Differential Equations

Question

Find the general solution of the differential equation
y″(t)+36y(t)=0

 

I found the general solution to be: Asin(6t)+Bcos(6t)

 

For each of the following initial conditions, find a particular solution.

(iii) y(0)=1, y(1)=0: y=?
(iv) y(0)=0, y(1)=1: y=?

 

 

 

2. Check by differentiation that y=2cos⁡2t+6sin⁡2t is a solution to y″+4y=0 by finding the terms in the sum.

y''=?

4y=?

y''+4y=?

 

 

 

3. Solve the following differential equation:

y″+9y=0; y=3, y′=3 at x=Π/3

 

I would appreciate help with these problems!

Answer 1

I will help you with the first problem.  If you understand how to work it, you should be able to work the second and third problems.

 

y″ + 36y = 0

y = Asin(6t) + Bcos(6t)

Given y(0) = 1 and y(1) = 0:

 

y(0) = B = 1 ⇒

y = Asin(6t) + cos(6t)

 

y(1) = Asin6 + cos6 = 0 ⇒ A = -cos6 / sin6 = -cot6  

[Note:  6 is radians]

 

y(t) = (-cot6)sin(6t) + cos(6t)

 

If given y(0) = 0 and y(1) = 1:

 

y(0) = B = 0 ⇒

 

y = Asin(6t)

 

y(1) = 1 ⇒ Asin6 = 1 ⇒ A = 1/sin6 = csc6

[Note: 6 is radians]

 

y(t) = (csc6)sin(6t)