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}
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}
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}
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}
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 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}
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}
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}
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 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}
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
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 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}
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}
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}
Hence c_{2}=\frac{1}{10}e^{9}
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
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
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}
Hence c_{2}=\frac{1}{2}e^{-1}+e^{-1}=\frac{3}{2}e^{-1}
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
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*}
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*}
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*}
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*}
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*}
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.