Solved using Laplace transform

These are solved using Laplace transform. These are only solved using this method if ’hint’="laplace" is given.

\begin{align*} y^{\prime \prime }+2y^{\prime }+y & =0\\ y\left ( 1\right ) & =2\\ y^{\prime }\left ( 0\right ) & =2 \end{align*}

\begin{align*} \left ( s^{2}Y-sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) +2\left ( sY-y\left ( 0\right ) \right ) +Y & =0\\ \left ( s^{2}Y-sy\left ( 0\right ) -2\right ) +\left ( 2sY-2y\left ( 0\right ) \right ) +Y & =0 \end{align*}

Since not all initial conditions are at zero, and we need to have them at zero to use Laplace, then one way is to let \(y\left ( 0\right ) =y_{0}\) as unknown (we could also have used \(y\left ( 0\right ) =c_{1}\)). Find the solution, then solve for \(y_{0}\) using the initial condition \(y\left ( 1\right ) =2\). This shows how it is done. The above becomes

\begin{align*} \left ( s^{2}Y-sy_{0}-2\right ) +\left ( 2sY-2y_{0}\right ) +Y & =0\\ Y\left ( s^{2}+2s+1\right ) -sy_{0}-2-2y_{0} & =0\\ Y & =\frac {sy_{0}+2+2y_{0}}{s^{2}+2s+1}\end{align*}

\begin{equation} y\left ( t\right ) =\left ( y_{0}+2t+y_{0}t\right ) e^{-t} \tag {1}\end{equation}

\begin{align*} 2 & =\left ( y_{0}+2+y_{0}\right ) e^{-1}\\ 2e & =2y_{0}+2\\ y_{0} & =e-1 \end{align*}

\begin{align*} y\left ( t\right ) & =\left ( e-1+2t+\left ( e-1\right ) t\right ) e^{-t}\\ & =e^{-t}\left ( -1+e+t+et\right ) \end{align*}

\begin{align*} y^{\prime \prime }-2y^{\prime }-3y & =0\\ y\left ( 4\right ) & =-3\\ y^{\prime }\left ( 4\right ) & =-17 \end{align*}

\[ \left ( s^{2}Y-sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) -2\left ( sY-y\left ( 0\right ) \right ) -3Y=0 \]

Since given initial conditions are not at \(t=0\), then let \(y\left ( 0\right ) =c_{1},y^{\prime }\left ( 0\right ) =c_{2}\) and the above becomes

\begin{align*} \left ( s^{2}Y-sc_{1}-c_{2}\right ) -2\left ( sY-c_{1}\right ) -3Y & =0\\ Y\left ( s^{2}-2s-3\right ) -sc_{1}-c_{2}+2c_{1} & =0\\ Y & =\frac {sc_{1}+c_{2}-2c_{1}}{s^{2}-2s-3}\end{align*}

\begin{equation} y\left ( t\right ) =\frac {1}{4}e^{-t}\left ( c_{2}\left ( e^{4t}-1\right ) +c_{1}\left ( 3+e^{4t}\right ) \right ) \tag {1}\end{equation}

\begin{equation} y^{\prime }\left ( t\right ) =\frac {1}{4}e^{-t}(4c_{1}e^{-4t}+4c_{2}e^{4t})-\frac {1}{4}e^{-t}(c_{2}(-1+e^{4t})+c_{1}(3+e^{4t})) \tag {2}\end{equation}

\begin{align*} -3 & =\frac {1}{4}e^{-4}\left ( c_{2}\left ( e^{16}-1\right ) +c_{1}\left ( 3+e^{16}\right ) \right ) \\ -17 & =\frac {1}{4}e^{-4}(4c_{1}e^{-16}+4c_{2}e^{16})-\frac {1}{4}e^{-4}(c_{2}(-1+e^{16})+c_{1}(3+e^{16})) \end{align*}

\begin{align*} c_{1} & =\frac {-5+2e^{16}}{e^{12}}\\ c_{2} & =\frac {-15-2e^{16}}{e^{12}}\end{align*}

\begin{align*} y\left ( t\right ) & =\frac {1}{4}e^{-t}\left ( \frac {-15-2e^{16}}{e^{12}}\left ( e^{4t}-1\right ) +\frac {-5+2e^{16}}{e^{12}}\left ( 3+e^{4t}\right ) \right ) \\ & =-e^{3t}\left ( 5e^{-12}-2e^{4}e^{-4t}\right ) \\ & =-5e^{3t-12}+2e^{4-t}\end{align*}

\begin{align*} y^{\prime \prime }+2y^{\prime }+5y & =50t-100\\ y\left ( 2\right ) & =-4\\ y^{\prime }\left ( 2\right ) & =14 \end{align*}

\[ \left ( s^{2}Y-sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) +2\left ( sY-y\left ( 0\right ) \right ) +5Y=\frac {50}{s^{2}}-\frac {100}{s}\]

