Find the general solution of \(y^{\prime \prime }-2y^{\prime }+y=0\)
Solution:
The characteristic equation is found by substituting \(y=e^{rt}\) into the ODE and simplifying, giving\begin{align*} r^{2}-2r+1 & =0\\ \left ( r-1\right ) \left ( r-1\right ) & =0 \end{align*}
Hence \(r=1\) double root. Therefore the two solutions are\begin{align*} y_{1} & =e^{t}\\ y_{2} & =te^{t} \end{align*}
And the general solution is linear combination of the above solutions\[ y=c_{1}e^{t}+c_{2}te^{t}\]
Find the general solution of \(9y^{\prime \prime }+6y^{\prime }+y=0\)
Solution:
The characteristic equation is found by substituting \(y=e^{rt}\) into the ODE and simplifying, giving\begin{align*} 9r^{2}+6r+1 & =0\\ r & =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-6+\sqrt{36-36}}{18}=-\frac{1}{3} \end{align*}
Hence \(r=-\frac{1}{3}\) double root. Therefore the two solutions are\begin{align*} y_{1} & =e^{\frac{-1}{3}t}\\ y_{2} & =te^{\frac{-1}{3}t} \end{align*}
And the general solution is linear combination of the above solutions\[ y=c_{1}e^{\frac{-1}{3}t}+c_{2}te^{\frac{-1}{3}t}\]
Find the general solution of \(4y^{\prime \prime }-4y^{\prime }-3y=0\)
Solution:
The characteristic equation is found by substituting \(y=e^{rt}\) into the ODE and simplifying, giving\begin{align*} 4r^{2}-4r-3 & =0\\ r & =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{4\pm \sqrt{16+48}}{8}=\frac{4\pm 8}{8}=\frac{1\pm 2}{2}=\frac{1}{2}\pm 1 \end{align*}
Hence \(r_{1}=\frac{3}{2},r_{2}=-\frac{1}{2}\). Therefore the two solutions are\begin{align*} y_{1} & =e^{\frac{3}{2}t}\\ y_{2} & =e^{-\frac{1}{2}t} \end{align*}
And the general solution is linear combination of the above solutions\[ y=c_{1}e^{\frac{3}{2}t}+c_{2}e^{-\frac{1}{2}t}\]
Find the general solution of \(4y^{\prime \prime }+12y^{\prime }+9y=0\)
Solution:
The characteristic equation is found by substituting \(y=e^{rt}\) into the ODE and simplifying, giving\begin{align*} 4r^{2}+12r+9 & =0\\ r & =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-12\pm \sqrt{144-144}}{8}=\frac{-3}{2} \end{align*}
Hence \(r=\frac{-3}{2}\) double root. Therefore the two solutions are\begin{align*} y_{1} & =e^{\frac{-3}{2}t}\\ y_{2} & =te^{\frac{-3}{2}t} \end{align*}
And the general solution is linear combination of the above solutions
\[ y=c_{1}e^{\frac{-3}{2}t}+c_{2}te^{\frac{-3}{2}t}\]
Find the general solution of \(y^{\prime \prime }-2y^{\prime }+10y=0\)
Solution:
The characteristic equation is found by substituting \(y=e^{rt}\) into the ODE and simplifying, giving\begin{align*} r^{2}-2r+10 & =0\\ r & =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{2\pm \sqrt{4-40}}{2}=\frac{2\pm \sqrt{-36}}{2}=\frac{2\pm 6i}{2}=1\pm 3i \end{align*}
