User talk:Egm6321.f10.team5.steinberg/hw5

Part 1
We can rearrange Eqn. 5.1 into the following form:

When in this form, it is apparent that the equation is exact since it can be represented as

where

The 2nd condition of exactness involves two relations which are as follows:

Computing the partial derivatives for (5.1.5) and (5.1.6),

$$ f_{tt} = f_{ty} = f_{yy} = f_{yp} = f_{y} = 0 \! $$

$$ g_{tp} = g_{yp} = g_{pp} = 0 \! $$

$$ g_{y} = a_0 \! $$

In order for (5.1.5) to be satisfied, $$ a_0 \! $$ must be zero.

$$ (5.1.3) \Rightarrow \phi(t,y,p) = h(t,y) + \int a_2 dp \! $$

$$ \phi_t = h_t \! $$

$$ \phi_y = h_y \! $$

Substituting these results into (5.1.4) yields

From (5.1.7) we can solve for $$ h \! $$ to arrive at

Part 2
Let $$ h(t) = exp(\alpha t), \alpha \in \Re \! $$.

Multiplying both sides of Eqn. 5.1 and integrating yields

Part 2.1
Assume that the LHS of (5.2.1) will be of the following form

We set $$ \bar{a}_2 = 0 $$ to reduce the order, differentiate (5.2.2) and set it equal to the integrand of (5.2.1) as follows

$$ \frac{d}{dt}\Big[exp(\alpha t)(\bar{a}_2y+\bar{a}_1y'+\bar{a}_0y) \Big] = exp(\alpha t)\Big(\bar{a}_1 y + (\alpha \bar{a}_1+\bar{a}_0)y' + \alpha \bar{a}_0 y \Big) = exp(\alpha t)\big[a_2y''+a_1y'+a_0y\big] \! $$

The remaining coefficients of (5.2.2) are thus given by

$$ \bar{a}_1 = a_2 \! $$

$$ \bar{a}_0 = a_1 - \alpha a_2 = a_0/\alpha \! $$

Part 2.2
With

$$ a_1 - \alpha a_2 = a_0/\alpha \! $$

we can rewrite it in quadratic form by multiplying both sides by $$ \alpha \! $$ to result in the following

Part 2.3
Returning now to (5.2.1), we substitute (5.2.2) as the result of the LHS integration yielding the following reduced order equation

Dividing through by $$ exp(\alpha t) \! $$ and then by $$ \bar{a}_1 $$ results in the following

Part 2.4
With the L1-ODE-CC found in (5.2.5), we choose the integrating factor as

$$ h(t) = exp \Big[ \int \frac{\bar{a}_0}{\bar{a}_1} dt \Big] = e^{\beta t} \! $$

And since we have (5.2.5) in the correct form (see (2) p. 10-3), we can solve for y using (6) on p. 10-3

Part 2.5
$$ \beta = \frac{\bar{a}_0}{\bar{a}_1} = \frac{a_0/\alpha}{a_2} \Rightarrow \alpha\beta = \frac{a_0}{a_2} $$

$$ \beta = \frac{\bar{a}_0}{\bar{a}_1} = \frac{a_1 - \alpha a_2}{a_2} = \frac{a_1}{a_2} - \alpha \Rightarrow \alpha + \beta = \frac{a_1}{a_2} $$

Part 2.6
Eqn. (5.2.6) is actually the particular solution of the L2-ODE-CC given. Since $$ \alpha $$ and $$ \beta $$ are the roots of the characteristic equation, the complementary solution (or homogeneous solution) is found to be

Because the general solution is the linear combination of the homogeneous solution and particular solution, it is as follows

Part 2.7
If we use the trial solution lookup table for a particular solution if $$ f(t) = t^2 $$, we come up with a trial solution of

where $$ C_2 $$,$$ C_1 $$ and $$ C_0 $$ are undetermined constant coefficients.

To compare this particular solution to the one found in (5.2.6), we substitute $$ f(t) = t^2 $$ into (5.2.6) and find that

The result shown in (5.2.10) is clearly the same form as the trial solution in (5.2.9).

Part 2.8
We will now solve (5.1) for the forcing function $$ f(t) = exp(-t^2) $$.

Part 2.8.1
To begin with, since the characteristic equation is $$ (r+1)(r-2) $$, we know that the homogeneous solution is

and we know that the coefficients in (5.1) are

$$a_2 = 1 \!$$

$$a_1 = -1 \!$$

$$a_0 = -2 \!$$

Our integrating factor (p. 29-3) becomes

Now we turn to (2) p. 30-1 to find the particular solution

Part 2.8.2
To begin with, since the characteristic equation is $$ (r-4)^2 $$, we know that the homogeneous solution is

and we know that the coefficients in (5.1) are

$$a_2 = 1 \!$$

$$a_1 = -8 \!$$

$$a_0 = 16 \!$$

Our integrating factor (p. 29-3) becomes

Now using (2) p. 30-1 again to find the particular solution