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2.5 HW5

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2.5.1 Section 3.1 problem 9

Find the solution to $y^{\prime \prime }+y^{\prime }-2y=0;y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =1$ and sketch the solution and describe its behavior as $t$ increases.

solution

The characteristic equation is found by substituting $y=e^{rt}$ into the ODE and simplifying, giving $$\begin{align*} r^{2}+r-2 & =0\\ \left ( r+2\right ) \left ( r-1\right ) & =0 \end{align*}$$

Hence the roots are $r_{1}=-2,r_{2}=1$. Roots are real and distinct. The two solutions are $$\begin{align*} y_{1} & =e^{-2t}\\ y_{2} & =e^{t} \end{align*}$$

The general solution is linear combination of the above two solutions $$\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{-2t}+c_{2}e^{t} \end{align*}$$

Now $c_{1},c_{2}$ are found from initial conditions.  Applying first initial condition ($y\left ( 0\right ) =1$) to the general solution gives $$\begin{equation} 1=c_{1}+c_{2} \tag{1} \end{equation}$$ Taking time derivative of the general solution gives $y^{\prime }\left ( t\right ) =-2c_{1}e^{-2t}+c_{2}e^{t}$. Applying second initial condition to this results in $$\begin{equation} 1=-2c_{1}+c_{2} \tag{2} \end{equation}$$ Equation (1,2) are now solved for $c_{1},c_{2}$. From (1), $c_{1}=1-c_{2}$. Substituting this into (2) gives $$\begin{align*} 1 & =-2\left ( 1-c_{2}\right ) +c_{2}\\ & =-2+2c_{2}+c_{2}\\ & =-2+3c_{2} \end{align*}$$

Hence $c_{2}=\frac{1+2}{3}=1$. Therefore $c_{1}=1-1=0$. Hence $$\begin{align*} c_{1} & =0\\ c_{2} & =1 \end{align*}$$

Substituting these back into the general solution gives $$y\left ( t\right ) =e^{t}$$ Since the solution is exponential, it will grow in time and blows up. Here is sketch of the solution.

2.5.2 Section 3.1 problem 10

Find the solution to $y^{\prime \prime }+4y^{\prime }+3y=0;y\left ( 0\right ) =2,y^{\prime }\left ( 0\right ) =-1$ and sketch the solution and describe its behavior as $t$ increases.

solution

The characteristic equation is found by substituting $y=e^{rt}$ into the ODE and simplifying, giving $$\begin{align*} r^{2}+4r+3 & =0\\ \left ( r+3\right ) \left ( r+1\right ) & =0 \end{align*}$$

Hence the roots are $r_{1}=-3,r_{2}=-1$. Roots are real and distinct. The two solutions are $$\begin{align*} y_{1} & =e^{-3t}\\ y_{2} & =e^{-t} \end{align*}$$

The general solution is linear combination of the above two solutions $$\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{-3t}+c_{2}e^{-t} \end{align*}$$

Now $c_{1},c_{2}$ are found from initial conditions.  Applying first initial condition ($y\left ( 0\right ) =2$) to the general solution gives $$\begin{equation} 2=c_{1}+c_{2} \tag{1} \end{equation}$$ Taking time derivative of the general solution gives $y^{\prime }\left ( t\right ) =-3c_{1}e^{-3t}-c_{2}e^{-t}$. Applying second initial condition to this results in $$\begin{equation} -1=-3c_{1}-c_{2} \tag{2} \end{equation}$$ Equation (1,2) are now solved for $c_{1},c_{2}$. From (1), $c_{1}=2-c_{2}$. Substituting this into (2) gives $$\begin{align*} -1 & =-3\left ( 2-c_{2}\right ) -c_{2}\\ & =-6+3c_{2}-c_{2}\\ & =-6+2c_{2} \end{align*}$$

Hence $c_{2}=\frac{-1+6}{2}=2.5$. Therefore $c_{1}=2-2.5=0.5$. Hence $$\begin{align*} c_{1} & =0.5\\ c_{2} & =2.5 \end{align*}$$

Substituting these back into the general solution gives $$y\left ( t\right ) =0.5e^{-3t}+2.5e^{-t}$$ At $t$ becomes large, both solutions decay to zero. So we expect the general solution to go to zero very fast. Here is a sketch.

2.5.3 Section 3.1 problem 11

Find the solution to $6y^{\prime \prime }-5y^{\prime }+y=0;y\left ( 0\right ) =4,y^{\prime }\left ( 0\right ) =0$ and sketch the solution and describe its behavior as $t$ increases.

solution

The characteristic equation is found by substituting $y=e^{rt}$ into the ODE and simplifying, giving $$6r^{2}-5r+1=0$$ Hence $r_{1,2}=\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$, where $\Delta =b^{2}-4ac=25-\left ( 4\right ) \left ( 6\right ) =1$. Since $\Delta >0$, the roots will be real and distinct. The roots are $$\begin{align*} r_{1,2} & =\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}\\ & =\frac{5}{12}\pm \frac{1}{12} \end{align*}$$

Hence the roots are $r_{1}=\frac{1}{2},r_{2}=\frac{1}{3}$. Roots are real and distinct. The two solutions are $$\begin{align*} y_{1} & =e^{\frac{1}{2}t}\\ y_{2} & =e^{\frac{1}{3}t} \end{align*}$$

The general solution is linear combination of the above two solutions $$\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}e^{\frac{1}{2}t}+c_{2}e^{\frac{1}{3}t} \end{align*}$$

Now $c_{1},c_{2}$ are found from initial conditions.  Applying first initial condition ($y\left ( 0\right ) =4$) to the general solution gives $$\begin{equation} 4=c_{1}+c_{2} \tag{1} \end{equation}$$ Taking time derivative of the general solution gives $y^{\prime }\left ( t\right ) =\frac{1}{2}c_{1}e^{\frac{1}{2}t}+\frac{1}{3}c_{2}e^{\frac{1}{3}t}$. Applying second initial condition to this results in $$\begin{equation} 0=\frac{1}{2}c_{1}+\frac{1}{3}c_{2} \tag{2} \end{equation}$$ Equation (1,2) are now solved for $c_{1},c_{2}$. From (1), $c_{1}=4-c_{2}$. Substituting this into (2) gives $$\begin{align*} 0 & =\frac{1}{2}\left ( 4-c_{2}\right ) +\frac{1}{3}c_{2}\\ & =2-\frac{1}{2}c_{2}+\frac{1}{3}c_{2}\\ & =2-\frac{1}{6}c_{2} \end{align*}$$

Hence $c_{2}=12$. Therefore $c_{1}=4-12=-8$. Hence $$\begin{align*} c_{1} & =-8\\ c_{2} & =12 \end{align*}$$

Substituting these back into the general solution gives $$y\left ( t\right ) =-8e^{\frac{1}{2}t}+12e^{\frac{1}{3}t}$$ Since $e^{\frac{1}{2}t}$ grows faster than $e^{\frac{1}{3}t}$ and since $e^{\frac{1}{2}t}$ has negative coefficient, then the solution will go to $-\infty$ as $t$ increases. Here is sketch of the solution

2.5.4 Section 3.1 problem 12

Find the solution to $y^{\prime \prime }+3y^{\prime }=0;y\left ( 0\right ) =-2,y^{\prime }\left ( 0\right ) =3$ and sketch the solution and describe its behavior as $t$ increases.

solution

The characteristic equation is found by substituting $y=e^{rt}$ into the ODE and simplifying, giving $$\begin{align*} r^{2}+3r & =0\\ r\left ( r+3\right ) & =0 \end{align*}$$

Hence the roots are $r_{1}=0,r_{2}=-3$. Roots are real and distinct. The two solutions are $$\begin{align*} y_{1} & =1\\ y_{2} & =e^{-3t} \end{align*}$$

