
  3. What if the differential equation had more than two terms in derivatives?
  4. How do we find the particular solution for non-homogeneous cases?
  5. What if the characteristic equation had complex conjugates with different coefficients?

Tip: Always check the discriminant b24acb^2 - 4ac of the characteristic equation to determine the type of roots!

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Characteristic Equation
Complex Roots
Homogeneous Equations

Formulas

Characteristic Equation: r^2 + a*r + b = 0
Quadratic Formula: r = [-b ± √(b² - 4ac)] / 2a

Theorems

General Solution of Homogeneous Linear Differential Equations
Quadratic Formula Theorem

Suitable Grade Level

Undergraduate Mathematics or Higher