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

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

(K 2011 pg.85 #13)
$$ 8y'' - 6y' + y = 6 \cosh x \! $$              (1)

Initial conditions are:

$$ y(0) = 0.2, y'(0) = 0.05 \! $$

Solution
The general solution of the homogeneous ordinary differential equation is

$$ 8y'' - 6y' + y = 0 \! $$

We can use this information to determine the characteristic equation:

$$ 8 \lambda^2 - 6 \lambda + 1 = 0 \! $$

And proceeding to find the roots,

$$ 4\lambda(2 \lambda - 1) - 1(2 \lambda - 1) = 0 \! $$

Thus, $$ (4 \lambda - 1)(2 \lambda - 1) = 0 \! $$.

Solving for the roots, we find that $$ \lambda = \frac{1}{4}, \frac{1}{2}, \! $$

where the general solution is

$$ y_k = c_1 e^{\frac{1}{4}x} + c_2 e^{\frac{1}{2}x} \! $$.

The solution of $$ y_p \! $$ of the non-homogeneous ordinary differential equation is

$$ x = \frac{e^x+e^{-x}}{2} \! $$.

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

$$ y_p = y_{p1} + y_{p2} \! $$, where Table 2.1 tells us that:

$$ y_{p1} = Ae^x \! $$ and $$ y_{p2} = Be^x \! $$.

Therefore, $$ y_{p} = Ae^x + Be^{-x} \! $$.

Now, we can substitute the values ($$ y_p, y_p', y_p'' \! $$) into (1) to get:

$$ 8(Ae^x+Be^{-x}) - 6(Ae^x - Be^{-x}) + Ae^x + Be^{-x} = 6(\frac{e^x + e^{-x}}{2}) \! $$

$$ = 3Ae^x + 15Be^{-x} \! $$

$$ = 3(e^x + e^{-x}) \! $$

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

$$ 3A = 3 \! $$

$$ \therefore A = 1 \! $$

$$ B = \frac{3}{15} = \frac{1}{5} \ $$

and thus, $$ y_p = e^x + \frac{1}{5}e^{-x} \! $$

We find that the general solution is in fact:

$$ y = y_k + y_p \! $$

$$ y = c_1 e^{\frac{1}{4}}x + c_2 e^{\frac{1}{2}}x + e^x + 3e^{-x} \! $$

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

$$ y = c_1 e^{\frac{1}{4}}x + c_2 e^{\frac{1}{2}}x + e^x + \frac{1}{5} e^{-x} \! $$

Solving for the initial conditions given and first plugging in $$ y(0) = 0.2 \! $$, we get that:

$$ 0.2 = c_1 e^{\frac{1}{4}(0)} + c_2 e^{\frac{1}{2}(0)} + e^0 + 3e^{(0)} \! $$

$$ 0.2 = c_1 e^{(0)} + c_2 e^{(0)} + e^{(0)} + 3e^{(0)} \! $$

$$ 0.2 = c_1 + c_2 + 1 + \frac{1}{5} \! $$

$$ \therefore c_1 + c_2 = - 1 \! $$.             (2)

And now we can determine the first order ODE :

$$ y' = \frac{1}{4}c_1 e^{\frac{1}{4}(x)} + \frac{1}{2}c_2 e^{\frac{1}{2}(x)} + e^{x} - \frac{1}{5}e^{x} \! $$

The second initial condition that was given to us, $$ y'(0) = 0.05 \! $$ can now be plugged in:

$$ 0.05 = \frac{1}{4}c_1 e^{\frac{1}{4}(0)} + \frac{1}{2}c_2 e^{\frac{1}{2}(0)} + e^{(0)} - \frac{1}{5}e^{(0)} \! $$

$$ 0.05 = \frac{1}{4}c_1 e^{(0)} + \frac{1}{2}c_2 e^{(0)} + e^{(0)} - \frac{1}{5}e^{(0)} \! $$

$$ 0.05 = \frac{1}{4}c_1 + \frac{1}{2}c_2 + 1 - \frac{1}{5} \! $$

$$ \frac{1}{4}c_1 + \frac{1}{2}c_2 = -0.75 \! $$

$$ \therefore c_1 + 2 c_2 = -3 \! $$             (3)

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

$$ c_1 = 1, c_2 = - 2 \! $$.