The general solution is linear combination of the above two solutions $$y=c_{1}+c_{2}e^{-3t}$$ Now $c_{1},c_{2}$ are found from initial conditions.  Applying first initial condition ($y\left ( 0\right ) =-2$) to the general solution gives $$\begin{equation} -2=c_{1}+c_{2} \tag{1} \end{equation}$$ Taking time derivative of the general solution gives $y^{\prime }\left ( t\right ) =-3c_{2}e^{-3t}$. Applying second initial condition to this results in $$\begin{equation} 3=-3c_{2} \tag{2} \end{equation}$$ Hence $c_{2}=-1$. Therefore $c_{1}=-1$. Substituting these back into the general solution gives $$y\left ( t\right ) =-1-e^{-3t}$$ As $t\rightarrow \infty$, the term $e^{-3t}\rightarrow 0$ and we are left with $-1$. Hence $\lim _{t-\infty }y\left ( t\right ) =-1$. Here is sketch of the solution

2.5.5 Section 3.1 problem 13

Find the solution to $y^{\prime \prime }+5y^{\prime }+3y=0;y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0$ and sketch the solution and describe its behavior as $t$ increases.

solution

The characteristic equation is found by substituting $y=e^{rt}$ into the ODE and simplifying, giving $$r^{2}+5r+3=0$$ Hence $r_{1,2}=\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$, where $\Delta =b^{2}-4ac=25-\left ( 4\right ) \left ( 3\right ) =13$. Since $\Delta >0$, the roots will be real and distinct. The roots are $$\begin{align*} r_{1,2} & =\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}\\ & =\frac{-5}{2}\pm \frac{\sqrt{13}}{2} \end{align*}$$

Hence the roots are $r_{1}=\frac{-5}{2}+\frac{\sqrt{13}}{2},r_{2}=\frac{-5}{2}-\frac{\sqrt{13}}{2}$. The two solutions are $$\begin{align*} y_{1} & =e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}\\ y_{2} & =e^{\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) t} \end{align*}$$

The general solution is linear combination of the above two solutions $$y=c_{1}e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}+c_{2}e^{\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) t}$$ Now $c_{1},c_{2}$ are found from initial conditions.  Applying first initial condition ($y\left ( 0\right ) =1$) to the general solution gives $$\begin{equation} 1=c_{1}+c_{2} \tag{1} \end{equation}$$ Taking time derivative of the general solution gives $$y^{\prime }\left ( t\right ) =c_{1}\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}+c_{2}\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) e^{\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) t}$$ Applying second initial condition to this results in $$\begin{equation} 0=c_{1}\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) +c_{2}\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) \tag{2} \end{equation}$$ From (1), $c_{1}=1-c_{2}$ and from (2) $$\begin{align*} 0 & =\left ( 1-c_{2}\right ) \left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) +c_{2}\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) \\ & =\frac{-5}{2}+\frac{\sqrt{13}}{2}+\frac{5}{2}c_{2}-\frac{\sqrt{13}}{2}c_{2}-\frac{5}{2}c_{2}-\frac{\sqrt{13}}{2}c_{2}\\ & =-\frac{5}{2}+\frac{\sqrt{13}}{2}-\sqrt{13}c_{2}\\ c_{2} & =\frac{-5}{2\sqrt{13}}+\frac{1}{2}\\ & =\frac{-5+\sqrt{13}}{2\sqrt{13}} \end{align*}$$

