Download Fall 2018 MATH 2930: Review of First & Second Order Differential Equations and more Summaries Differential Equations in PDF only on Docsity!
MATH 2930, Fall 2018 TA: Aleksandra Niepla Sections: 212, 217 (^) Name:
Chapter 2.7, 3.1, 3.3 Review
Objectives: (1) Approximate solutions to first order differential equations using Euler’s method (2) Practice solving second order linear homogeneous differential equations with constant coefficients
Part 1- Numerical Approximations: Euler’s Method
- Consider the initial value problem
y′^ = 0. 5 − t + 2y, y(0) = 1
(i) Find approximate values of the solution of the given initial value problem at t = 0. 1 , 0. 2 , 0. 3 , and 0. 4 using Euler’s method with h = 0. 1. (ii) Find the solution y(t) of the given problem and evaluate y(t) at t = 0. 1 , 0. 2 , 0. 3 , and 0. 4. Compare your results with those of part (i).
(i) We have the following approximations:
(ii) Rewriting the equation we have:
y′^ − 2 y = 0. 5 − t. Let μ(t) = e
∫ (^) − 2 dt = e−^2 t. Then (e−^2 ty(t))′^ = 0. 5 e−^2 t^ − te−^2 t. Integrating both sides and dividing by e−^2 t^ we get:
y(t) = t 2
- Ce^2 t. The initial condition implies that C = 1. So, the solutions is:
y(t) = t/2 + e^2 t. Moreover, using a calculator we have:
y(0.1) = 1. 27 , y(0.2) = 1. 59 , y(0.3) = 1. 97 , y(0.4) = 2. 43.
Part 2- Second Order Linear Differential Equations
- Find the general solution of the given differential equation
y′′^ + 2y′^ − 3 y = 0.
The characteristic equation is
r^2 + 2r − 3 = 0. We factor to find its roots:
r^2 + 2r − 3 = 0 =⇒ (r + 3)(r − 1) = 0 =⇒ r 1 = − 3 , r 2 = 1. So, the general solutions is y(t) = c 1 e−^3 t^ + c 2 et.
- Find the solution of the given initial value problem. Describe its behavior as t increases:
y′′^ + 8y′^ − 9 y = 0 y(1) = 1, y′(1) = 0.
The characteristic equation is
r^2 + 8r − 9 − 0. We factor to find its roots
r^2 + 8r − 9 = 0 =⇒ (r + 9)(r − 1) = 0 =⇒ r 1 = − 9 , r 2 = 1. Therefore, the general solutions is
y(t) = c 1 e−^9 t^ + c 2 et. Using the two initial conditions provided we obtain the following system of equations:
c 1 e−^9 + c 2 e = 1 and − 9 c 1 e−^9 + c 2 e = 0 Multiplying the first by 9 and adding to the second gives:
10 ec 2 = 9 =⇒ c 2 =
e−^1.
Next, we can plug c 2 into the first equation and solve for c 1 :
c 1 e−^9 +^9 10 = 1 =⇒ c 1 = 1 10 e^9.
So, the final solutions is:
y(t) =
10 e
−9(t−1) (^) +^9 10 e
t− (^1) ; y → ∞ as t → ∞.
- Find the general solution of the given differential equation
y′′^ + 6y′^ + 13y = 0.
The characteristic equation is
r^2 + 6r + 13 = 0. The roots are:
r =
= − 3 − 2 i, −3 + 2i.
So, the general solutions is y(t) = c 1 e−^3 t^ cos(2t) + c 2 sin(2t)e−^3 t.
d^2 y dt^2
d dt
dy dx
t
dy dx
t^2
d^2 y dx^2
t^2
dy dx
t^2
t^2
d^2 y dx^2
dy dx
Plugging this into the differential equation we obtain:
d^2 dx^2
dy dx
dy dx
This is now a homogeneous second order differential equation with characteristic equation:
r^2 + 3r + 2 = 0. The roots are r = − 1 , − 2. So the general solution is:
y(x) = c 1 e−x^ + c 2 e−^2 x.
Finally, using that x = ln(t), we get
y(t) = c 1 t−^1 + c 2 t−^2.