Hence \(r_{1}=1+3i,r_{2}=1-3i\). Therefore the two solutions are\begin{align*} y_{1} & =e^{\left ( 1+3i\right ) t}=e^{t}e^{i3t}\\ y_{2} & =e^{t}e^{i3t} \end{align*}
And the general solution is linear combination of the above solutions, the complex exponential can be converted to trig functions \(\cos ,\sin \) using the standard Euler identities, resulting in\[ y=e^{t}\left ( c_{1}\cos 3t+c_{2}\sin 3t\right ) \]
Find the general solution of \(y^{\prime \prime }-6y^{\prime }+9y=0\)
Solution:
The characteristic equation is found by substituting \(y=e^{rt}\) into the ODE and simplifying, giving\begin{align*} r^{2}-6r+9 & =0\\ \left ( r-3\right ) ^{2} & =0 \end{align*}
Hence \(r=3\). Double root. Therefore the two solutions are\begin{align*} y_{1} & =e^{3t}\\ y_{2} & =te^{3t} \end{align*}
And the general solution is linear combination of the above solutions\[ y=c_{1}e^{3t}+c_{2}te^{3t}\]
Find the general solution of \(4y^{\prime \prime }+17y^{\prime }+4y=0\)
Solution:
The characteristic equation is found by substituting \(y=e^{rt}\) into the ODE and simplifying, giving\begin{align*} 4r^{2}+17r+4 & =0\\ r & =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-17\pm \sqrt{289-64}}{8}=\frac{-17\pm \sqrt{225}}{8}=\frac{-17\pm 15}{8} \end{align*}
Hence \(r_{1}=\frac{-17-15}{8}=\allowbreak -4,r_{2}=\frac{-17+15}{8}=-\frac{1}{4}\). Therefore the two solutions are\begin{align*} y_{1} & =e^{-4t}\\ y_{2} & =e^{-\frac{1}{4}t} \end{align*}
And the general solution is linear combination of the above solutions\[ y=c_{1}e^{-4t}+c_{2}e^{-\frac{1}{4}t}\]
Find the general solution of \(16y^{\prime \prime }+24y^{\prime }+9y=0\)
Solution:
The characteristic equation is found by substituting \(y=e^{rt}\) into the ODE and simplifying, giving\begin{align*} 16r^{2}+24r+9 & =0\\ r & =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-24\pm \sqrt{576-4\left ( 16\right ) \left ( 9\right ) }}{32}=\frac{-24}{32}=-\frac{3}{4} \end{align*}
Hence \(r=-\frac{3}{4}\). Double root. Therefore the two solutions are\begin{align*} y_{1} & =e^{-\frac{3}{4}t}\\ y_{2} & =te^{-\frac{3}{4}t} \end{align*}
And the general solution is linear combination of the above solutions\[ y=c_{1}e^{-\frac{3}{4}t}+c_{2}te^{-\frac{3}{4}t}\]
Find the general solution of \(25y^{\prime \prime }-20y^{\prime }+4y=0\)
Solution:
The characteristic equation is found by substituting \(y=e^{rt}\) into the ODE and simplifying, giving\begin{align*} 25r^{2}-20r+4 & =0\\ r & =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{20\pm \sqrt{400-4\left ( 25\right ) \left ( 4\right ) }}{50}=\frac{20}{50}=\frac{2}{5} \end{align*}
Hence \(r=\frac{2}{5}\). Double root. Therefore the two solutions are\begin{align*} y_{1} & =e^{\frac{2}{5}t}\\ y_{2} & =te^{\frac{2}{5}t} \end{align*}
And the general solution is linear combination of the above solutions\[ y=c_{1}e^{\frac{2}{5}t}+c_{2}te^{\frac{2}{5}t}\]