Therefore $c_{1}=1+\frac{5-\sqrt{13}}{2\sqrt{13}}$ and the solution becomes $$\begin{align*} y\left ( t\right ) & =\left ( 1+\frac{5-\sqrt{13}}{2\sqrt{13}}\right ) e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}+\left ( \frac{-5+\sqrt{13}}{2\sqrt{13}}\right ) e^{\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) t}\\ & =e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}+\frac{5-\sqrt{13}}{2\sqrt{13}}e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}+\left ( \frac{-5+\sqrt{13}}{2\sqrt{13}}\right ) e^{\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) t}\\ & =e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}+\frac{5\sqrt{13}-13}{26}e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}+\left ( \frac{-5\sqrt{13}+13}{26}\right ) e^{\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) t}\\ & =\frac{1}{26}\left ( 26e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}+\left ( 5\sqrt{13}-13\right ) e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}+\left ( -5\sqrt{13}+13\right ) e^{\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) t}\right ) \\ & =\frac{1}{26}\left ( 26e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}+5\sqrt{13}e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}-13e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}-5\sqrt{13}e^{\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) t}+13e^{\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) t}\right ) \\ & =\frac{1}{26}\left ( 13e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}+5\sqrt{13}e^{\left ( \frac{-5}{2}+\frac{\sqrt{13}}{2}\right ) t}-5\sqrt{13}e^{\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) t}+13e^{\left ( \frac{-5}{2}-\frac{\sqrt{13}}{2}\right ) t}\right ) \end{align*}$$

Here is sketch of the solution showing that $y\rightarrow 0$ as $t\rightarrow \infty$

2.5.6 Section 3.1 problem 14

Find the solution to $2y^{\prime \prime }+y^{\prime }-4y=0;y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =1$ and sketch the solution and describe its behavior as $t$ increases.

solution

The characteristic equation is found by substituting $y=e^{rt}$ into the ODE and simplifying, giving $$2r^{2}+r-4=0$$ Hence $r_{1,2}=\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}$, where $\Delta =b^{2}-4ac=1-\left ( 4\right ) \left ( 2\right ) \left ( -4\right ) =33$. Since $\Delta >0$, the roots will be real and distinct. The roots are $$\begin{align*} r_{1,2} & =\frac{-b}{2a}\pm \frac{\sqrt{b^{2}-4ac}}{2a}\\ & =\frac{-1}{4}\pm \frac{\sqrt{33}}{4} \end{align*}$$

Hence the roots are $r_{1}=\frac{1}{4}+\frac{\sqrt{33}}{4},r_{2}=\frac{1}{4}-\frac{\sqrt{33}}{4}$. The two solutions are $$\begin{align*} y_{1} & =e^{\left ( -\frac{1}{4}+\frac{\sqrt{33}}{4}\right ) t}\\ y_{2} & =e^{\left ( -\frac{1}{4}-\frac{\sqrt{33}}{4}\right ) t} \end{align*}$$

The general solution is linear combination of the above two solutions $$y=c_{1}e^{\left ( -\frac{1}{4}+\frac{\sqrt{33}}{4}\right ) t}+c_{2}e^{\left ( -\frac{1}{4}-\frac{\sqrt{33}}{4}\right ) t}$$ Now $c_{1},c_{2}$ are found from initial conditions.  Applying first initial condition ($y\left ( 0\right ) =0$) to the general solution gives $$\begin{equation} 0=c_{1}+c_{2} \tag{1} \end{equation}$$ Taking time derivative of the general solution gives $$y^{\prime }\left ( t\right ) =c_{1}\left ( -\frac{1}{4}+\frac{\sqrt{33}}{4}\right ) e^{\left ( \frac{1}{4}+\frac{\sqrt{33}}{4}\right ) t}+c_{2}\left ( -\frac{1}{4}-\frac{\sqrt{33}}{4}\right ) e^{\left ( \frac{1}{4}-\frac{\sqrt{33}}{4}\right ) t}$$ Applying second initial condition to this results in $$\begin{equation} 1=c_{1}\left ( -\frac{1}{4}+\frac{\sqrt{33}}{4}\right ) +c_{2}\left ( -\frac{1}{4}-\frac{\sqrt{33}}{4}\right ) \tag{2} \end{equation}$$ From (1), $c_{1}=-c_{2}$ and from (2) $$\begin{align*} 1 & =-c_{2}\left ( -\frac{1}{4}+\frac{\sqrt{33}}{4}\right ) +c_{2}\left ( -\frac{1}{4}-\frac{\sqrt{33}}{4}\right ) \\ & =\frac{1}{4}c_{2}-\frac{\sqrt{33}}{4}c_{2}-\frac{1}{4}c_{2}-\frac{\sqrt{33}}{4}c_{2}\\ & =\frac{-\sqrt{33}}{2}c_{2}\\ c_{2} & =\frac{-2}{\sqrt{33}} \end{align*}$$