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

$$ y = e^{\frac{1}{4}x} - 2 e^{\frac{1}{2}x} + e^{x} + \frac{1}{5} e^{-x} \! $$

(K 2011 pg.85 #14)
$$ y'' + 4y' + 4y = e^{-2x} sin2x \! $$              (1)

Initial conditions are:

$$ y(0)=1, y'(0) = -1.5 \! $$

Solution
The general solution of the homogeneous ordinary differential equation is

$$ y'' + 4y' + 4y = 0 \! $$

We can use this information to determine the characteristic equation:

$$ \lambda^2 + 4 \lambda + 4 = 0 \! $$

And proceeding to find the roots,

$$ (\lambda + 2) (\lambda + 2) = 0 \! $$

Solving for the roots, we find that $$ \lambda = -2, -2 \! $$

where the general solution is:

$$ y_k = c_1 e^{-2x} + c_2 e^{-2x} x \! $$, or:

$$ y_k = (c_1 + c_2 x) e^{-2x} \! $$

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:

$$ y_p = e^{-2x}(K x cos2x + Mx sin2x) \! $$, since the solution of $$ y_k \! $$ is a double root of the characteristic equation.

We can then derive to get $$ y_p' \! $$:

$$ y_p' = -2 e ^{-2x} (K x cos2x + M x sin2x) + e^{-2x} (Kcos2x-2Kxsin2x+Msin2x+2Mxcos2x) \! $$

$$ y_p' = (-2K+2M) e^{-2x} x cos2x + (-2M-2K) e^-2x xsin2x + Ke^{-2x} cos2x + Me^{-2x} sinx \! $$

Deriving once again to solve for $$ y_p'' \! $$, we get the following:

$$ y_p'' = 4e^{-2x} (Kxcos2x + Mxsin2x) - 2e^{-2x} (Kcos2x - 2Kxsin2x + Msin2x + 2Mxcos2x) - 2e^{-2x} (Kcos2x - 2Kxsin2x + Msin2x + 2Mxcos2x) \! $$

$$ y_p'' = 4e^{-2x} (Kxcos2x + Mxsin2x) - 4e^{-2x} (Kcos2x - 2Kxsin2x + Msin2x + 2Mxcos2x) + e^{-2x} (-4Ksin2x - 4Kxcos2x + 4Mcos2x - 4Mxsin2x) \! $$

$$ y_p'' = (-4K + 4M) e^{-2x} cos2x + (-4K - 4M)e^{-2x} sin2x \ $$

Now, we can substitute the values ($$ y_p, y_p', y_p'' \! $$) into (1) to get:

$$ (-4K + 4M) e^{-2x}cos2x+(-4M-4K)e^{-2x}sin2x+4((-2K+2M)e^{-2x}xcos2x+(-2M-2K)e^{-2x}xsin2x+Ke^{-2x}cos2x+Me^{-2x}sinx) + 4(e^{-2x}(Kxcos2x + Mxsin2x)) = e^{2x}sin2x \! $$

$$ \therefore (-3K+4M)e^{-2x}cos2x + (-3M-4K)e^{-2x}sin2x = e^{-2x}sin2x \! $$

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

$$ -3K + 4M =0 \! $$ and $$ -4K - 3M = 1 \! $$

and finally discover that:

$$ M = - \frac{3}{25} \! $$ and $$ K = - \frac{4}{25} \! $$.

Plugging in these values in $$ y_p \! $$, we find that:

$$ y_p = e^{-2x}( - \frac{4}{25}xcos2x - \frac{3}{25}xsin2x) \! $$

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

$$ y = y_k + y_p \! $$

$$ y = (c_1 + c_2 x) e^{-2x}+e^{-2x} (-\frac{4}{25}xcos2x - \frac{3}{25}xsin2x) \! $$

Solving for the initial conditions given and first plugging in $$ y(0) = 1 \! $$, we get that:

$$ 1 = (c_1 + c_2 (0)) e^{-2(0)}+e^{-2(0)} (-\frac{4}{25}(0)cos2(0) - \frac{3}{25}(0)sin2(0)) \! $$

$$ 1 = (c_1 + c_2(0))e^0 \! $$

$$ \therefore c_1 = 1 \! $$

The second initial condition that was given to us, $$ y'(0) = -1.5 \! $$ can now be plugged in:

$$ y' = - \frac{1}{5} e^-2x (10c_1 + 10c_2x - 5c_2 + (3-14x)sin2x+(4-2x)cos(2x)) \! $$

$$ -1.5 = -\frac{1}{5}(10c_1 - 5c_2 + 4) \! $$

$$ \therefore c_2 = -3.5 \! $$

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

$$ y = (1 - 3.5 x) e^{-2x}+e^{-2x} (-\frac{4}{25}xcos2x - \frac{3}{25}xsin2x) \! $$