Let y_{1}\left ( t\right ) =t^{2} and y_{2}\left ( t\right ) =t\left \vert t\right \vert
Solution
On the interval 0\leq t\leq 1, then \left \vert t\right \vert =t since t is positive. Hence y_{2}\left ( t\right ) =t^{2}, which is the same as y_{1}\left ( t\right ) =t^{2}. Therefore they are linearly dependent (same solution). In other words, y_{1}\left ( t\right ) =c_{1}y_{2}\left ( t\right ) where c_{1}=1.
When t\leq 0 now y_{2}\left ( t\right ) =-t^{2}. Hence we have y_{1}=y_{2} for 0\leq t\leq 1 and y_{1}=-y_{2} for -1\leq t<0. Therefore it is not possible to find the same constant c such that y_{1}=cy_{2} which will work for all t regions. This implies that y_{1}\left ( t\right ) and y_{2}\left ( t\right ) are linearly independent on -1\leq t\leq 1.
W\left [ y_{1},y_{2}\right ] \left ( t\right ) =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} =y_{1}y_{2}^{\prime }-y_{2}y_{1}^{\prime } If W\left ( t\right ) =0 is in some region or at some point, then it must be zero anywhere. Therefore let us pick the interval 0\leq t\leq 1 to calculate W\left ( t\right ) . This way we avoid having to deal with the \left \vert t\right \vert when taking derivatives since on this interval, y_{1}=t^{2} and also y_{2}=t^{2}. Now W\left ( t\right ) becomes\begin{align*} W\left ( t\right ) & =t^{2}\left ( 2t\right ) -t^{2}\left ( 2t\right ) \\ & =0 \end{align*}
Therefore W\left ( t\right ) =0 everywhere.
Since p,q are continuous on -1<t<1, then by uniqueness theorem, we know there are two fundamental solutions y_{1},y_{2}, which must be linearly independent that their linear combination give the general solution y\left ( t\right ) =c_{1}y_{1}\left ( t\right ) +c_{2}y_{2}\left ( t\right ) .
But from part(b) above we found that the given functions y_{1},y_{2} are not linearly independent on -1<t<1, hence these can never be the fundamental solutions to y^{\prime \prime }+p\left ( t\right ) y^{\prime }+q\left ( t\right ) y=0.
Solve y^{\prime \prime }+2y^{\prime }+5y=0 with y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =2
Solution
Let y=e^{\lambda t}. Substituting in the above ODE gives\begin{align*} \lambda ^{2}e^{\lambda t}+2\lambda e^{\lambda t}+5e^{\lambda t} & =0\\ e^{\lambda t}\left ( \lambda ^{2}+2\lambda +5\right ) & =0 \end{align*}
Since e^{\lambda t}\neq 0, the above simplifies to \lambda ^{2}+2\lambda +5=0. The roots are \lambda =\frac{-b}{2a}\pm \frac{1}{2a}\sqrt{b^{2}-4ac}=\frac{-2}{2}\pm \frac{1}{2}\sqrt{4-4\left ( 5\right ) } or \lambda =-1\pm \frac{1}{2}\sqrt{-16}. Hence \lambda =-1\pm 2i Therefore the general solution is linear combination of\begin{align*} y\left ( t\right ) & =c_{1}e^{\lambda _{1}t}+c_{2}e^{\lambda _{2}t}\\ & =c_{1}e^{\left ( -1+2i\right ) t}+c_{2}e^{\left ( -1-2i\right ) t}\\ & =e^{-t}\left ( c_{1}e^{2it}+c_{2}e^{-2it}\right ) \end{align*}