Therefore $c_{1}=\frac{2}{\sqrt{33}}$ and the solution becomes $$y=\frac{2}{\sqrt{33}}e^{\left ( -\frac{1}{4}+\frac{\sqrt{33}}{4}\right ) t}-\frac{2}{\sqrt{33}}e^{\left ( -\frac{1}{4}-\frac{\sqrt{33}}{4}\right ) t}$$ Since $-\frac{1}{4}+\frac{\sqrt{33}}{4}=1.186$ and $-\frac{1}{4}-\frac{\sqrt{33}}{4}=-1.686$ then the above can be written as $$y=\frac{2}{\sqrt{33}}e^{1.186t}-\frac{2}{\sqrt{33}}e^{-1.186t}$$ Then we see that as $t\rightarrow \infty$ the second term $e^{-1.186t}\rightarrow 0$ and we are left with $e^{1.186t}$ which will go to $\infty$ for large $t$. Hence $$\lim _{t\rightarrow \infty }y\left ( t\right ) =\infty$$ Here is sketch of the solution

2.5.7 Section 3.1 problem 15

Find the solution to $y^{\prime \prime }+8y^{\prime }-9y=0;y\left ( 1\right ) =1,y^{\prime }\left ( 1\right ) =0$ and sketch the solution and describe its behavior as $t$ increases.

solution

The characteristic equation is found by substituting $y=e^{rt}$ into the ODE and simplifying, giving $$\begin{align*} r^{2}+8r-9 & =0\\ \left ( r-1\right ) \left ( r+9\right ) & =0 \end{align*}$$

Hence the roots are $r_{1}=1,r_{2}=-9$. The two solutions are $$\begin{align*} y_{1} & =e^{t}\\ y_{2} & =e^{-9t} \end{align*}$$

The general solution is linear combination of the above two solutions $$y=c_{1}e^{t}+c_{2}e^{-9t}$$ Now $c_{1},c_{2}$ are found from initial conditions.  Applying first initial condition ($y\left ( 1\right ) =1$) to the general solution gives $$\begin{equation} 1=c_{1}e^{1}+c_{2}e^{-9} \tag{1} \end{equation}$$ Taking time derivative of the general solution gives $$y^{\prime }\left ( t\right ) =c_{1}e^{t}-9c_{2}e^{-9t}$$ Applying second initial condition to this results in $$\begin{equation} 0=c_{1}e^{1}-9c_{2}e^{-9} \tag{2} \end{equation}$$ From (1), $c_{1}=\frac{1-c_{2}e^{-9}}{e^{1}}=e^{-1}-c_{2}e^{-10}$ and from (2) $$\begin{align*} 0 & =\left ( e^{-1}-c_{2}e^{-10}\right ) e^{1}-9c_{2}e^{-9}\\ & =1-c_{2}e^{-9}-9c_{2}e^{-9}\\ & =1+c_{2}\left ( -e^{-9}-9e^{-9}\right ) \\ 0 & =1+c_{2}\left ( -10e^{-9}\right ) \end{align*}$$

Hence $$c_{2}=\frac{1}{10}e^{9}$$ Therefore $c_{1}=e^{-1}-c_{2}e^{-10}=e^{-1}-\frac{1}{10}e^{9}e^{-10}=e^{-1}-\frac{1}{10}e^{-1}=\frac{9}{10}e^{-1}$ and the solution becomes $$\begin{align*} y & =\frac{9}{10}e^{-1}e^{t}+\frac{1}{10}e^{9}e^{-9t}\\ & =\frac{9}{10}e^{t-1}+\frac{1}{10}e^{9-9t} \end{align*}$$

