Differential Equations SecondOrder Linear DEs Prepared by Vince

Differential Equations Second-Order Linear DEs Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

A Second-Order Linear Differential Equation can always be put into the form: The general solution will always have the form: yh is the solution to the corresponding homogeneous equation, where g(t)=0 yp is a particular solution to the original DE. There should be two independent solutions to the homogeneous equation, and yh will be a linear combination of these two. In fact, the two independent solutions form a basis for a 2 dimensional solution space (an actual vector space like we saw last quarter). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Here a few examples of linear, second order DEs with constant coefficients. These equations are all homogeneous. We will solve some inhomogeneous equations later. Also, we will add some initial conditions to these problems. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

For these constant-coefficient equations we will essentially turn this into an algebra problem. Start by writing down the CHARACTERISTIC EQUATION. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

For these constant-coefficient equations we will essentially turn this into an algebra problem. Start by writing down the CHARACTERISTIC EQUATION. Solve for the roots of this equation, then each root will correspond to a solution of the DE. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

For these constant-coefficient equations we will essentially turn this into an algebra problem. Start by writing down the CHARACTERISTIC EQUATION. Solve for the roots of this equation, then each root will correspond to a solution of the DE. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

For these constant-coefficient equations we will essentially turn this into an algebra problem. Start by writing down the CHARACTERISTIC EQUATION. Solve for the roots of this equation, then each root will correspond to a solution of the DE. The general solution is a linear combination of these: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Now add some initial conditions. We will need to determine the constants in our general solution that match up with the given conditions. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Now add some initial conditions. We will need to determine the constants in our general solution that match up with the given conditions. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Now add some initial conditions. We will need to determine the constants in our general solution that match up with the given conditions. Here is our final solution. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Start with the characteristic equation: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Start with the characteristic equation: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Start with the characteristic equation: This time there is only one repeated root, so it seems like there is only one solution: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Start with the characteristic equation: This time there is only one repeated root, so it seems like there is only one solution: This is one of the solutions, but we need to find another independent solution. The trick is to multiply this solution by t: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Start with the characteristic equation: This time there is only one repeated root, so it seems like there is only one solution: This is one of the solutions, but we need to find another independent solution. The trick is to multiply this solution by t: Now the general solution is a linear combination of these: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Now add some initial conditions. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Now add some initial conditions. Calculate the derivative, then plug in to find the constants. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Now add some initial conditions. Calculate the derivative, then plug in to find the constants. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Now add some initial conditions. Calculate the derivative, then plug in to find the constants. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

This time the roots are complex numbers. The complex exponential solutions are perfectly fine, but we can find real-valued solutions as well by using Euler’s formula. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

This time the roots are complex numbers. The complex exponential solutions are perfectly fine, but we can find real-valued solutions as well by using Euler’s formula. This is Euler’s formula. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

This time the roots are complex numbers. The complex exponential solutions are perfectly fine, but we can find real-valued solutions as well by using Euler’s formula. This is Euler’s formula. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

This time the roots are complex numbers. The complex exponential solutions are perfectly fine, but we can find real-valued solutions as well by using Euler’s formula. This is Euler’s formula. cos(-t)=cos(t) and sin(-t)=-sin(t) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

This time the roots are complex numbers. The complex exponential solutions are perfectly fine, but we can find real-valued solutions as well by using Euler’s formula. This is Euler’s formula. cos(-t)=cos(t) and sin(-t)=-sin(t) Any linear combination of these solutions will also be a solution to the DE, specifically: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Plug in the given values to find the solution that matches up. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Plug in the given values to find the solution that matches up. The solution is simply an oscillation because the roots of the characteristic equation were completely imaginary. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

These roots have both a real and imaginary part. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

These roots have both a real and imaginary part. The corresponding general solution is: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Now add some initial values Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Now add some initial values The general solution is: Compute the first derivative, then plug in the given values: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Now add some initial values The solution is Using some trigonometry this can be re-written as a single cosine function. Here is the general formula: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Now add some initial values The solution is Using some trigonometry this can be re-written as a single cosine function. The plot of the solution shows the function oscillating between the exponential “envelope” and eventually approaching 0. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB