Question

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

Short answer

This is a third-order linear homogeneous differential equation. Its 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 Differential Equation

The given differential equation is $y^{\prime \prime \prime }+2y^{\prime \prime }+2y^{\prime }=0$. This equation is a third-order linear homogeneous differential equation with constant coefficients.

Write the Characteristic Equation

For a linear homogeneous differential equation with constant coefficients, solve the characteristic equation. The characteristic equation for $y^{\prime \prime \prime }+2y^{\prime \prime }+2y^{\prime }=0$ is \lambda^3 + 2\lambda^2 + 2\lambda = 0\.

Solve the Characteristic Equation

Factor out \lambda\ from the characteristic equation ${\lambda }^3+2{\lambda }^2+2\lambda =0$: \lambda(\lambda^2 + 2\lambda + 2) = 0\. This gives us \lambda = 0 \ and the quadratic equation \lambda^2 + 2\lambda + 2 = 0\.

Solve the Quadratic Equation

Solve the quadratic equation ${\lambda }^2+2\lambda +2=0$ using the quadratic formula: $\lambda =\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. Here, $a=1,b=2,c=2$. Substituting these values in gives: \lambda = \frac{-2 \pm \sqrt{4 - 8}}{2}\, which simplifies to \lambda = -1 \pm i\.

Write the General Solution

The characteristic roots are ${\lambda }_1=0$, ${\lambda }_2=-1+i$, and ${\lambda }_3=-1-i$. The corresponding general solution of the differential equation is \y(t) = C_1 + e^{-t}(C_2 \cos(t) + C_3 \sin(t))\.

Key concepts

homogeneous differential equations

Homogeneous differential equations are a specific type of differential equation. These equations have terms that all involve the unknown function or its derivatives. In other words, there are no constant or independent functions added or subtracted.
Consider the given equation: $y^{\text{'''}}+2y^{\text{''}}+2y^{\text{'}}=0$. This is a homogeneous differential equation because all the terms involve the function $y$ and its derivatives. There are no extra terms.
Homogeneous differential equations often result in simpler calculations and solvable forms compared to non-homogeneous ones, where terms between constants and functions unrelated to $y$ or its derivatives appear. In this context, recognizing that our equation is homogeneous helps us decide our approach to solving it.

characteristic equation

The characteristic equation is a crucial tool for solving linear homogeneous differential equations with constant coefficients. This equation transforms the differential equation into an algebraic equation whose roots help determine the solution.
For our given equation, $y^{\text{'''}}+2y^{\text{''}}+2y^{\text{'}}=0$, the characteristic equation is formed by substituting $y$ with $e^{\text{lambda t}}$, resulting in the polynomial:
$${\text{lambda}}^3+2{\text{lambda}}^2+2\text{lambda}=0$$.
Once we have this polynomial, we can find its roots, which play a key role in determining the general solution to the differential equation.

quadratic formula

The quadratic formula is a method used to find the roots of quadratic equations of the form $a{x}^2+bx+c=0$. The general form of the quadratic formula is:
$$\text{lambda}=\frac{-b\text{±}\text{√}(b^2-4ac)}{2a}$$.
In our problem, we need the roots of the quadratic equation ${\text{lambda}}^2+2\text{lambda}+2=0$.
Here, $a=1,b=2,c=2$. Substituting these values into the quadratic formula gives:
$$\text{lambda}=\frac{-2\text{±}\text{√}(2^2-4\text{∙}1\text{∙}2)}{2\text{∙}1}=\frac{-2\text{±}\text{√}(4-8)}{2}=\frac{-2\text{±}\text{√}(-4)}{2}=\frac{-2\text{±}2i}{2}=-1\text{±}i$$.
Therefore, the roots are $\text{lambda}=-1+i$ and $\text{lambda}=-1-i$. Here, the real part is $-1$ and the imaginary part $i$ represents oscillatory behavior in the solution.

general solutions of differential equations

The general solution to a linear differential equation combines the solutions corresponding to each root of its characteristic equation. For the equation $y^{\text{'''}}+2y^{\text{''}}+2y^{\text{'}}=0$, we identified roots ${\text{text{lambda}}}_1=0,{\text{text{lambda}}}_2=-1+i,{\text{text{lambda}}}_3=-1-i$.
For simple roots like ${\text{text{lambda}}}_1=0$, the solution is a constant $C_1$. For complex roots $-1\text{±}i$, the solutions include exponential and trigonometric functions. Hence, the terms are combinations of $e^{-1\text{t}}\text{cos(t)}$ and $e^{-1\text{t}}\text{sin(t)}$.
Thus, the general solution combines these:
$$y(t)=C_1+e^{\text{-t}}(C_2\text{cos(t)}+C_3\text{sin(t)})$$.
Here, $C_1$, $C_2$, and $C_3$ are arbitrary constants determined by initial conditions set for the specific problem context.