Then we see that as $t\rightarrow \infty$ the second term $e^{9-9t}\rightarrow 0$ and we are left with $e^{t-1}$ which will go to $\infty$ for large $t$. Hence $$\lim _{t\rightarrow \infty }y\left ( t\right ) =\infty$$ Here is sketch of the solution.

2.5.8 Section 3.1 problem 16

Find the solution to $4y^{\prime \prime }-y=0;y\left ( -2\right ) =1,y^{\prime }\left ( -2\right ) =-1$ and sketch the solution and describe its behavior as $t$ increases.

solution

The characteristic equation is found by substituting $y=e^{rt}$ into the ODE and simplifying, giving $$4r^{2}-1=0$$ Hence the roots are $r_{1}=\pm \frac{1}{2}$. The two solutions are $$\begin{align*} y_{1} & =e^{\frac{1}{2}t}\\ y_{2} & =e^{-\frac{1}{2}t} \end{align*}$$

The general solution is linear combination of the above two solutions $$y=c_{1}e^{\frac{1}{2}t}+c_{2}e^{-\frac{1}{2}t}$$ Now $c_{1},c_{2}$ are found from initial conditions.  Applying first initial condition ($y\left ( -2\right ) =1$) to the general solution gives $$\begin{equation} 1=c_{1}e^{-1}+c_{2}e \tag{1} \end{equation}$$ Taking time derivative of the general solution gives $$y^{\prime }\left ( t\right ) =\frac{1}{2}c_{1}e^{\frac{1}{2}t}-\frac{1}{2}c_{2}e^{-\frac{1}{2}t}$$ Applying second initial condition to this results in $$\begin{equation} -1=\frac{1}{2}c_{1}e^{-1}-\frac{1}{2}c_{2}e \tag{2} \end{equation}$$ From (1), $c_{1}=\frac{1-c_{2}e}{e^{-1}}=e-c_{2}e^{2}$ and from (2) $$\begin{align*} -1 & =\frac{1}{2}\left ( e-c_{2}e^{2}\right ) e^{-1}-\frac{1}{2}c_{2}e\\ & =\frac{1}{2}-\frac{1}{2}c_{2}e-\frac{1}{2}c_{2}e\\ & =\frac{1}{2}-c_{2}e \end{align*}$$

Hence $$c_{2}=\frac{1}{2}e^{-1}+e^{-1}=\frac{3}{2}e^{-1}$$ Therefore $c_{1}=e-\left ( \frac{3}{2}e^{-1}\right ) e^{2}=e-\frac{3}{2}e=-\frac{1}{2}e$ and the solution becomes $$\begin{align*} y & =c_{1}e^{\frac{1}{2}t}+c_{2}e^{-\frac{1}{2}t}\\ & =-\frac{1}{2}ee^{\frac{1}{2}t}+\frac{3}{2}e^{-1}e^{\frac{-1}{2}t}\\ & =-\frac{1}{2}e^{1+\frac{t}{2}}+\frac{3}{2}e^{-1-\frac{t}{2}} \end{align*}$$

Then we see that as $t\rightarrow \infty$ the second term $e^{-1-\frac{t}{2}}\rightarrow 0$ and we are left with $-\frac{1}{2}e^{1+\frac{t}{2}}$ which will go to $-\infty$ for large $t$. Hence $$\lim _{t\rightarrow \infty }y\left ( t\right ) =-\infty$$ Here is sketch of the solution.

