Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A sin(rt)+Bcos(rt). The amplitide is given by √ A 2 + B 2 and the period is given by 2 π r

Step-by-step explanation:

This is a homogeneous linear equation. So, assume a solution will be proportional to:

Now, substitute into the differential equation:

Using the characteristic equation:

Factor out

Where:

Therefore the zeros must come from the polynomial:

Solving for :

These roots give the next solutions:

Where and are arbitrary constants. Now, the general solution is the sum of the previous solutions:

Using Euler's identity:

Redefine:

Since these are arbitrary constants

Now, let's find its derivative in order to find and

Evaluating     :

Evaluating     :

Finally, the solution is given by:

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3