Section 2.1 2nd order differential equation

Objective:

1. Definition of 2nd order differential equations, homogeneous and non-homogeneous

2. Applications of 2nd order differential equations

3. Unique solution of a 2nd order differential equation

Definition: The second order differential equation is a differential equation such that its highest derivative is the second derivative. The differential equation is linear if it can be presented as $y^{″}+p_1(t)y^{\prime }+p_2(t)y=q(t)$. When $q(t)=0$, we call the ODE is homogeneous and if $q(t)\ne 0$, then it is a nonhomogeneous ODE.

In this chapter, we are going to solve 2nd order ODE. The application of 2nd order ODE is the spring question. Suppose a mass $m$ hangs from a vertical spring. Assume the the mass is at rest then gravity and spring force is equal, $mg=kL$ where $L$ is the extra length of the spring from its natural length. Let $s(t)$ denote the displacement of the mass from its equilibrium position at time $t$, measured downward.

There are forces on the spring: (a) the spring force coming from Hook’s law which is proportional to the displacement of the mass from its natural length: $F_s(t)=k(s(t)+L)$; (b) the gravity force coming from the weight of the mass: $F_g(t)=mg$; (c) the damping force coming from the environment which is proportional to the velocity of the mass: $F_d(t)=r{s}^{\prime }(t)$; the possible external force acting on the mass $F(t)$. According to Newton’s law, the net force is $ma=m{s}^{″}(t)$. The differential equation of the spring system is

$$ms”(t)=mg-k(s(t)+L)-r{s}^{\prime }(t)+F(t).$$

Hence the ODE becomes 

$$ms”(t)+r{s}^{\prime }(t)+ks(t)=F(t).$$

Theorem (Existence and Uniqueness) Consider the initial value problem $y^{″}+p_1(t)y^{\prime }+p_2(t)y=q(t),y(t_0)=y_0,y^{\prime }(t_0)=y_1$ where $p_1(t)$, $p_2(t)$ and $q(t)$ are continuous on an open interval $I$ that contains $t_0$. Then there exists a unique solution $y=f(t)$ on $I$.

Example 1: $(x+2)y^{″}+\text{cot}(x)y^{\prime }+(x-2)y=9$,$y(-1)=5$, $y^{\prime }(-1)=6$. Determine the longest interval which the given IVP is certain to have a unique solution. 

Exercise 1: $(x-2)y^{″}+y^{\prime }+(x+3)\text{tan}(x)y=9$,$y(3)=4$, $y^{\prime }(3)=2$. Determine the longest interval which the given IVP is certain to have a unique solution. 

Example 2: $(t^2-4)y^{″}+\text{ln}(t)y^{\prime }+\sqrt{1-t}y=9$,$y(-3)=5$, $y^{\prime }(-3)=6$. Determine the longest interval which the given IVP is certain to have a unique solution. 

Exercise 2: $(9-t^2)y^{″}+\sqrt{t-2}{y}^{\prime }+(t+3)y=9$,$y(4)=4$, $y^{\prime }(4)=2$. Determine the longest interval which the given IVP is certain to have a unique solution. 

Theorem  (Principle of Superposition) If $y_1$ and $y_2$ are solutions to the equation $y^{″}+p(t)y^{\prime }+q(t)y=0$ then the linear combination $c_1{y}_1+c_2{y}_2$ is also a solution, for all constants $c_1$ and $c_2$.

Proof:

Example 3: Show $e^{2t}$, $e^{-3t}$ and $c_1{e}^{2t}+c_2{e}^{-3t}$ are solutions of $y^{″}+y^{\prime }-6y=0$ for any $c_1$ and $c_2$ constants. Verify $2$ and $-3$ are roots of ${\lambda }^2+\lambda -6=0$. 

Exercise 3: Show $e^{-4t}$, $e^{-t}$ and $c_1{e}^{-4t}+c_2{e}^{-t}$ are solutions of $y^{″}+5y^{\prime }+4y=0$ for any $c_1$ and $c_2$ constants. Verify $-4$ and $-1$ are roots of ${\lambda }^2+5\lambda +4=0$. 

Definition: The characteristic equation of the ODE $a{y}^{″}+b{y}^{\prime }+cy=0$ is $a{\lambda }^2+b\lambda +c=0$. 

Remark: If ${\lambda }_1$ is a root of the characteristic equation of the ODE, $a{y}^{″}+b{y}^{\prime }+cy=0$ then $e^{{\lambda }_{1}t}$ is a solution of the ODE, $a{y}^{″}+b{y}^{\prime }+cy=0$.

Exercise 4: Find a root ${\lambda }_1$ of the characteristic equation of the ODE, $y^{″}+5y^{\prime }+6y=0$ then show $e^{{\lambda }_{1}t}$ is a solution of the ODE, $y^{″}+5y^{\prime }+6y=0$.

Group Work:

1. Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution.

$$(x+1)y”+y^{\prime }+(x-2)(\text{tan}(x))y=0,y(0)=1,y^{\prime }(0)=2.$$

2. Use principle of superposition and characteristic equation to find a general solution of $y^{″}-4y^{\prime }-12y=0$. 

3. Use principle of superposition and characteristic equation to find a general solution of $y^{″}+6y^{\prime }=0$. 

4. Use principle of superposition and characteristic equation to find a general solution of $y^{″}-9y=0$. 

5. Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution. Do not attempt to find the solution.

$$(x^2-1)y”+y^{\prime }+(\text{ln}(x))y=0,y(2)=1,y\prime (2)=2.$$