Since given initial conditions are not at \(t=0\), then let \(y\left ( 0\right ) =c_{1},y^{\prime }\left ( 0\right ) =c_{2}\) and the above becomes

\begin{align*} \left ( s^{2}Y-sc_{1}-c_{2}\right ) +2\left ( sY-c_{1}\right ) +5Y & =\frac {50}{s^{2}}-\frac {100}{s}\\ Y\left ( s^{2}+2s+5\right ) -sc_{1}-c_{2}-2c_{1} & =\frac {50}{s^{2}}-\frac {100}{s}\\ Y & =\frac {sc_{1}+c_{2}+2c_{1}+\frac {50}{s^{2}}-\frac {100}{s}}{s^{2}+2s+5}\end{align*}

\begin{equation} y\left ( t\right ) =-24+10t+\left ( 24+c_{1}\right ) e^{-t}\cos \left ( 2t\right ) +\left ( 14+c_{1}+c_{2}\right ) e^{-t}\cos t\sin t \tag {1}\end{equation}

\begin{equation} y^{\prime }\left ( t\right ) =e^{-t}\left ( 10e^{t}+\left ( c_{2}-10\right ) \cos \left ( 2t\right ) -\left ( 110+5c_{1}+c_{2}\right ) \cos t\sin t\right ) \tag {2}\end{equation}

\begin{align*} -4 & =-24+20+\left ( 24+c_{1}\right ) e^{-2}\cos \left ( 4\right ) +\left ( 14+c_{1}+c_{2}\right ) e^{-2}\cos 2\sin 2\\ 14 & =e^{-2}\left ( 10e^{2}+\left ( c_{2}-10\right ) \cos \left ( 4\right ) -\left ( 110+5c_{1}+c_{2}\right ) \cos 2\sin 2\right ) \end{align*}

\begin{align*} c_{1} & =-2\left ( 12+e^{2}\sin 4\right ) \\ c_{2} & =2\left ( 5+e^{2}\left ( 2\cos 4+\sin 4\right ) \right ) \end{align*}

\[ y\left ( t\right ) =-24+10t+\left ( 24-2\left ( 12+e^{2}\sin 4\right ) \right ) e^{-t}\cos \left ( 2t\right ) +\left ( 14-2\left ( 12+e^{2}\sin 4\right ) +2\left ( 5+e^{2}\left ( 2\cos 4+\sin 4\right ) \right ) \right ) e^{-t}\cos t\sin t \]

\begin{align*} y^{\prime \prime }+2y^{\prime }+10y & =\delta \left ( t\right ) \\ y\left ( 0\right ) & =0\\ y^{\prime }\left ( 0\right ) & =0 \end{align*}

\[ \left ( s^{2}Y-sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) +2\left ( sY-y\left ( 0\right ) \right ) +10Y=1 \]

\begin{align*} s^{2}Y+2sY+10Y & =1\\ Y & =\frac {1}{s^{2}+2s+10}\\ & =\frac {1}{\left ( s+2\right ) \left ( s+5\right ) }\end{align*}

\begin{align*} y & =\frac {1}{6}ie^{\left ( -1-3i\right ) t}-\frac {1}{6}ie^{\left ( -1+3i\right ) t}\\ & =\frac {1}{6}ie^{-t}e^{-3it}-\frac {1}{6}ie^{-t}e^{3it}\\ & =\frac {1}{6}ie^{-t}\left ( e^{-3it}-e^{3it}\right ) \\ & =\frac {1}{6}ie^{-t}\left ( \cos 3t-i\sin 3t-\left ( \cos 3t+i\sin 3t\right ) \right ) \\ & =\frac {1}{6}ie^{-t}\left ( -i\sin 3t-i\sin 3t\right ) \\ & =\frac {1}{6}ie^{-t}\left ( -2i\sin 3t\right ) \\ & =\frac {1}{3}e^{-t}\sin 3t \end{align*}

Where \(U\left ( t\right ) \) is Heaviside function which is one for \(t>0\). Note that it seems one should not give IC at same point of application of \(\delta \left ( t\right ) \) as in this problem. So this problem might be ill posed. Need to look more into this.

\begin{align*} y^{\prime \prime }+2y^{\prime }+y & =0\\ y^{\prime }\left ( 0\right ) & =2 \end{align*}

This problem shows what to do when one IC is missing. Basically, if an IC is missing, it is just kept unknown. Taking Laplace transform gives