But c_{1}e^{2it}+c_{2}e^{-2it} can be rewritten, using Euler relation, as C_{1}\cos 2t+C_{2}\sin 2t. The above solution becomes\begin{equation} y\left ( t\right ) =e^{-t}\left ( C_{1}\cos 2t+C_{2}\sin 2t\right ) \tag{1} \end{equation} C_{1},C_{2} are now found from initial conditions. At t=0 0=C_{1} The solution (1) simplifies to\begin{equation} y\left ( t\right ) =C_{2}e^{-t}\sin 2t \tag{2} \end{equation} Taking time derivative gives y^{\prime }\left ( t\right ) =C_{2}\left ( -e^{-t}\sin 2t+2e^{-t}\cos 2t\right ) At t=0 the above becomes\begin{align*} 2 & =2C_{2}\\ C_{2} & =1 \end{align*}
Substituting the above in (2) gives the final general solution y\left ( t\right ) =e^{-t}\sin 2t
Solve the following initial-value problems y^{\prime \prime }+2y^{\prime }+y=0 with y\left ( 2\right ) =1,y^{\prime }\left ( 2\right ) =-1
Solution
Let y=e^{\lambda t}. Substituting in the above ODE gives\begin{align*} \lambda ^{2}e^{\lambda t}+2\lambda e^{\lambda t}+e^{\lambda t} & =0\\ e^{\lambda t}\left ( \lambda ^{2}+2\lambda +1\right ) & =0 \end{align*}
Since e^{\lambda t}\neq 0, the above simplifies to \lambda ^{2}+2\lambda +1=0 or \left ( \lambda +1\right ) ^{2}=0. Hence there is a double root \lambda =-1. One fundamental solution is y_{1}=e^{-t} To find the second solution, reduction of order is used. Let the second solution be \begin{align} y_{2}\left ( t\right ) & =y_{1}\left ( t\right ) u\left ( t\right ) \nonumber \\ & =e^{-t}u \tag{1} \end{align}
Hence \begin{align} y_{2}^{\prime } & =-e^{-t}u+e^{-t}u^{\prime }\tag{2}\\ y_{2}^{\prime \prime } & =e^{-t}u-e^{-t}u^{\prime }-e^{-t}u^{\prime }+e^{-t}u^{\prime \prime } \tag{3} \end{align}
Substituting (1,2,3) into the ODE gives (since y_{2} is assumed to be a solution)\begin{align*} \left ( e^{-t}u-e^{-t}u^{\prime }-e^{-t}u^{\prime }+e^{-t}u^{\prime \prime }\right ) +2\left ( -e^{-t}u+e^{-t}u^{\prime }\right ) +\left ( e^{-t}u\right ) & =0\\ \left ( u-u^{\prime }-u^{\prime }+u^{\prime \prime }\right ) +2\left ( -u+u^{\prime }\right ) +u & =0\\ u^{\prime \prime }-2u^{\prime }+u-2u+2u^{\prime }+u & =0\\ u^{\prime \prime } & =0 \end{align*}
Hence the solution is u=C_{1}t+C_{2}. Therefore from (1) the second solution is\begin{align*} y_{2}\left ( t\right ) & =y_{1}\left ( t\right ) u\left ( t\right ) \\ & =e^{-t}\left ( C_{1}t+C_{2}\right ) \end{align*}
Therefore the general solution is \begin{align*} y\left ( t\right ) & =C_{3}y_{1}+C_{4}y_{2}\\ & =C_{3}e^{-t}+C_{4}e^{-t}\left ( C_{1}t+C_{2}\right ) \end{align*}
Combining constants gives\begin{align*} y\left ( t\right ) & =C_{3}e^{-t}+e^{-t}\left ( C_{1}t+C_{2}\right ) \\ & =\left ( C_{3}+C_{2}\right ) e^{-t}+C_{1}te^{-t} \end{align*}