Find the general solution of \(2y^{\prime \prime }+2y^{\prime }+y=0\)
Solution:
The characteristic equation is found by substituting \(y=e^{rt}\) into the ODE and simplifying, giving\begin{align*} 2r^{2}+2r+1 & =0\\ r & =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-2\pm \sqrt{4-4\left ( 2\right ) \left ( 1\right ) }}{4}=\frac{-2\pm \sqrt{-4}}{4}=\frac{-2\pm 2i}{4}=\frac{-1}{2}\pm \frac{i}{2} \end{align*}
Hence \(r_{1}=\frac{-1}{2}+\frac{i}{2},r_{2}=\frac{-1}{2}-\frac{i}{2}\). Therefore the two solutions are\begin{align*} y_{1} & =e^{\left ( \frac{-1}{2}-\frac{i}{2}\right ) t}=e^{\frac{-1}{2}t}e^{\frac{-i}{2}t}\\ y_{2} & =e^{\left ( \frac{-1}{2}+\frac{i}{2}\right ) t}=e^{\frac{-1}{2}t}e^{\frac{i}{2}t} \end{align*}
And the general solution is linear combination of the above solutions, the complex exponential can be converted to trig functions \(\cos ,\sin \) using the standard Euler identities, resulting in\[ y=e^{\frac{-1}{2}t}\left ( c_{1}\cos \frac{t}{2}+c_{2}\sin \frac{t}{2}\right ) \]
Find the general solution of \(y^{\prime \prime }-2y^{\prime }-3y=3e^{2t}\)
Solution:
The first step is to solve the homogenous ODE and find \(y_{h}\), then find a particular solution \(y_{p}\) to the inhomogeneous ODE, then add both solutions \(y_{h}+y_{p}\) in order to find the complete solution.
Finding \(y_{h}\)
We need to solve homogenous ODE \[ y^{\prime \prime }-2y^{\prime }-3y=0 \] The characteristic equation is found by substituting \(y=e^{rt}\) into the above ODE and simplifying, giving\begin{align*} r^{2}-2r-3 & =0\\ \left ( r+1\right ) \left ( r-3\right ) & =0 \end{align*}
Hence \(r_{1}=-1,r_{2}=3\). Therefore the two solution are\begin{align*} y_{1} & =e^{-t}\\ y_{2} & =e^{3t} \end{align*}
And the homogeneous solution is linear combination of the above solutions\[ y_{h}=c_{1}e^{-t}+c_{2}e^{3t}\]
Finding \(y_{p}\)
Now we need to find one particular solution to \[ y^{\prime \prime }-2y^{\prime }-3y=3e^{2t}\] We guess \(y_{p}=Ae^{2t}\). Hence\begin{align*} y_{p}^{\prime } & =2Ae^{2t}\\ y_{p}^{\prime \prime } & =4Ae^{2t} \end{align*}
Substituting this into the original ODE in order to solve for \(A\) gives\begin{align*} 4Ae^{2t}-2\left ( 2Ae^{2t}\right ) -3\left ( Ae^{2t}\right ) & =3e^{2t}\\ -3Ae^{2t} & =3e^{2t} \end{align*}
Hence \(A=-1\) and therefore\[ y_{p}=-e^{2t}\] Therefore the general solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}e^{-t}+c_{2}e^{3t}-e^{2t} \end{align*}
Find the general solution of \(y^{\prime \prime }+2y^{\prime }+5y=3\sin 2t\)
Solution:
The first step is to solve the homogenous ODE and find \(y_{h}\), then find a particular solution \(y_{p}\) to the inhomogeneous ODE, then add both solutions \(y_{h}+y_{p}\) in order to find the complete solution.
