Back to the class
Problems #15 pg.184 and #9 pg.190 are graded.
#15,pg.184: Find the solution of the initial value problem
$$y''+y'-2y=2t;y(0)=0,y'(0)=1.$$
Solution: We will use the method of undetermined coefficients. First we solve the homogeneous equation
$$y''+y'-2y=0.$$
We use the characteristic equation $r^2+r-2=0$. This equation has roots at $r=-2,1$. Thus the homogeneous solution is
$$y_h(t) = c_1 e^{-2t} + c_2e^t.$$
Now we must find the particular solution $y_p(t)$ of the nonhomogeneous equation. We will proceed using the method of undetermined coefficients. The table tells us to guess
$$y_p(t) = t^s \left( A_0 + A_1 t \right).$$
In this case $\alpha=\beta=0$ and $\alpha + \beta i$ is not a root of the characteristic equation, so we take $s=0$. This yields the guess
$$y_p(t)=A_0+A_1t.$$
We will plug this into the differential equation. Compute $y_p'(t)=A_1$ and $y_p''(t)=0$. Plugging in yields
$$0+A_1-2(A_0+A_1t)=2t,$$
or equivalently
$$(A_1-2A_0) -2 A_1 t = 2t.$$
Equating coefficients leads to the system of equations
$$\left\{ \begin{array}{ll}
A_1-2A_0 &= 0\\
-2A_1 &= 2,
\end{array} \right.$$
yielding $A_1=-1$ and $A_0=\dfrac{A_1}{2}=-\dfrac{1}{2}$. Thus our particular solution is
$$y_p(t) = -\dfrac{1}{2}-t.$$
Thus the general solution is
$$y(t)=c_1e^{-2t} + c_2e^t -\dfrac{1}{2}-t.$$
Compute $y'(t) = -2c_1 e^{-2t} + c_2e^t - 1$. Now we may find the values of $c_1$ and $c_2$ using the initial conditions:
$$\left\{ \begin{array}{ll}
0=y(0)&=c_1 + c_2 - \dfrac{1}{2} \\
1=y'(0)&=-2c_1+c_2-1
\end{array} \right.$$
This system has solution $c_1=-\dfrac{1}{2}$ and $c_2=1$. Thus the solution of the initial value problem is
$$y(t) =-\dfrac{1}{2}e^{-2t} + e^t - \dfrac{1}{2}-t.$$
#9,pg.190: Find the general solution of the differential equation
$$4y''+y=2\sec \left( \dfrac{t}{2} \right); -\pi < t < \pi.$$
Solution: We solve the homogeneous equation
$$4y''+y=0$$
by solving the characteristic equation $4r^2+1=0$. We see that the solution of the characteristic equation is $\pm i\sqrt{ \dfrac{1}{2}}$ and hence the homogeneous solution is
$$y_h(t) = c_1 \cos \left( \dfrac{1}{2}t \right) + c_2 \sin \left( \dfrac{1}{2} t \right).$$
We will find the particular solution using variation of parameters. To do so, we "guess"
$$y_p(t) = u_1(t)y_1(t) + u_2(t)y_2(t) = u_1(t) \cos \left( \dfrac{t}{2} \right) + u_2(t) \sin \left( \dfrac{t}{2} \right).$$
First we calculate the Wronskian of the homogeneous solutions:
$$W\{y_1,y_2\}(t) = \det \begin{bmatrix} \cos(t/2) & \sin(t/2) \\ -\frac{1}{2}\sin(t/2) & \frac{1}{2}\cos(t/2) \end{bmatrix} = \dfrac{1}{2}\cos^2(t/2) - \left(-\dfrac{1}{2} \right)\sin(t/2) = \dfrac{1}{2}.$$
Before we proceed, first put the differential equation in standard form by dividing by $4$:
$$y''+\dfrac{y}{4} = \dfrac{1}{2} \sec \left( \dfrac{t}{2} \right).$$
Now we may apply variation of parameters to see
$$u_1(t)=-\displaystyle\int \dfrac{g(t)y_2(t)}{W\{y_1,y_2\}(t)}= \displaystyle\int \dfrac{\frac{1}{2}\sec(t/2)\sin(t/2)}{\frac{1}{2}} = \displaystyle\int \tan(t/2) dt=2 \log | \cos(t/2)|,$$
and
$$u_2(t) = \displaystyle\int \dfrac{g(t)y_1(t)}{W\{y_1,y_2\}(t)} = -\displaystyle\int \dfrac{\frac{1}{2}\sec(t/2)\cos(t/2)}{\frac{1}{2}}=-\displaystyle\int 1 dt = t$$
Thus the general solution is
$$y(t)=c_1 \cos \left( \dfrac{t}{2} \right) + c_2\sin \left( \dfrac{t}{2} \right) + 2 \log \left( \left|\cos \left( \dfrac{t}{2} \right) \right| \right)\cos \left( \dfrac{t}{2} \right) + t \sin \left( \dfrac{t}{2} \right).$$
