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 [latex]y''+p_{1}(t)y'+p_{2}(t)y=q(t)[/latex]. When [latex]q(t)=0[/latex], we call the ODE is homogeneous and if [latex]q(t)\neq0[/latex], 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 [latex]m[/latex] hangs from a vertical spring. Assume the the mass is at rest then gravity and spring force is equal, [latex]mg=kL[/latex] where [latex]L[/latex] is the extra length of the spring from its natural length. Let [latex]s(t)[/latex] denote the displacement of the mass from its equilibrium position at time [latex]t[/latex], 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: [latex]F_{s}(t)=k(s(t)+L)[/latex]; (b) the gravity force coming from the weight of the mass: [latex]F_{g}(t)=mg[/latex]; (c) the damping force coming from the environment which is proportional to the velocity of the mass: [latex]F_{d}(t)=rs'(t)[/latex]; the possible external force acting on the mass [latex]F(t)[/latex]. According to Newton’s law, the net force is [latex]ma=ms''(t)[/latex]. The differential equation of the spring system is

\[ms”(t)=mg-k(s(t)+L)-rs'(t)+F(t).\]

Hence the ODE becomes 

\[ms”(t)+rs'(t)+ks(t)=F(t).\]

Theorem (Existence and Uniqueness) Consider the initial value problem [latex]y''+p_{1}(t)y'+p_{2}(t)y=q(t),y(t_{0})=y_{0},y'(t_{0})=y_{1}[/latex] where [latex]p_{1}(t)[/latex], [latex]p_{2}(t)[/latex] and [latex]q(t)[/latex] are continuous on an open interval [latex]I[/latex] that contains [latex]t_{0}[/latex]. Then there exists a unique solution [latex]y=f(t)[/latex] on [latex]I[/latex].

Example 1: [latex](x+2)y''+\text{cot}(x)y'+(x-2)y=9[/latex],[latex]y(-1)=5[/latex], [latex]y'(-1)=6[/latex]. Determine the longest interval which the given IVP is certain to have a unique solution. 

Exercise 1: [latex](x-2)y''+y'+(x+3)\text{tan}(x)y=9[/latex],[latex]y(3)=4[/latex], [latex]y'(3)=2[/latex]. Determine the longest interval which the given IVP is certain to have a unique solution. 

Example 2: [latex](t^{2}-4)y''+\text{ln}(t)y'+\sqrt{1-t}y=9[/latex],[latex]y(-3)=5[/latex], [latex]y'(-3)=6[/latex]. Determine the longest interval which the given IVP is certain to have a unique solution. 

Exercise 2: [latex](9-t^{2})y''+\sqrt{t-2}y'+(t+3)y=9[/latex],[latex]y(4)=4[/latex], [latex]y'(4)=2[/latex]. Determine the longest interval which the given IVP is certain to have a unique solution. 

Theorem  (Principle of Superposition) If [latex]y_{1}[/latex] and [latex]y_{2}[/latex] are solutions to the equation [latex]y''+p(t)y'+q(t)y=0[/latex] then the linear combination [latex]c_{1}y_{1}+c_{2}y_{2}[/latex] is also a solution, for all constants [latex]c_{1}[/latex] and [latex]c_{2}[/latex].

Proof:

Example 3: Show [latex]e^{2t}[/latex], [latex]e^{-3t}[/latex] and [latex]c_{1}e^{2t}+c_{2}e^{-3t}[/latex] are solutions of [latex]y''+y'-6y=0[/latex] for any [latex]c_{1}[/latex] and [latex]c_{2}[/latex] constants. Verify [latex]2[/latex] and [latex]-3[/latex] are roots of [latex]\lambda^{2}+\lambda-6=0[/latex]. 

Exercise 3: Show [latex]e^{-4t}[/latex], [latex]e^{-t}[/latex] and [latex]c_{1}e^{-4t}+c_{2}e^{-t}[/latex] are solutions of [latex]y''+5y'+4y=0[/latex] for any [latex]c_{1}[/latex] and [latex]c_{2}[/latex] constants. Verify [latex]-4[/latex] and [latex]-1[/latex] are roots of [latex]\lambda^{2}+5\lambda+4=0[/latex]. 

Definition: The characteristic equation of the ODE [latex]ay''+by'+cy=0[/latex] is [latex]a\lambda^{2}+b\lambda+c=0[/latex]. 

Remark: If [latex]\lambda_{1}[/latex] is a root of the characteristic equation of the ODE, [latex]ay''+by'+cy=0[/latex] then [latex]e^{\lambda_{1}t}[/latex] is a solution of the ODE, [latex]ay''+by'+cy=0[/latex].

Exercise 4: Find a root [latex]\lambda_{1}[/latex] of the characteristic equation of the ODE, [latex]y''+5y'+6y=0[/latex] then show [latex]e^{\lambda_{1}t}[/latex] is a solution of the ODE, [latex]y''+5y'+6y=0[/latex].

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.

\[(x+1)y”+y’+(x-2)(\text{tan}(x))y  =0, y(0)=1, y'(0)=2.\]

2. Use principle of superposition and characteristic equation to find a general solution of [latex]y''-4y'-12y=0[/latex]. 

3. Use principle of superposition and characteristic equation to find a general solution of [latex]y''+6y'=0[/latex]. 

4. Use principle of superposition and characteristic equation to find a general solution of [latex]y''-9y=0[/latex]. 

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.

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