2.5.9 Section 3.2 problem 1

Find the Wronskian of the given pair of functions $e^{2t},e^{-\frac{3t}{2}}$

solution

We are given $y_{1}\left ( t\right ) =e^{2t},y_{2}\left ( t\right ) =e^{\frac{-3}{2}t}$, hence by definition, the Wronskian is $$\begin{align*} W & =\begin{vmatrix} y_{1}\left ( t\right ) & y_{2}\left ( t\right ) \\ y_{1}^{\prime }\left ( t\right ) & y_{2}^{\prime }\left ( t\right ) \end{vmatrix} \\ & =\begin{vmatrix} e^{2t} & e^{-\frac{3t}{2}}\\ 2e^{2t} & -\frac{2}{3}e^{-\frac{3t}{2}}\end{vmatrix} \\ & =\frac{-3}{2}e^{\frac{t}{2}}-2e^{\frac{t}{2}}\\ & =\frac{-7}{2}e^{\frac{t}{2}} \end{align*}$$

2.5.10 Section 3.2 problem 2

Find the Wronskian of the given pair of functions $\cos t,\sin t$

solution

We are given $y_{1}\left ( t\right ) =\cos t,y_{2}\left ( t\right ) =\sin t$, hence by definition, the Wronskian is $$\begin{align*} W & =\begin{vmatrix} y_{1}\left ( t\right ) & y_{2}\left ( t\right ) \\ y_{1}^{\prime }\left ( t\right ) & y_{2}^{\prime }\left ( t\right ) \end{vmatrix} \\ & =\begin{vmatrix} \cos t & \sin t\\ -\sin t & \cos t \end{vmatrix} \\ & =\cos ^{2}t+\sin ^{2}t\\ & =1 \end{align*}$$

2.5.11 Section 3.2 problem 3

Find the Wronskian of the given pair of functions $e^{-2t},te^{-2t}$

solution

We are given $y_{1}\left ( t\right ) =e^{-2t},y_{2}\left ( t\right ) =te^{-2t}$, hence by definition, the Wronskian is $$\begin{align*} W & =\begin{vmatrix} y_{1}\left ( t\right ) & y_{2}\left ( t\right ) \\ y_{1}^{\prime }\left ( t\right ) & y_{2}^{\prime }\left ( t\right ) \end{vmatrix} \\ & =\begin{vmatrix} e^{-2t} & te^{-2t}\\ -2e^{-2t} & e^{-2t}-2te^{-2t}\end{vmatrix} \\ & =\left ( e^{-2t}\right ) \left ( e^{-2t}-2te^{-2t}\right ) +2e^{-2t}te^{-2t}\\ & =e^{-4t}-2te^{-4t}+2te^{-4t}\\ & =e^{-4t} \end{align*}$$

2.5.12 Section 3.2 problem 4

Find the Wronskian of the given pair of functions $x,xe^{x}$

solution

We are given $y_{1}\left ( x\right ) =x,y_{2}\left ( x\right ) =xe^{x}$, hence by definition, the Wronskian is $$\begin{align*} W & =\begin{vmatrix} y_{1}\left ( x\right ) & y_{2}\left ( x\right ) \\ y_{1}^{\prime }\left ( x\right ) & y_{2}^{\prime }\left ( x\right ) \end{vmatrix} \\ & =\begin{vmatrix} x & xe^{x}\\ 1 & e^{x}+xe^{x}\end{vmatrix} \\ & =\left ( x\right ) \left ( e^{x}+xe^{x}\right ) -xe^{x}\\ & =xe^{x}+x^{2}e^{x}-xe^{x}\\ & =x^{2}e^{x} \end{align*}$$

