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

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Problem Statement[edit]

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

Given[edit]

(K 2011 pg.85 #13)[edit]

(1)

Initial conditions are:

Solution[edit]

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[edit]

(K 2011 pg.85 #14)[edit]

(1)

Initial conditions are:

Solution[edit]

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: