Question

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$y^{\prime \prime \prime}+2 y^{\prime \prime}+2 y^{\prime}=0$$

Short answer

The given differential equation is a third-order homogeneous linear differential equation. The general solution is \[ y(t) = C_1 + e^{-t }( C_2 cos(t) + C_3 sin(t) )\]

Step-by-step solution

Identify the type of the differential equation

The given differential equation is a third-order homogeneous linear differential equation: \[ y^{\text{'''} } + 2 y^{\text{''} } + 2 y^{\text{'}} = 0 \]

Form the characteristic equation

For a homogeneous linear differential equation of the form \[ a_n y^{(n)} + a_{n-1} y^{(n-1)} + \tdots + a_1 y' + a_0 y = 0 \], the characteristic equation is obtained by replacing each derivative term with a power of \(r\). Thus, the characteristic equation for \[ y^{\text{'''}} + 2 y^{\text{''}} + 2 y^{\text{'}} = 0 \] is: \[ r^3 + 2r^2 + 2r = 0 \]

Factor the characteristic equation

To find the roots, first factor out the common factor:\[ r (r^2 + 2r + 2) = 0 \]

Solve the factored equation

Set each factor equal to zero to find the roots:\[ r = 0 \] and solve the quadratic equation: \[ r^2 + 2r + 2 = 0 \] Using the quadratic formula \[ r = \frac{-b ± √{b^2 - 4ac}}{2a} \] where \( a = 1, b = 2, c = 2 \):\[ r = \frac{-2 ±√{2^2 - 4 (1)(2)}}{2 (1)} = \frac{-2 ±√{4 - 8}}{2} = \frac{-2 ± √{-4}}{2} = \frac{-2 ± 2i}{2} = -1 ± i \] So the roots are: \[ r = 0, -1 + i, -1 - i \]

Write the general solution

Given that the roots are real and complex, the general solution form for the differential equation with roots \(r_1, \alpha ± i \beta\): \[ y = C_1 e^{r_1 t} + e^{ α t} ( C_2 cos( β t ) + C_3 sin( β t ) ) \] For the roots \( 0, -1+i\), and \( -1-i \): \[ y(t) = C_1 + e^{-t} ( C_2 cos( t ) + C_3 sin( t ) ) \]}

Key concepts

Homogeneous Linear Differential Equation

A homogeneous linear differential equation is a type of differential equation where the dependent variable and all its derivatives appear linearly. This means you won't see any powers or products of the dependent variable or its derivatives in the equation.

The general form is: