University of Florida/Egm4313/s12.team11.perez.gp/R3.9

Problem Statement

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Solve the initial value problem. State which rule you are using. Show each step of your calculation in detail.

Given

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(K 2011 pg.85 #13)

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  (1)

Initial conditions are:

 

Solution

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The general solution of the homogeneous ordinary differential equation is

 

We can use this information to determine the characteristic equation:

 

And proceeding to find the roots,

 

Thus,  .

Solving for the roots, we find that  

where the general solution is

 .

The solution of   of the non-homogeneous ordinary differential equation is

 .

Using the Sum rule as described in Section 2.7, the above function translates into the following:

 , where Table 2.1 tells us that:

  and  .

Therefore,  .

Now, we can substitute the values ( ) into (1) to get:

 

 

 

Now that we have this equation, we can equate coefficients to find that:

 

 

 

and thus,  

We find that the general solution is in fact:

 

 

whereas the general solution of the given ordinary differential equation is actually:

 

Solving for the initial conditions given and first plugging in  , we get that:

 

 

 

 . (2)

And now we can determine the first order ODE :

 

The second initial condition that was given to us,   can now be plugged in:

 

 

 

 

  (3)

Once we solve (2) and (3), we can get the values:

 .

And once we substitute these values, we get the following solution for this IVP:


 

Given

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(K 2011 pg.85 #14)

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  (1)

Initial conditions are:

 

Solution

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The general solution of the homogeneous ordinary differential equation is

 

We can use this information to determine the characteristic equation:

 

And proceeding to find the roots,

 

Solving for the roots, we find that  

where the general solution is:

 , or:

 

Now, according to the Modification Rule and Table 2.1 in Section 2.7, we know that we have to multiply by x to get:

 , since the solution of   is a double root of the characteristic equation.

We can then derive to get  :

 

 

Deriving once again to solve for  , we get the following:

 

 

 

Now, we can substitute the values ( ) into (1) to get:

 

 

Now that we have this equation, we can equate coefficients to find that:

  and  

and finally discover that:

  and  .

Plugging in these values in  , we find that:

 

And finally, we arrive at the general solution of the given ordinary differential equation:

 

 

Solving for the initial conditions given and first plugging in  , we get that:

 

 

 

The second initial condition that was given to us,   can now be plugged in:

 

 

 

And once we substitute these values, we get the following solution for this IVP: