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MAT 204 Practice Exam 2

Please work out each of the given problems on your own paper.  Credit will be based on the steps that you show towards the final answer.  Show your work.

Problem 1  (15 Points each)

Solve the following differential equations.

A.  y''' - 3y'' + 3y' - y  =  0

Solution

The characteristic equation is 

        r³ - 3r² + 3r - 1

We can factor this either by a calculator or by noting that r  =  1 is a root and dividing by r - 1.  Either way we get

        (r - 1)³ =  0

which has a root of order 3 at x  =  1.

The solution is therefore

        y  =  c₁e^t + c₂te^t + c₃t²e^t 

B.  y^(iv) - y  =  0  

Solution

  The characteristic equation is 

        r⁴ - 1  =  0

This factors as 

        (r - 1)(r + 1)(r² + 1)  =  0

So the roots are 

        r  =  1,      r  =  -1,      r  =  i,      r  =  -i

The general solution is thus

        y  =  c₁e^t + c₂e^-t + (c₁cos t  +  c₂ sin t)

Problem 2  (15 Points each)
Consider the differential equation

        y''' - y'  =  e^2t + e^3t

A.     Solve this differential equation using the method of UC functions  

Solution

First find the homogeneous solution

        y''' - y'  =  0

The characteristic equation is given by 

        r³ - r  =  0

This factors as 

        r(r + 1)(r - 1)  =  0

        r  =  0,     r  =  1,      r  =  -1

The homogeneous solution is 

        y_h  =  c₁ + c₂e^t + c₃e^-t 

Next find the particular solution using UC functions.  We have

        y_p  =  Ae^2t + Be^3t Notice that these do not coincide with y_h 

        y_p'  =  2Ae^2t + 3Be^3t 

        y_p''  =  4Ae^2t + 9Be^3t 

        y_p'''  =  8Ae^2t + 27Be^3t 

Plugging into the original differential equation gives

         (8Ae^2t + 27Be^3t ) - (2Ae^2t + 3Be^3t)  =  e^2t + e^3t       

         (8A - 2A)e^2t + (27B - 3B)e^3t ) =  e^2t + e^3t       

        6A  =  1    and     24A  =  1

        A  =  1/6    and    B  =  1/24

The final solution is 

        y  =  y_h + y_p  =  c₁ + c₂e^t + c₃e^-t + 1/6 e^2t + 1/24 e^3t

B.     Solve this differential equation using the method of variation of parameters.

  Solution

We have already solved the homogeneous system

        y_h  =  c₁ + c₂e^t + c₃e^-t 

Thus the final solution is 

        y  =  v₁ + v₂e^t + v₃e^-t 

Now compute the Wronskian 

The third column of the adjoint is 

Divide by the Wronskian to arrive at the third column of the inverse:

 We get the three differential equations

        v₁'  =  (-1)(e^2t + e^3t)    v₂'  =  1/2 e^-t (e^2t + e^3t)    v₃'  =  1/2 e^t (e^2t + e^3t)

        v₁'  =  -e^2t - e^3t    v₂'  =  1/2 (e^t + e^2t)    v₃'  =  1/2 (e^3t + e^4t)

Now integrate to get

        v₁  =  -1/2 e^2t - 1/3 e^3t + C₁

        v₂  =  1/2 (e^t + 1/4 e^2t)  + C₂

          v₃  =  1/2 (1/3e^3t + 1/4e^4t) + C₃

We have

        y  =  (-1/2 e^2t - 1/3 e^3t + c₁) + (1/2 (e^t + 1/4 e^2t)  + c₂)e^t +(1/2 (1/3e^3t + 1/4e^4t) + c₃)e^-t

        =  c₁ + c₂e^t + c₃e^-t + (-1/2 + 1/2 + 1/6)e^2t + (-1/3 + 1/4 + 1/8)e^4t

        =  c₁ + c₂e^t + c₃e^-t + 1/6 e^2t + 1/24 e^4t

Problem 3  (15 Points)

A 2 kg mass stretches a spring 1.96 meters.  The mass is attached to a viscous damper that exerts a force of 4N when the velocity is 0.1 m/sec.  The mass is then pulled down 0.5m and released. 

A.     Determine the equation of motion for this system.

Solution

We use the formula

mu'' + gu' + ku  =  F  

We have 

        m  =  2    and    F  =  0

Use hooks law to get

        (2)(9.8)  =  k(1.96)

Which produces

        k  =  10

To find g, we use

        4  =  g(.1)

        g  =  40

This gives us the differential equation

        2u'' + 40u' + 10u  =  0    u(0)  =  0.5

or

        u'' + 20u' + 5u  =  0    u(0)  =  0.5

The characteristic equation for this differential equation is 

        r² + 20r + 5  =  0

and has solution approximately equal to -19.7 and -0.25.  The general solution is 

       y  =  c₁e^-19.7t + c₂e^-0.25t

Now use the initial conditions to find the constants.  We have 

        0.5  =  c₁ + c₂ 

Taking the derivative and plugging in 0, we get

        0  =  -19.7c₁ - 0.25c₂

These two equations give

        c₁  =  -0.01            c₂  =  0.51       

The final solution is 

       y  =  -0.01e^-18.66t + 0.51e^-1.34t

B.  Describe (qualitatively) the difference between replacing the viscous damper with an external force of F  =  3sin( t) and replacing the damper with an external force of F  =  2cos(3t).  

Solution

If there is no viscous damper and the external force is 

        F  =  3sin( t)

Then the differential equation becomes

        2u''  + 10u  =  3sin( t)        u(0)  =  0.5

The homogenous equation is 

        2u'' + 10u  =  0 

or

        u'' + 5u  =  0

Which has solution 

        y_h  =  c₁cos( t) + c₂ sin( t)  

Since the external force is part of the homogeneous solution, hence the motion exhibits resonance and the motion is unbounded.  In reality, the spring will eventually break.

For the second force, the equation is 

        2u''  + 10u  =  2cos(3t)        u(0)  =  0.5

Since the force is not part of the homogeneous solution, there is no resonance.  Instead the motion will have a beat.

Problem 4 
Solve the given differential equation by means of a power series about x = 0.  Find a recurrence relation and write the final solution in the form

        y'' + xy' + 2y  =  0,        y(0)  =  0,     y'(0)  =  1

Solution 

We have

Plugging back into the original differential equation gives

Now adjust the sums so that they all contain the same powers of x.

Now pull out the n = 0 term and combine the sums

Next use the initial conditions to get

        a₀  =  0        a₁  =  1

        2a₂ + 2a₀  =  0  

Hence

         a₂  =  0

The recursion relationship gives

        (n + 2)(n + 1)a_n+2 + (n + 2)a_n  =  0

or

                           -a_n
            a_n+2  =                                 
                           n + 1

Hence all the even coefficients are 0.  The odds are

                                      -1                      1                        -1                         1
        a₁  =  1        a₃  =                a₅  =                  a₇  =                  a₉  =                    
                                        2                    3^. 4                    3^. 4^. 6                 3^. 4^. 6^. 8

In general

                             (-1)ⁿ                    (-1)ⁿ
    a_2n+1  =                                  =                           
                         2^. 4^. 6^. ...^. (2n)            2ⁿ n!

The general solution is 

Problem 5 

Determine the general solution of the differential equation that is valid in any interval not including the singular point. 

        x²y'' - xy' + y  =  0      

Solution

We let

        y  =  x^r

Then 

        y'  =  rx^r-1        y''  =  r(r - 1)x^r-2

Substituting back into the original equation gives

        r(r - 1)x^r - rx^r + x^r  =  0

        r² - r - r + 1  =  r² - 2r + 1  =  0

        (r - 1)²  =  0

        r  =  1 (repeated twice)

Hence

        y  =  (c₁ + c₂ ln|x|) |x^r|

Problem 6 

Solve the following differential equation

            y(0)  =  y'(0)  =  0

Solution

Since this is a piecewise defined differential equation, we use the method of Laplace Transforms.  

        L(y'') + 4L(y)  =  L(g)  =  L[sin t (1 - u_2p(t))  =  L(sin t) - L[(sin t) u_2p(t)]

                                          1                     1
        s²L(y) + 4L(y)  =              - e^-2ps               
                                       s² + 1              s² + 1

Solving for L(y), we get

                                                1
        L(y)  =  (1 - e^-2ps)                              
                                        (s² + 1)(s² + 4)

Now use partial fractions 

                   1                        As + B            Cs + D
                                      =                      +                        
           (s² + 1)(s² + 4)            s² + 1             s² + 4

or

        (As + B)(s² + 1) + (Cs + D)(s² + 4)  =  1

        As³ + Bs² + As + B + Cs³ + Ds² + 4Cs + 4D  =  0

Equating coefficients gives

        A + C  =  0        B + D  =  0        A + 4C  =  0        B + 4D  = 1

This has solution 

        A  =  0        B  =  -1/3        C  =  0        D  =  1/3

Putting this together gives

                       1                           1                   1
         L(y)  =        (1 - e^-2ps)                     -                        
                       3                       s² + 4             s² + 1

                       1        1                   1                     e^-2ps           e^-2ps
         L(y)  =        (                   -                  -                  +                )
                       3       s² + 4            s² + 1            s² + 4          s² + 1 

Now take the inverse Laplace Transform to get

        y  =  1/3 ( 1/2 sin(2t) - sin t - 1/2 u_2p sin(2(t-2PI)) + u_2p sin t-2PI)

Problem 7  Please answer the following true or false.  If true, explain why.  If false, explain why or provide a counter-example.

 Solution