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2.5 HW5

  2.5.1 Section 3.1 problem 9
  2.5.2 Section 3.1 problem 10
  2.5.3 Section 3.1 problem 11
  2.5.4 Section 3.1 problem 12
  2.5.5 Section 3.1 problem 13
  2.5.6 Section 3.1 problem 14
  2.5.7 Section 3.1 problem 15
  2.5.8 Section 3.1 problem 16
  2.5.9 Section 3.2 problem 1
  2.5.10 Section 3.2 problem 2
  2.5.11 Section 3.2 problem 3
  2.5.12 Section 3.2 problem 4
  2.5.13 Section 3.2 problem 5
  2.5.14 Section 3.2 problem 6
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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.

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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.

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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

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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

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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

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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

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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.

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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.

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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.