Since not all initial conditions are given, then we let the missing IC be some unknown. In this case \(y\left ( 0\right ) =c_{1}\). And continue as before. The above becomes

\begin{align*} \left ( s^{2}Y-sc_{1}-2\right ) +\left ( 2sY-2c_{1}\right ) +Y & =0\\ Y\left ( s^{2}+2s+1\right ) -sc_{1}-2-2c_{1} & =0\\ Y & =\frac {sc_{1}+2+2c_{1}}{s^{2}+2s+1}\end{align*}

\begin{equation} y\left ( t\right ) =\left ( c_{1}+2t+c_{1}t\right ) e^{-t} \tag {1}\end{equation}

We can if we want, now replace \(c_{1}=y\left ( 0\right ) \) to make it more clear what the \(c_{1}\) represents.

\begin{equation} y\left ( t\right ) =\left ( y\left ( 0\right ) +2t+y\left ( 0\right ) t\right ) e^{-t} \tag {2}\end{equation}

\begin{align*} y^{\prime \prime }+2y^{\prime }+y & =0\\ y\left ( 0\right ) & =0 \end{align*}

\begin{align*} \left ( s^{2}Y-sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) +2\left ( sY-y\left ( 0\right ) \right ) +Y & =0\\ \left ( s^{2}Y-y^{\prime }\left ( 0\right ) \right ) +2sY+Y & =0 \end{align*}

Since one IC is missing, then let \(y^{\prime }\left ( 0\right ) =c_{2}\). The above becomes

\begin{align*} \left ( s^{2}Y-c_{2}\right ) +2sY+Y & =0\\ Y\left ( s^{2}+2s+1\right ) -c_{2} & =0\\ Y & =\frac {c_{2}}{s^{2}+2s+1}\end{align*}

\begin{equation} y\left ( t\right ) =c_{2}te^{-t} \tag {1}\end{equation}

We can if we want, now replace \(c_{2}=y^{\prime }\left ( 0\right ) \) to make it more clear what the \(c_{2}\) represents.

\begin{equation} y\left ( t\right ) =y^{\prime }\left ( 0\right ) te^{-t} \tag {2}\end{equation}

4.3.1.3.7 Example 7 This example is for higher order ode, showing how to easily handle IC if at zero or not or if some missing or not, all using same process. Given

\begin{equation} y^{\prime \prime \prime }+y^{\prime \prime }+y^{\prime }+y=0 \tag {1}\end{equation}

\begin{align*} y\left ( 1\right ) & =a\\ y^{\prime \prime }\left ( 0\right ) & =b \end{align*}

The idea is to always use \(c_{0},c_{1},c_{2}\) for \(y\left ( 0\right ) ,y^{\prime }\left ( 0\right ) ,y^{\prime \prime }\left ( 0\right ) \) and then at the very end solve for these from the given initial conditions. We will get two equations (since we only have 2 IC) and 3 unknowns. So some of the \(c_{0},c_{1},c_{2}\) will remain in the solution as unknowns which is OK. Applying Laplace transform on (1) gives

\[ s^{3}Y-y^{\prime \prime }\left ( 0\right ) -sy^{\prime }\left ( 0\right ) -s^{2}y\left ( 0\right ) +s^{2}Y-y^{\prime }\left ( 0\right ) -sy\left ( 0\right ) +sY-y\left ( 0\right ) +Y=0 \]

We now replace \(y^{\prime \prime }\left ( 0\right ) =c_{2},y^{\prime }\left ( 0\right ) =c_{1},y\left ( 0\right ) =c_{0}\) and simplify the above which becomes

\begin{align*} Y\left ( s^{3}+s^{2}+s+1\right ) -c_{2}-sc_{1}-s^{2}c_{0}-c_{1}-sc_{0}-c_{0} & =0\\ Y\left ( s^{3}+s^{2}+s+1\right ) -s^{2}c_{0}-s\left ( c_{1}+c_{0}\right ) -c_{2}-c_{1}-c_{0} & =0\\ Y & =\frac {c_{2}+c_{1}+c_{0}+s\left ( c_{1}+c_{0}\right ) +s^{2}c_{0}}{s^{3}+s^{2}+s+1}\end{align*}

\begin{equation} y\left ( t\right ) =\frac {1}{2}\left ( c_{0}-c_{2}\right ) \cos \left ( t\right ) +\frac {1}{2}e^{-t}\left ( c_{0}+c_{2}\right ) +\frac {1}{2}\sin \left ( t\right ) \left ( c_{0}+2c_{1}+c_{2}\right ) \tag {2}\end{equation}

