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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 $$y(t)=c_1 \cos \left( \dfrac{t}{2} \right) + c_2\sin \left( \dfrac{t}{2} \right) + 2 \log \left( \left|\cos \left( \dfrac{t}{2} \right) \right| \right)\cos \left( \dfrac{t}{2} \right) + t \sin \left( \dfrac{t}{2} \right).$$