Let A=\left ( C_{3}+C_{2}\right ) ,B=C_{1}, then the final solution is\begin{equation} y\left ( t\right ) =Ae^{-t}+Bte^{-t} \tag{4} \end{equation} Now A,B are found from initial conditions y\left ( 2\right ) =1,y^{\prime }\left ( 2\right ) =-1. First initial condition gives from (4)\begin{equation} 1=Ae^{-2}+2Be^{-2} \tag{5} \end{equation} Taking derivative of (4) gives y^{\prime }\left ( t\right ) =-Ae^{-t}+B\left ( e^{-t}-te^{-t}\right ) Applying second initial condition on the above gives\begin{align} -1 & =-Ae^{-2}+B\left ( e^{-2}-2e^{-2}\right ) \nonumber \\ & =-Ae^{-2}-Be^{-2} \tag{6} \end{align}
Now we need to solve (5,6) for (A,B). Adding (5,6) gives 0=Be^{-2} Hence B=0. Therefore from (5) we can now solve for A\begin{align*} 1 & =Ae^{-2}\\ A & =e^{2} \end{align*}
Hence (4) now becomes\begin{align*} y\left ( t\right ) & =e^{2}e^{-t}\\ & =e^{2-t} \end{align*}
Solve the following initial-value problems y^{\prime \prime }+4y^{\prime }+4y=t^{\frac{5}{2}}e^{-2t} with y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =0
Solution
The first step is to solve the homogenous ODE y^{\prime \prime }+4y^{\prime }+4y=0. The characteristic equation is \begin{align*} \lambda ^{2}+4\lambda +4 & =0\\ \left ( \lambda +2\right ) \left ( \lambda +2\right ) & =0 \end{align*}
Hence a double root at \lambda =-2. The first solution is y_{1}=e^{=2t}. Therefore the second solution is y_{2}=te^{-2t} (obtained using reduction of order as was done in the above problem with equal roots). Therefore the homogenous y_{h}\left ( t\right ) is y_{h}\left ( t\right ) =C_{1}e^{-2t}+C_{2}te^{-2t} To find the particular solution y_{p}\left ( t\right ) , Variation of parameters will be used. Assuming the particular solution is y_{p}\left ( t\right ) =u_{1}\left ( t\right ) y_{1}\left ( t\right ) +u_{2}\left ( t\right ) y_{2}\left ( t\right ) Where \begin{equation} u_{1}\left ( t\right ) =-\int \frac{y_{2}\left ( t\right ) f\left ( t\right ) }{W\left ( t\right ) }dt \tag{1} \end{equation} And\begin{equation} u_{2}\left ( t\right ) =\int \frac{y_{1}\left ( t\right ) f\left ( t\right ) }{W\left ( t\right ) }dt \tag{2} \end{equation} Where in the above f\left ( t\right ) =t^{\frac{5}{2}}e^{-2t} and y_{1}=e^{=2t},y_{2}=te^{-2t}. We now need to find W\left ( t\right ) \begin{align*} W\left ( t\right ) & =\begin{vmatrix} y_{1} & y_{2}\\ y_{1}^{\prime } & y_{2}^{\prime }\end{vmatrix} \\ & =y_{1}y_{2}^{\prime }-y_{2}y_{1}^{\prime }\\ & =e^{=2t}\left ( e^{-2t}-2te^{-2t}\right ) +2te^{-2t}e^{-2t}\\ & =e^{-4t}-2te^{-4t}+2te^{-4t}\\ & =e^{-4t} \end{align*}
Therefore (1) becomes\begin{align*} u_{1}\left ( t\right ) & =-\int \frac{te^{-2t}t^{\frac{5}{2}}e^{-2t}}{e^{-4t}}dt\\ & =-\int t^{\frac{5}{2}+1}dt\\ & =-\int t^{\frac{7}{2}}dt\\ & =-\frac{t^{\frac{9}{2}}}{\frac{9}{2}}\\ & =\frac{-2}{9}t^{\frac{9}{2}} \end{align*}