Now we only need to solve for the constants using the given initial conditions. This results in these two equations (since we have 2 IC only). Using \(y\left ( 1\right ) =a\) gives

\begin{equation} a=\frac {1}{2}\left ( c_{0}-c_{2}\right ) \cos \left ( 1\right ) +\frac {1}{2}e^{-1}\left ( c_{0}+c_{2}\right ) +\frac {1}{2}\sin \left ( 1\right ) \left ( c_{0}+2c_{1}+c_{2}\right ) \tag {3}\end{equation}

Taking derivatives twice of (2) and using \(y^{\prime \prime }\left ( 0\right ) =b\) gives the second equation

\[ y^{\prime \prime }\left ( t\right ) =\frac {1}{2}e^{-t}\left ( c_{0}+c_{2}\right ) +\frac {1}{2}\left ( -c_{0}-2c_{1}-c_{2}\right ) \sin \left ( t\right ) -\frac {1}{2}\left ( c_{0}-c_{2}\right ) \cos \left ( t\right ) \]

\begin{align} b & =\frac {1}{2}\left ( c_{0}+c_{2}\right ) -\frac {1}{2}\left ( c_{0}-c_{2}\right ) \nonumber \\ & =c_{2} \tag {4}\end{align}

Now we need to solve (3,4) for \(c_{0},c_{1},c_{2}\). From (4) we see that \(c_{2}=b\). Substituting this into (3) gives

\begin{equation} a=\frac {1}{2}\left ( c_{0}-b\right ) \cos \left ( 1\right ) +\frac {1}{2}e^{-1}\left ( c_{0}+b\right ) +\frac {1}{2}\sin \left ( 1\right ) \left ( c_{0}+2c_{1}+b\right ) \tag {5}\end{equation}

\[ c_{1}=-\frac {1}{2\sin \left ( 1\right ) }\left ( \cos \left ( 1\right ) +e^{-1}+\sin \left ( 1\right ) \right ) c_{0}+\frac {1}{2\sin \left ( 1\right ) }\left ( b\cos \left ( 1\right ) -be^{-1}-b\sin \left ( 1\right ) +2a\right ) \]

We are done. The solution (2) is now found by replacing \(c_{2},c_{1}\) into it. \(c_{0}\) remains are the only unknown. This method works for any combination of IC given even if some at zero or not.

\begin{equation} tx^{\prime \prime }\left ( t\right ) +x^{\prime }+tx=0 \tag {1}\end{equation}

Assuming \(x\left ( 0\right ) =1,x^{\prime }\left ( 0\right ) =0\). In solving ode using Laplace where the coeﬃcient in time varying, we will be using the relation

\begin{equation} L\left ( t^{n}f\left ( t\right ) \right ) =\left ( -1\right ) ^{n}F^{\left ( n\right ) }\left ( s\right ) \tag {2}\end{equation}

Where \(F\left ( s\right ) \) is the laplace transform of \(f\left ( t\right ) \). For example, if the input is \(tx\left ( t\right ) \) the Laplace tranaform is \(-X^{\prime }\left ( s\right ) =-\frac {dX\left ( s\right ) }{ds}\) and if the input is \(t^{2}x\left ( t\right ) \) then the Laplace transform is \(\frac {d^{2}X\left ( s\right ) }{s^{2}}\) and so on. This will generate an ODE in \(X\left ( s\right ) \) which we have to solve for \(X\left ( s\right ) \). Applying this to (1) gives

\begin{align*} L\left ( x^{\prime \prime }\right ) & =s^{2}X\left ( s\right ) -sx\left ( 0\right ) -x^{\prime }\left ( 0\right ) \\ L\left ( x^{\prime }\right ) & =sX\left ( s\right ) -x\left ( 0\right ) \\ L\left ( x\right ) & =X\left ( s\right ) \end{align*}

\[ -\frac {d}{ds}\left ( s^{2}X\left ( s\right ) -sx\left ( 0\right ) -x^{\prime }\left ( 0\right ) \right ) +\left ( sX\left ( s\right ) -x\left ( 0\right ) \right ) -\frac {d}{ds}X\left ( s\right ) =0 \]

\begin{align*} -\frac {d}{ds}\left ( s^{2}X\left ( s\right ) -s\right ) +\left ( sX\left ( s\right ) -1\right ) -\frac {d}{ds}X\left ( s\right ) & =0\\ -\left ( 2sX+s^{2}X^{\prime }-1\right ) +sX-1-X^{\prime }\left ( s\right ) & =0\\ X^{\prime }\left ( -s^{2}-1\right ) +X\left ( -2s+s\right ) & =0\\ \left ( s^{2}+1\right ) X^{\prime }+sX & =0 \end{align*}

This diﬀerential equation is now solved for \(X\left ( s\right ) \) which gives

But \(\operatorname {BesselJ}_{0}\left ( 0\right ) =1\), hence \(c_{1}=1\) and the solution is

4.3.1.3 Solved using Laplace transform