Finding \(y_{h}\)
We need to solve homogenous ODE \[ y^{\prime \prime }+2y^{\prime }+5y=0 \] The characteristic equation is found by substituting \(y=e^{rt}\) into the above ODE and simplifying, giving\begin{align*} r^{2}+2r+5 & =0\\ r & =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}=\frac{-2\pm \sqrt{4-20}}{2}=\frac{-2\pm \sqrt{-16}}{2}=\frac{-2\pm 4i}{2}=-1\pm 2i \end{align*}
From before, we know the solution is of the form\[ y_{h}=e^{-t}\left ( c_{1}\cos 2t+\sin 2t\right ) \] Where\begin{align*} y_{1} & =e^{-t}\cos 2t\\ y_{2} & =e^{-t}\sin 2t \end{align*}
Finding \(y_{p}\)
Now we need to find one particular solution to \[ y^{\prime \prime }+2y^{\prime }+5y=3\sin 2t \] We guess \(y_{p}=A\cos 2t+B\sin 2t\) hence\begin{align*} y_{p}^{\prime } & =-2A\sin 2t+2B\cos 2t\\ y_{p}^{\prime \prime } & =-4A\cos 2t-4B\sin 2t \end{align*}
Substituting these back into the original ODE in order to solve for \(A,B\) gives\begin{align*} y_{p}^{\prime \prime }+2y_{p}^{\prime }+5y_{p} & =3\sin 2t\\ -4A\cos 2t-4B\sin 2t+2\left ( -2A\sin 2t+2B\cos 2t\right ) +5\left ( A\cos 2t+B\sin 2t\right ) & =3\sin 2t\\ -4A\cos 2t-4B\sin 2t-4A\sin 2t+4B\cos 2t+5A\cos 2t+5B\sin 2t & =3\sin 2t\\ \left ( A+4B\right ) \cos 2t+\left ( B-4A\right ) \sin 2t & =3\sin 2t \end{align*}
Hence\begin{align*} A+4B & =0\\ B-4A & =3 \end{align*}
From first equation, \(A=-4B,\) and the second equation becomes \(B-4\left ( -4B\right ) =3\) or \(B+16B=3\) or \(B=\frac{3}{17}\), hence \(A=\frac{-12}{17}\), therefore\[ y_{p}=\frac{-12}{17}\cos 2t+\frac{3}{17}\sin 2t \] Therefore the general solution is\begin{align*} y & =y_{h}+y_{p}\\ & =e^{-t}\left ( c_{1}\cos 2t+\sin 2t\right ) -\frac{12}{17}\cos 2t+\frac{3}{17}\sin 2t \end{align*}
Find the general solution of \(y^{\prime \prime }-y^{\prime }-2y=-2t+4t^{2}\)
Solution:
The first step is to solve the homogenous ODE and find \(y_{h}\), then find a particular solution \(y_{p}\) to the inhomogeneous ODE, then add both solutions \(y_{h}+y_{p}\) in order to find the complete solution.
Finding \(y_{h}\)
We need to solve homogenous ODE \[ y^{\prime \prime }-y^{\prime }-2y=0 \] The characteristic equation is found by substituting \(y=e^{rt}\) into the above ODE and simplifying, giving\begin{align*} r^{2}-r-2 & =0\\ \left ( r+1\right ) \left ( r-2\right ) & =0 \end{align*}
Hence \(r_{1}=-1,r_{2}=2\) and therefore\[ y_{h}=c_{1}e^{-t}+c_{2}e^{2t}\] Finding \(y_{p}\)
Now we need to find one particular solution to \[ y^{\prime \prime }-y^{\prime }-2y=-2t+4t^{2}\] We guess \(y_{p}=A_{0}+A_{1}t+A_{2}t^{2}\). Therefore\begin{align*} y_{p}^{\prime } & =A_{1}+2A_{2}t\\ y_{p}^{\prime \prime } & =2A_{2} \end{align*}
Substituting these back into the original ODE gives\begin{align*} 2A_{2}-\left ( A_{1}+2A_{2}t\right ) -2\left ( A_{0}+A_{1}t+A_{2}t^{2}\right ) & =-2t+4t^{2}\\ t^{0}\left ( 2A_{2}-A_{1}-2A_{0}\right ) +t\left ( -2A_{2}-2A_{1}\right ) +t^{2}\left ( -2A_{2}\right ) & =-2t+4t^{2} \end{align*}
Hence\begin{align*} 2A_{2}-A_{1}-2A_{0} & =0\\ -2A_{2}-2A_{1} & =-2\\ -2A_{2} & =4 \end{align*}
From the last equation, \(A_{2}=-2\), and from the second equation \(A_{1}=\frac{-2+2\left ( -2\right ) }{-2}=3\) and from the first equation \(2\left ( -2\right ) -3-2A_{0}=0\) hence \(A_{0}=\frac{4+3}{-2}=-\frac{7}{2}\), Therefore\begin{align*} y_{p} & =A_{0}+A_{1}t+A_{2}t^{2}\\ & =-\frac{7}{2}+3t-2t^{2} \end{align*}
Therefore the general solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}e^{-t}+c_{2}e^{2t}-\frac{7}{2}+3t-2t^{2} \end{align*}
Find the general solution of \(y^{\prime \prime }+y^{\prime }-6y=12e^{3t}+12e^{-2t}\)
Solution:
The first step is to solve the homogenous ODE and find \(y_{h}\), then find a particular solution \(y_{p}\) to the inhomogeneous ODE, then add both solutions \(y_{h}+y_{p}\) in order to find the complete solution.