And (2) becomes\begin{align*} u_{2}\left ( t\right ) & =\int \frac{e^{-2t}t^{\frac{5}{2}}e^{-2t}}{e^{-4t}}dt\\ & =\int t^{\frac{5}{2}}dt\\ & =\frac{t^{\frac{7}{2}}}{\frac{7}{2}}\\ & =\frac{2}{7}t^{\frac{7}{2}} \end{align*}
Since y_{p}\left ( t\right ) =u_{1}\left ( t\right ) y_{1}\left ( t\right ) +u_{2}\left ( t\right ) y_{2}\left ( t\right ) , then using the above results we obtain the particular solution\begin{align*} y_{p}\left ( t\right ) & =\left ( \frac{-2}{9}t^{\frac{9}{2}}\right ) e^{-2t}+\left ( \frac{2}{7}t^{\frac{7}{2}}\right ) te^{-2t}\\ & =e^{-2t}\left ( \frac{-2}{9}t^{\frac{9}{2}}+\frac{2}{7}t^{\frac{9}{2}}\right ) \\ & =\frac{4}{63}e^{-2t}t^{\frac{9}{2}} \end{align*}
Since y\left ( t\right ) =y_{h}\left ( t\right ) +y_{p}\left ( t\right ) then the final solution is\begin{align} y\left ( t\right ) & =\left ( C_{1}e^{-2t}+C_{2}te^{-2t}\right ) +\frac{4}{63}e^{-2t}t^{\frac{9}{2}}\nonumber \\ & =e^{-2t}\left ( C_{1}+C_{2}t+\frac{4}{63}t^{\frac{9}{2}}\right ) \tag{3} \end{align}
Now initial conditions are applied to find C_{1},C_{2}. From y\left ( 0\right ) =0, then (3) becomes 0=C_{1} Hence the solution (3) simplifies to\begin{equation} y\left ( t\right ) =e^{-2t}\left ( C_{2}t+\frac{4}{63}t^{\frac{9}{2}}\right ) \tag{4} \end{equation} Taking derivatives\begin{align*} y^{\prime }\left ( t\right ) & =-2e^{-2t}\left ( C_{2}t+\frac{4}{63}t^{\frac{9}{2}}\right ) +e^{-2t}\left ( C_{2}+\left ( \frac{4}{63}\right ) \left ( \frac{7}{2}\right ) t^{\frac{7}{2}}\right ) \\ & =-2e^{-2t}\left ( C_{2}t+\frac{4}{63}t^{\frac{9}{2}}\right ) +e^{-2t}\left ( C_{2}+\frac{2}{9}t^{\frac{7}{2}}\right ) \end{align*}
Applying the second BC y^{\prime }\left ( 0\right ) =0 to the above gives 0=C_{2} The solution (4) now reduces to y\left ( t\right ) =\frac{4}{63}t^{\frac{9}{2}}e^{-2t} Which is just the particular solution. This makes sense, since both initial conditions are zero, then the homogenous solution will be zero.
Find the particular solution for y^{\prime \prime }+2y^{\prime }=1+t^{2}+e^{-2t}
Solution
The first step is to solve the homogeneous solution y_{h}\left ( t\right ) of the ODE y^{\prime \prime }+2y^{\prime }=0. Let u=y^{\prime }. Then the ODE becomes u^{\prime }+2u=0 The integrating factor is I=e^{\int 2dt}=e^{2t}. The above becomes\begin{align*} \frac{d}{dt}\left ( ue^{2t}\right ) & =0\\ ue^{2t} & =C_{1}\\ u & =C_{1}e^{-2t} \end{align*}
But y^{\prime }=u. Integrating gives \begin{align*} y_{h}\left ( t\right ) & =\int C_{1}e^{-2t}dt+C_{2}\\ & =\frac{-1}{2}C_{1}e^{-2t}+C_{2}\\ & =C_{3}e^{-2t}+C_{2} \end{align*}
Hence the fundamental solutions are \begin{align*} y_{1} & =e^{-2t}\\ y_{2} & =1 \end{align*}
