Second Order Differential Equations

Lineal second order differential equations are of the form:

p (t) y^{″} + q (t) y^{'} + r (t) y = g (t)

Usually we look at constant coefficient second order linear differential equations:

a y^{″} + b y^{'} + c y = g (t)

If $g (t) = 0$ then the D.E is homogeneous and if $g (t) \neq 0$ the D.E. is non-homogeneous

Homogeneous , constant coefficient, linear, 2nd order D.E

a y^{″} + b y^{'} + c y = 0

Principle of superposition

If $y_{1} (t)$ and $y_{2} (t)$ are two solutions to a linear, homogeneous differential equation then so is any linear combination:$$y(t) = c_1 y_1(t)+c_2 y_2(t)$$

If we further assume that the D.E. is 2nd order then the general solution will be this superposition.

Let's assume that solutions will be of the form: $$y(t) =e^{rt}$$
and its derivatives : $y^{'} (t) = r e^{r t}$ and $y^{″} (t) = r^{2} e^{r t}$ . We plug this into the differential:

a y^{″} + b y^{'} + c y = 0

a * r^{2} e^{r t} + b * r e^{r t} + c e^{r t} = 0

e^{r t} * (a * r^{2} + b * r + c) = 0

Exponentials can never be zero, so if indeed the solutions are of this form then it will be true that:

a * r^{2} + b * r + c = 0

This is the Characteristic equation for the initial D.E. It is a quadratic equation, meaning it will have two root $r_{1}$ and $r_{2}$ and therefore two solutions: $$y_1 =e^{r_1t}$$

y_{2} = e^{r_{2} t}

These two roots have 3 possible forms :

Real distinct roots, r1≠r2

a y^{″} + b y^{'} + c y = 0

a * r^{2} + b * r + c = 0

y (t) = c_{1} e^{r_{1} t} + c_{2} e^{r_{2} t}

And solve for initial conditions or boundary conditions

Complex roots r1,2 = λ±μ i

If the roots are of the form

r_{1, 2} = λ \pm μ i

and the corresponding solutions

y_{1} (t) = e^{λ + μ} & y_{2} (t) = e^{λ - μ}

Since we started only with real numbers the solutions must also be real. We can find 2 new real solutions by adding and subtracting y1 and y2 and then superpose those to get the general solution
Euler's Formula :

e^{i θ} = c o s θ + i s i n θ

e^{- i θ} = c o s θ - i s i n θ

We split the solutions to exponentials with real and imaginary components:

y_{1} = e^{λ t} e^{i μ t} = e^{λ t} (c o s (μ t) + i s i n (μ t))

y_{2} = e^{λ t} e^{- i μ t} = e^{λ t} (c o s (μ t) - i s i n (μ t))

Linear combination of the two solutions to arrive at a real solution:

u_{1} (t) = \frac{1}{2} y_{1} + \frac{1}{2} y_{2} = \frac{1}{2} 2 e^{λ t} c o s (μ t) = e^{λ t} c o s (μ t)

u_{2} (t) = \frac{1}{2 i} y_{1} - \frac{1}{2 i} y_{2} = \frac{1}{2 i} 2 i e^{λ t} s i n (μ t) = e^{λ t} s i n (μ t)

So the General solution can be written as the linear combination of u1 and u2:

y (t) = c_{1} e^{λ t} c o s (μ t) + c_{2} e^{λ t} s i n (μ t)

Double roots, r1=r2=r

r_{1} = r_{2} = r

y_{1} (t) = y_{2} (t) = e^{r t}

We have only one solution and we need a second one (2nd order). The roots of the quadratic are:

r_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Since we have a double root :

b^{2} - 4 a c = 0

r_{1, 2} = \frac{- b}{2 a}

The first solution will be :

y_{1} (t) = e^{\frac{- b t}{2 a}}

Let's suppose that there is a second solution in the form of :

y_{2} = u (t) y_{1} (t) = u (t) e^{\frac{- b t}{2 a}}

If we plug y2 and its derivatives on the initial D.E we will end up with this equation:

e^{- \frac{b t}{2 a}} (a u^{″} (t) - \frac{1}{4 a} (b^{2} - 4 a c) u (t)) = 0

but the second part is 0 and exponentials and the constant a cannot be zero (otherwise it would not be 2nd order):

u^{″} (t) = 0

u (t) = c t + k

The c and k constants can be integrated into the constant of the general solution and we are left with:

y (t) = c_{1} e^{r t} + c_{2} t e^{r t}