Finding \(y_{h}\)
We need to solve homogenous ODE \[ y^{\prime \prime }+y^{\prime }-6y=0 \] The characteristic equation is found by substituting \(y=e^{rt}\) into the above ODE and simplifying, giving\begin{align*} r^{2}+r-6 & =0\\ \left ( r+3\right ) \left ( r-2\right ) & =0 \end{align*}
Hence \(r_{1}=-3,r_{2}=2\) and therefore\[ y_{h}=c_{1}e^{-3t}+c_{2}e^{2t}\] Finding \(y_{p}\)
Now we need to find one particular solution to \[ y^{\prime \prime }+y^{\prime }-6y=12e^{3t}+12e^{-2t}\] We guess \(y_{p}=Ae^{3t}+Be^{-2t}\). Therefore
\begin{align*} y_{p}^{\prime } & =3Ae^{3t}-2Be^{-2t}\\ y_{p}^{\prime \prime } & =9Ae^{3t}+4Be^{-2t} \end{align*}
Substituting these back into the original ODE gives\begin{align*} y_{p}^{\prime \prime }+y_{p}^{\prime }-6y_{p} & =12e^{3t}+12e^{-2t}\\ 9Ae^{3t}+4Be^{-2t}+3Ae^{3t}-2Be^{-2t}-6\left ( Ae^{3t}+Be^{-2t}\right ) & =12e^{3t}+12e^{-2t}\\ e^{3t}\left ( 9A+3A-6A\right ) +e^{-2t}\left ( 4B-2B-6B\right ) & =12e^{3t}+12e^{-2t}\\ 6Ae^{3t}-4Be^{-2t} & =12e^{3t}+12e^{-2t} \end{align*}
Comparing coefficients gives\begin{align*} A & =2\\ B & =-3 \end{align*}
Hence\[ y_{p}=2e^{3t}-3e^{-2t}\] And the final solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}e^{-3t}+c_{2}e^{2t}+2e^{3t}-3e^{-2t} \end{align*}
Find the general solution of \(y^{\prime \prime }-2y^{\prime }-3y=-3te^{-t}\)
Solution:
The first step is to solve the homogenous ODE and find \(y_{h}\), then find a particular solution \(y_{p}\) to the inhomogeneous ODE, then add both solutions \(y_{h}+y_{p}\) in order to find the complete solution.
