Answer to Question #154989 in Differential Equations for ali

Question

Solve Second Order (linear and homogenous) Differential Equation 𝑑²𝑦 /𝑑𝑥² + 𝑝(𝑥) 𝑑𝑦/ 𝑑𝑥 + 𝑄(𝑥)𝑦 = 0 by writing it in terms of quadratic equation and solving for its roots, mainly the complex roots. 

Answer 1

$$\dfrac{d^2y}{dx^2}+p(x)\dfrac{dy}{dx}+Q(x)y=0$$

Trivial Solution: For the homogeneous equation above, note that the function $y(x)=0$ always satisfies the given equation, regardless what $p(x)$ and $Q(x)$ are. This constant zero solution is called the trivial solution of such an equation. 

The Principle of Superposition: If $y_1$ and $y_2$ are any two solutions of the homogeneous equation $\dfrac{d^2y}{dx^2}+p(x)\dfrac{dy}{dx}+Q(x)y=0,$ where $p(x)$ and $Q(x)$ are continuous on an open interval $I.$ Then the Wronskian $W(y_1,y_2)(x)$ is given by 

$$W(y_1,y_2)(x)=Ce^{-\int p(x)dx}$$

where $C$ is a constant that depends on $y_1$ and $y_2,$ but not on $x.$

Then any function of the form $y=C_1y_1+C_2y_2$ is also a solution of the equation, for any pair of constants $C_1$ and $C_2.$  

Given a second order linear equation with constant coefficients

$$\dfrac{d^2y}{dx^2}+p\dfrac{dy}{dx}+qy=0$$

Solve its characteristic equation $r^2+pr+q=0.$ The general solution depends on the type of roots obtained (use the quadratic formula to find the roots if you are unable to factor the polynomial!):

1. When $p^2-4q>0,$ there are two distinct real roots $r_1, r_2$

$$y=C_1e^{r_1x}+C_2e^{r_2x}$$

2. When $p^2-4q<0,$ there are two complex conjugate roots $r=\lambda\pm\mu i$

$$y=C_1e^{\lambda x}\cos\mu x+C_2e^{\lambda x}\sin \mu x$$

3. When $p^2-4q=0,$ there is one repeated real root $r$

$$y=C_1e^{rx}+C_2xe^{rx}$$

Since $p(x)=p$ and $Q(x)=q,$ being constants, are continuous for every real number, therefore, according to the Existence and Uniqueness Theorem, in each case above there is always a unique solution valid on (−∞, ∞) for any pair of initial conditions $y(x_0)=y_0, y'(x_0)=y'_0.$