Internal problem ID [15567]
Internal file name [OUTPUT/15568_Tuesday_May_14_2024_10_48_03_PM_99659885/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Chapter 3. Section 24.2. Solving the Cauchy problem for linear differential equation with
constant coefficients. Exercises page 249
Problem number: 845.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_laplace", "exact linear second order ode", "second_order_integrable_as_is", "second_order_ode_missing_y", "second_order_linear_constant_coeff"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {2 x^{\prime \prime }-2 x^{\prime }={\mathrm e}^{t} \left (t +1\right )} \] With initial conditions \begin {align*} \left [x \left (0\right ) = {\frac {1}{2}}, x^{\prime }\left (0\right ) = {\frac {1}{2}}\right ] \end {align*}
This is a linear ODE. In canonical form it is written as \begin {align*} x^{\prime \prime } + p(t)x^{\prime } + q(t) x &= F \end {align*}
Where here \begin {align*} p(t) &=-1\\ q(t) &=0\\ F &=\frac {{\mathrm e}^{t} \left (t +1\right )}{2} \end {align*}
Hence the ode is \begin {align*} x^{\prime \prime }-x^{\prime } = \frac {{\mathrm e}^{t} \left (t +1\right )}{2} \end {align*}
The domain of \(p(t)=-1\) is \[
\{-\infty Solving using the Laplace transform method. Let \begin {align*} \mathcal {L}\left (x\right ) =Y(s) \end {align*}
Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (x^{\prime }\right ) &= s Y(s) - x \left (0\right )\\ \mathcal {L}\left (x^{\prime \prime }\right ) &= s^2 Y(s) - x'(0) - s x \left (0\right ) \end {align*}
The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} 2 s^{2} Y \left (s \right )-2 x^{\prime }\left (0\right )-2 s x \left (0\right )-2 s Y \left (s \right )+2 x \left (0\right ) = \frac {s}{\left (s -1\right )^{2}}\tag {1} \end {align*}
But the initial conditions are \begin {align*} x \left (0\right )&={\frac {1}{2}}\\ x'(0) &={\frac {1}{2}} \end {align*}
Substituting these initial conditions in above in Eq (1) gives \begin {align*} 2 s^{2} Y \left (s \right )-s -2 s Y \left (s \right ) = \frac {s}{\left (s -1\right )^{2}} \end {align*}
Solving the above equation for \(Y(s)\) results in \begin {align*} Y(s) = \frac {s^{2}-2 s +2}{2 \left (s -1\right )^{3}} \end {align*}
Applying partial fractions decomposition results in \[ Y(s)= \frac {1}{2 \left (s -1\right )^{3}}+\frac {1}{2 s -2} \] The inverse Laplace of each term above
is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (\frac {1}{2 \left (s -1\right )^{3}}\right ) &= \frac {t^{2} {\mathrm e}^{t}}{4}\\ \mathcal {L}^{-1}\left (\frac {1}{2 s -2}\right ) &= \frac {{\mathrm e}^{t}}{2} \end {align*}
Adding the above results and simplifying gives \[ x=\frac {{\mathrm e}^{t} \left (t^{2}+2\right )}{4} \] Simplifying the solution gives \[
x = \frac {{\mathrm e}^{t} \left (t^{2}+2\right )}{4}
\]
The solution(s) found are the following \begin{align*}
\tag{1} x &= \frac {{\mathrm e}^{t} \left (t^{2}+2\right )}{4} \\
\end{align*} Verification of solutions
\[
x = \frac {{\mathrm e}^{t} \left (t^{2}+2\right )}{4}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [2 x^{\prime \prime }-2 x^{\prime }={\mathrm e}^{t} \left (t +1\right ), x \left (0\right )=\frac {1}{2}, x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=\frac {1}{2}\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & x^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & x^{\prime \prime }=\frac {t \,{\mathrm e}^{t}}{2}+x^{\prime }+\frac {{\mathrm e}^{t}}{2} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} x\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & x^{\prime \prime }-x^{\prime }=\frac {{\mathrm e}^{t} \left (t +1\right )}{2} \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}-r =0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & r \left (r -1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (0, 1\right ) \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )=1 \\ \bullet & {} & \textrm {2nd solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )={\mathrm e}^{t} \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & x=c_{1} x_{1}\left (t \right )+c_{2} x_{2}\left (t \right )+x_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & x=c_{1} +c_{2} {\mathrm e}^{t}+x_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} x_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} x_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [x_{p}\left (t \right )=-x_{1}\left (t \right ) \left (\int \frac {x_{2}\left (t \right ) f \left (t \right )}{W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )}d t \right )+x_{2}\left (t \right ) \left (\int \frac {x_{1}\left (t \right ) f \left (t \right )}{W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )}d t \right ), f \left (t \right )=\frac {{\mathrm e}^{t} \left (t +1\right )}{2}\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )=\left [\begin {array}{cc} 1 & {\mathrm e}^{t} \\ 0 & {\mathrm e}^{t} \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (x_{1}\left (t \right ), x_{2}\left (t \right )\right )={\mathrm e}^{t} \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} x_{p}\left (t \right ) \\ {} & {} & x_{p}\left (t \right )=-\frac {\left (\int {\mathrm e}^{t} \left (t +1\right )d t \right )}{2}+\frac {{\mathrm e}^{t} \left (\int \left (t +1\right )d t \right )}{2} \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & x_{p}\left (t \right )=\frac {t^{2} {\mathrm e}^{t}}{4} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & x=c_{1} +c_{2} {\mathrm e}^{t}+\frac {t^{2} {\mathrm e}^{t}}{4} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} x=c_{1} +c_{2} {\mathrm e}^{t}+\frac {t^{2} {\mathrm e}^{t}}{4} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} x \left (0\right )=\frac {1}{2} \\ {} & {} & \frac {1}{2}=c_{1} +c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & x^{\prime }=c_{2} {\mathrm e}^{t}+\frac {t \,{\mathrm e}^{t}}{2}+\frac {t^{2} {\mathrm e}^{t}}{4} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} x^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=\frac {1}{2} \\ {} & {} & \frac {1}{2}=c_{2} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =0, c_{2} =\frac {1}{2}\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & x=\frac {{\mathrm e}^{t} \left (t^{2}+2\right )}{4} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & x=\frac {{\mathrm e}^{t} \left (t^{2}+2\right )}{4} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 13
\[
x = \frac {{\mathrm e}^{t} \left (t^{2}+2\right )}{4}
\]
✓ Solution by Mathematica
Time used: 0.059 (sec). Leaf size: 17
\[
x(t)\to \frac {1}{4} e^t \left (t^2+2\right )
\]
33.16.2 Maple step by step solution
`Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
trying high order exact linear fully integrable
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = _b(_a)+(1/2)*exp(_a)*_a+(1/2)*exp(_a), _b(_a)` *** Sublevel 2 ***
Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful
<- high order exact linear fully integrable successful`
dsolve([2*diff(x(t),t$2)-2*diff(x(t),t)=(1+t)*exp(t),x(0) = 1/2, D(x)(0) = 1/2],x(t), singsol=all)
DSolve[{2*x''[t]-2*x'[t]==(1+t)*Exp[t],{x[0]==1/2,x'[0]==1/2}},x[t],t,IncludeSingularSolutions -> True]