Finding \(y_{h}\)
We need to solve homogenous ODE \[ y^{\prime \prime }-2y^{\prime }-3y=0 \] The characteristic equation is found by substituting \(y=e^{rt}\) into the above ODE and simplifying, giving\begin{align*} r^{2}-2r-3 & =0\\ \left ( r-3\right ) \left ( r+1\right ) & =0 \end{align*}
Hence \(r_{1}=3,r_{2}=-1\) and therefore\[ y_{h}=c_{1}e^{3t}+c_{2}e^{-t}\] Finding \(y_{p}\)
Now we need to find one particular solution to \[ y^{\prime \prime }-2y^{\prime }-3y=-3te^{-t}\] Guess for \(t\) is \(A_{0}+B_{0}t\) and the guess for \(e^{-t}\) is \(Cte^{-t}\) (where we multiplied by \(t\) since \(e^{-t}\) shows up in the homogenous solution. Therefore the product is \begin{align*} y_{p} & =\left ( A_{0}+B_{0}t\right ) Cte^{-t}\\ & =A_{0}Cte^{-t}+CB_{0}t^{2}e^{-t} \end{align*}
Let \(A_{0}C=A,CB_{0}=B\), and the above becomes\begin{align*} y_{p} & =Ate^{-t}+Bt^{2}e^{-t}\\ & =\left ( A+Bt\right ) te^{-t} \end{align*}
Substituting these back into the ODE and solving for \(A,B\) gives \(B=\frac{3}{8}\) and \(A=\frac{3}{16}\), hence\begin{align*} y_{p} & =\left ( At+Bt^{2}\right ) e^{-t}\\ & =\left ( \frac{3}{16}t+\frac{3}{8}t^{2}\right ) e^{-t} \end{align*}
And the final solution is\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}e^{3t}+c_{2}e^{-t}+\left ( \frac{3}{16}t+\frac{3}{8}t^{2}\right ) e^{-t} \end{align*}
Find the general solution of \(y^{\prime \prime }+2y^{\prime }=3+4\sin 2t\)
Solution:
The first step is to solve the homogenous ODE and find \(y_{h}\), then find a particular solution \(y_{p}\) to the inhomogeneous ODE, then add both solutions \(y_{h}+y_{p}\) in order to find the complete solution.
Finding \(y_{h}\)
We need to solve homogenous ODE \[ y^{\prime \prime }+2y^{\prime }=0 \] The characteristic equation is found by substituting \(y=e^{rt}\) into the above ODE and simplifying, giving\begin{align*} r^{2}+2r & =0\\ r\left ( r+2\right ) & =0 \end{align*}
Hence \(r_{1}=0,r_{2}=-2\) and therefore\[ y_{h}=c_{1}+c_{2}e^{2t}\] Finding \(y_{p}\)
Now we need to find one particular solution to \[ y^{\prime \prime }+2y^{\prime }=3+4\sin 2t \] Guess that \(y_{p}=At+B\cos 2t+C\sin 2t\), hence\begin{align*} y_{p}^{\prime } & =A-2B\sin 2t+2C\cos 2t\\ y_{p}^{\prime \prime } & =-4B\cos 2t-4C\sin 2t \end{align*}
Substituting back into \begin{align*} y_{p}^{\prime \prime }+2y_{p}^{\prime } & =3+4\sin 2t\\ -4B\cos 2t-4C\sin 2t+2\left ( A-2B\sin 2t+2C\cos 2t\right ) & =3+4\sin 2t\\ -4B\cos 2t-4C\sin 2t+2A-4B\sin 2t+4C\cos 2t & =3+4\sin 2t\\ \left ( -4B+4C\right ) \cos 2t+2A+\left ( -4C-4B\right ) \sin 2t & =3+4\sin 2t \end{align*}
Hence\begin{align*} 2A & =3\\ 1 & =-C-B\\ 0 & =-B+C \end{align*}
From first equation, \(A=\frac{3}{2},\) From third equation, \(B=C\) and from the second equation \(1=-2B\) or \(B=\frac{-1}{2}\), hence \(C=\frac{-1}{2}\), and the particular solution is\[ y_{p}=\frac{3}{2}t+\frac{-1}{2}\cos 2t-\frac{1}{2}\sin 2t \] Hence the complete solution is\[ y=c_{1}+c_{2}e^{2t}+\frac{3}{2}t-\frac{1}{2}\cos 2t-\frac{1}{2}\sin 2t \]