2.5.13 Section 3.2 problem 5

Find the Wronskian of the given pair of functions $e^{t}\sin t,e^{t}\cos t$

solution

We are given $y_{1}\left ( t\right ) =e^{t}\sin t,y_{2}\left ( t\right ) =e^{t}\cos t$, hence by definition, the Wronskian is $$\begin{align*} W & =\begin{vmatrix} y_{1}\left ( t\right ) & y_{2}\left ( t\right ) \\ y_{1}^{\prime }\left ( t\right ) & y_{2}^{\prime }\left ( t\right ) \end{vmatrix} \\ & =\begin{vmatrix} e^{t}\sin t & e^{t}\cos t\\ e^{t}\sin t+e^{t}\cos t & e^{t}\cos t-e^{t}\sin t \end{vmatrix} \\ & =\left ( e^{t}\sin t\right ) \left ( e^{t}\cos t-e^{t}\sin t\right ) -e^{t}\cos t\left ( e^{t}\sin t+e^{t}\cos t\right ) \\ & =e^{2t}\sin t\cos t-e^{2t}\sin ^{2}t-e^{2t}\cos t\sin t-e^{2t}\cos ^{2}t\\ & =-e^{2t}\sin ^{2}t-e^{2t}\cos ^{2}t\\ & =-2e^{2t}\left ( \sin ^{2}t+\cos ^{2}t\right ) \\ & =-2e^{2t} \end{align*}$$

2.5.14 Section 3.2 problem 6

Find the Wronskian of the given pair of functions $\cos ^{2}\theta ,1+\cos 2\theta$

solution

We are given $y_{1}\left ( \theta \right ) =\cos ^{2}\theta ,y_{2}\left ( \theta \right ) =1+\cos 2\theta$, hence by definition, the Wronskian is $$\begin{align*} W & =\begin{vmatrix} y_{1}\left ( \theta \right ) & y_{2}\left ( \theta \right ) \\ y_{1}^{\prime }\left ( \theta \right ) & y_{2}^{\prime }\left ( \theta \right ) \end{vmatrix} \\ & =\begin{vmatrix} \cos ^{2}\theta & 1+\cos 2\theta \\ -2\cos \theta \sin \theta & -2\sin 2\theta \end{vmatrix} \\ & =-2\cos ^{2}\theta \sin 2\theta -\left ( 1+\cos 2\theta \right ) \left ( -2\cos \theta \sin \theta \right ) \\ & =-2\cos ^{2}\theta \sin 2\theta -\left ( -2\cos \theta \sin \theta -2\cos \theta \sin \theta \cos 2\theta \right ) \\ & =-2\cos ^{2}\theta \sin 2\theta +2\cos \theta \sin \theta +2\cos \theta \sin \theta \cos 2\theta \end{align*}$$

Using $\cos 2\theta =2\cos ^{2}\theta -1$ And $\sin 2\theta =2\sin \theta \cos \theta$ the above becomes $$\begin{align*} W & =-2\cos ^{2}\theta \left ( 2\sin \theta \cos \theta \right ) +2\cos \theta \sin \theta +2\cos \theta \sin \theta \left ( 2\cos ^{2}\theta -1\right ) \\ & =-4\cos ^{3}\theta \sin \theta +2\cos \theta \sin \theta +4\cos ^{3}\theta \sin \theta -2\cos \theta \sin \theta \\ & =-4\cos ^{3}\theta \sin \theta +4\cos ^{3}\theta \sin \theta \\ & =0 \end{align*}$$

We could also see that $W=0$ more directly, by noticing that $y_{1}=\cos ^{2}\theta =1-\sin ^{2}\theta$ and since $\sin ^{2}\theta =\frac{1}{2}-\frac{1}{2}\cos 2\theta$ then

$$\begin{align*} y_{1} & =\cos ^{2}\theta \\ & =1-\left ( \frac{1}{2}-\frac{1}{2}\cos 2\theta \right ) \\ & =\frac{1}{2}+\frac{1}{2}\cos 2\theta \\ & =\frac{1}{2}\left ( 1+\cos 2\theta \right ) \end{align*}$$

Therefore, $y_{1}=\frac{1}{2}y_{2}$. Hence $y_{2}$ is just a scaled version of $y_{1}$ and so these are two solutions are not linearly independent functions, (parallel to each others in vector space view) and so we expect that the Wronskian to be zero.