We now go back to the original ODE and find the particular solution y_{p}. Since the RHS is p\left ( t\right ) +e^{-2t} where p\left ( t\right ) =1+t^{2}, we can use linearity and find particular solution y_{p_{1}}\left ( t\right ) associated with p\left ( t\right ) only and then find y_{p_{2}}\left ( t\right ) associated with e^{-2t} only and then add them together to obtain y_{p}\left ( t\right ) . In other words y_{p}\left ( t\right ) =y_{p_{1}}\left ( t\right ) +y_{p_{2}}\left ( t\right ) To find y_{p_{1}}\left ( t\right ) associated with 1+t^{2} we guess y_{p_{1}}\left ( t\right ) =C_{0}+C_{1}t+C_{2}t^{2}. But because the ODE is missing the y term in it, then we have to multiply this guess by an extra t. Therefore it becomes y_{p_{1}}\left ( t\right ) =t\left ( C_{0}+C_{1}t+C_{2}t^{2}\right ) To find y_{p_{2}}\left ( t\right ) associated with e^{-2t} we guess y_{p_{2}}=Ae^{-2t}. But because e^{-2t} is also a fundamental solution of the homogenous solution found above, we have to again adjust this and multiply the guess by t. Hence it becomes y_{p_{2}}\left ( t\right ) =Ate^{-2t} Therefore the full guess for particular solution becomes\begin{align} y_{p}\left ( t\right ) & =y_{p_{1}}\left ( t\right ) +y_{p_{2}}\left ( t\right ) \nonumber \\ & =t\left ( C_{0}+C_{1}t+C_{2}t^{2}\right ) +Ate^{-2t}\nonumber \\ & =tC_{0}+C_{1}t^{2}+C_{2}t^{3}+Ate^{-2t}\tag{1A} \end{align}
Now\begin{equation} y_{p}^{\prime }\left ( t\right ) =C_{0}+2C_{1}t+3C_{2}t^{2}+Ae^{-2t}-2Ate^{-2t}\tag{1} \end{equation} And\begin{equation} y_{p}^{\prime \prime }\left ( t\right ) =2C_{1}+6C_{2}t-2Ae^{-2t}-2Ae^{-2t}+4Ate^{-2t}\tag{2} \end{equation} Substituting (1,2) into LHS of y^{\prime \prime }+2y^{\prime }=1+t^{2}+e^{-2t} gives\begin{align*} \left ( 2C_{1}+6C_{2}t-2Ae^{-2t}-2Ae^{-2t}+4Ate^{-2t}\right ) +2\left ( C_{0}+2C_{1}t+3C_{2}t^{2}+Ae^{-2t}-2Ate^{-2t}\right ) & =1+t^{2}+e^{-2t}\\ 2C_{0}+2C_{1}+4tC_{1}+6tC_{2}-2Ae^{-2t}+6t^{2}C_{2} & =1+t^{2}+e^{-2t}\\ e^{-2t}\left ( -2A\right ) +t\left ( 4C_{1}+6C_{2}\right ) +t^{2}\left ( 6C_{2}\right ) +\left ( 2C_{0}+2C_{1}\right ) & =1+t^{2}+e^{-2t} \end{align*}
Comparing coefficients gives\begin{align*} -2A & =1\\ 4C_{1}+6C_{2} & =0\\ 6C_{2} & =1\\ 2C_{0}+2C_{1} & =1 \end{align*}
Solving gives A=-\frac{1}{2},C_{0}=\frac{3}{4},C_{1}=-\frac{1}{4},C_{2}=\frac{1}{6}. Substituting the above in (1A) gives the particular solution as\begin{align*} y_{p}\left ( t\right ) & =t\left ( \frac{3}{4}-\frac{1}{4}t+\frac{1}{6}t^{2}\right ) -\frac{1}{2}te^{-2t}\\ & =\frac{3}{4}t-\frac{1}{4}t^{2}+\frac{1}{6}t^{3}-\frac{1}{2}te^{-2t} \end{align*}
Therefore the general solution is\begin{align*} y\left ( t\right ) & =y_{h}\left ( t\right ) +y_{p}\left ( t\right ) \\ & =C_{3}e^{-2t}+C_{2}+\left ( \frac{3}{4}t-\frac{1}{4}t^{2}+\frac{1}{6}t^{3}-\frac{1}{2}te^{-2t}\right ) \end{align*}