1. What is meant by homogeneous? What should you look for to determine the order

Question

1. What is meant by homogeneous? What should you look for to determine the order of a differential equation?

2.   Given y''-4y'-12y=0 replace y'' with m2, y' with m, and y with 1 to create the characteristic equation.

Characteristic equation is ______________________________            

3. Solve the characteristic equation above for the roots m1 and m2. Hint: you may need to solve by factoring or by using the quadratic formula.

4. Write the solution of the differential equation based on the roots found in question 3.

      a) Are m1 and m2 real numbers and not equal to each the other? Then the solution is

y=C1em1x+C2em2x

     b) Are m1 and m2 real numbers and equal to each the other? Then the solution is

          y=C1em1x+C2em2x

     c) Are m1 and m2 complex numbers of the form a + bi and a - bi? Then the solution is

          y=C1eaxsin(bx)+C2eaxcos(bx)

Solution is _____________________

5. Solve the following homogeneous differential equation by repeating the process above.

3y''+5y'+7y=0

Create the characteristic equation.

Solve the characteristic equation.

Write the solution of the differential equation.

Solution is_________________________

6. Solve the following homogeneous differential equation by repeating the process above.

16y''+40y'+25y=0

Create the characteristic equation.

Solve the characteristic equation.

Write the solution of the differential equation.

Solution is_________________________

7. Take the result of question 6 and find the particular solution given that y(0) = 1 and y'(0)=0 . Why do we need two initial conditions for second order differential equations? How are the initial conditions used to find a particular solution? What must be done in order to use the second initial condition?

Particular Solution is___________________________

Part II: Second Order Non-Homogeneous Differential Equations

1. Follow the six steps below to solve the following: y''+6y'-16=6x-1

Remember that the final solution form is y=yh+yp

Step 1: Find yh. This is the homogeneous solution. Use the process outlined in part I above to find it.

yh=

Step 2: Write the generic form of yp.   The right side of our equation is f(x) = 6x

Explanation / Answer

Part I:

1.)Homogenous means right hand side of the differential equation is zero like y'' + y' + y =0

We should look for the highest order of y to determine the order of a differential equation like y'' in y'' + y' + y=0, order is 2

2.) m^2 - 4m - 12=0

3.) (m-6)(m+2)=0

So m1 = -2, m2 = 6

4.) Solution is y = C1e^(-2x) + C2e^(6x)

5.) 3y'' + 5y' + 7y =0

3m^2 + 5m + 7=0

m = [-5 +/- sqrt(25-4(21))]/6 = [-5 +/- sqrt(59)]/6

So m1 = [-5 + sqrt(59)]/6

m2 = [-5 - sqrt(59)]/6

Since roots are complex with a = -5/6 and b = sqrt(59)/6

Solution is y= C1e^(-5x/6)sin(sqrt(59)x/6) + C2e^(-5x/6)cos(sqrt(59)x/6)

6.) 16y'' + 40y' + 25y=0

16m^2 + 40m + 25 =0

(4m+5)^2 = 0

So m1=m2 = -5/4

Solution is y = C1e^(-5x/4) + C2*xe^(-5x/4)

7.) y(0)=1, y'(0)=0

y(0)=1

So C1 = 1

y'(0)=0

So -5C1/4 + C2 =0

So C2 = 5/4

So particular solution is yp = ((5/4)x + 1)e^(-5x/4)

Part II:

1.) m^2 + 6m -16=0

(m+8)(m-2)=0

m1 = -8, m2 =2

So yh = C1e^(-8x) + C2e^(2x)

2.) yp = Ax + B

3.) y'p = A
y''p = 0

4.)y''p + 6(y'p) -16(yp) = 6x -1

0 + 6A - 16(Ax+B) = 6x-1

So -16A = 6

A = -6/16 = -3/8

6A - 16B = -1

6(-3/8) - 16B = -1

So B = -5/64

5.) A = -3/8 , B = -5/64

yp = Ax + B = -3x/8 -5/64 = (-24x-5)/64 = -(24x+5)/64

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