2.1.55 Problem 55

Internal problem ID [10041]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 55
Date solved : Monday, March 09, 2026 at 03:06:29 AM
CAS classification : [[_2nd_order, _quadrature]]

2.1.55.1 second order ode quadrature

0.032 (sec)

\begin{align*} y^{\prime \prime }&=1 \\ \end{align*}

Entering second order ode quadrature solverThe ODE can be written as

\[ y^{\prime \prime } = 1 \]

Integrating once gives

\[ y^{\prime }= t + c_1 \]

Integrating again gives

\[ y= \frac {t^{2}}{2} + c_1 t + c_2 \]

Summary of solutions found

\begin{align*} y &= \frac {1}{2} t^{2}+c_1 t +c_2 \\ \end{align*}

Figure 2.40: Phase plot for \(y^{\prime \prime } = 1\)

2.1.55.2 second order linear constant coeff

0.360 (sec)

\begin{align*} y^{\prime \prime }&=1 \\ \end{align*}

Entering second order linear constant coefficient ode solverThis is second order non-homogeneous ODE. In standard form the ODE is

\[ A y''(t) + B y'(t) + C y(t) = f(t) \]

Where \(A=1, B=0, C=0, f(t)=1\). Let the solution be

\[ y = y_h + y_p \]

Where \(y_h\) is the solution to the homogeneous ODE

\begin{align*} A y''(t) + B y'(t) + C y(t) &= 0 \end{align*}

And \(y_p\) is a particular solution to the non-homogeneous ODE

\begin{align*} A y''(t) + B y'(t) + C y(t) &= f(t) \end{align*}

Where \(y_h\) is the solution to

\[ y^{\prime \prime } = 0 \]

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A y''(t) + B y'(t) + C y(t) = 0 \]

Where in the above \(A=1, B=0, C=0\). Let the solution be \(y=e^{\lambda t}\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{t \lambda } = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda t}\) gives

\[ \lambda ^{2} = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form. Using the quadratic formula the roots are

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=0, C=0\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {\left (0\right )^2 - (4) \left (1\right )\left (0\right )}\\ &= 0 \end{align*}

Hence this is the case of a double root \(\lambda _{1,2} = 0\). Therefore the solution is

\[ y= c_1 1 + c_2 t \tag {1} \]

Therefore the homogeneous solution \(y_h\) is

\[ y_h = c_2 t +c_1 \]

The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

\[ 1 \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{1\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \{1, t\} \]

Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes

\[ [\{t\}] \]

Since \(t\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes

\[ [\{t^{2}\}] \]

Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set.

\[ y_p = A_{1} t^{2} \]

The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

\[ 2 A_{1} = 1 \]

Solving for the unknowns by comparing coefficients results in

\[ \left [A_{1} = {\frac {1}{2}}\right ] \]

Substituting the above back in the above trial solution \(y_p\), gives the particular solution

\[ y_p = \frac {t^{2}}{2} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_2 t +c_1\right ) + \left (\frac {t^{2}}{2}\right ) \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {1}{2} t^{2}+c_2 t +c_1 \\ \end{align*}

Figure 2.41: Phase plot for \(y^{\prime \prime } = 1\)

2.1.55.3 second order linear exact ode

0.101 (sec)

\begin{align*} y^{\prime \prime }&=1 \\ \end{align*}

Entering second order linear exact ode solverAn ode of the form

\begin{align*} p \left (t \right ) y^{\prime \prime }+q \left (t \right ) y^{\prime }+r \left (t \right ) y&=s \left (t \right ) \end{align*}

is exact if

\begin{align*} p''(t) - q'(t) + r(t) &= 0 \tag {1} \end{align*}

For the given ode the above values are

\begin{align*} p(x) &= 1\\ q(x) &= 0\\ r(x) &= 0\\ s(x) &= 1 \end{align*}

Hence

\begin{align*} p''(x) &= 0\\ q'(x) &= 0 \end{align*}

Therefore (1) becomes

\begin{align*} 0- \left (0\right ) + \left (0\right )&=0 \end{align*}

This shows the ode is exact. Since the ode is exact, it can be written as

\begin{align*} \left (p \left (t \right ) y^{\prime }+\left (q \left (t \right )-p^{\prime }\left (t \right )\right ) y\right )' &= s(x) \end{align*}

Integrating the above gives

\begin{align*} p \left (t \right ) y^{\prime }+\left (q \left (t \right )-p^{\prime }\left (t \right )\right ) y&=\int {s \left (t \right )\, dt} \end{align*}

Substituting the values of \(p,q,r,s\) into the above results in

\begin{align*} y^{\prime }&=\int {1\, dt} \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dy} &= \int {t +c_1\, dt}\\ y &= \frac {1}{2} t^{2}+c_1 t + c_2 \end{align*}

\begin{align*} y&= \frac {1}{2} t^{2}+c_1 t +c_2 \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {1}{2} t^{2}+c_1 t +c_2 \\ \end{align*}

Figure 2.42: Phase plot for \(y^{\prime \prime } = 1\)

2.1.55.4 second order ode missing y

0.117 (sec)

\begin{align*} y^{\prime \prime }&=1 \\ \end{align*}

Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(y\). Let

\begin{align*} u(t) &= y^{\prime } \end{align*}

Then

\begin{align*} u'(t) &= y^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} u^{\prime }\left (t \right )-1 = 0 \end{align*}

Which is now solved for \(u(t)\) as first order ode.

Entering first order ode quadrature solverSince the ode has the form \(u^{\prime }\left (t \right )=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {du} &= \int {1\, dt}\\ u \left (t \right ) &= t + c_1 \end{align*}

\begin{align*} u \left (t \right )&= t +c_1 \end{align*}

In summary, these are the solution found for \(y\)

\begin{align*} u \left (t \right ) &= t +c_1 \\ \end{align*}

For solution \(u \left (t \right ) = t +c_1\), since \(u=y^{\prime }\) then the new first order ode to solve is

\begin{align*} y^{\prime } = t +c_1 \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dy} &= \int {t +c_1\, dt}\\ y &= \frac {1}{2} t^{2}+c_1 t + c_2 \end{align*}

\begin{align*} y&= \frac {1}{2} t^{2}+c_1 t +c_2 \end{align*}

In summary, these are the solution found for \((y)\)

\begin{align*} y &= \frac {1}{2} t^{2}+c_1 t +c_2 \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {1}{2} t^{2}+c_1 t +c_2 \\ \end{align*}

Figure 2.43: Phase plot for \(y^{\prime \prime } = 1\)

2.1.55.5 second order integrable as is

0.060 (sec)

\begin{align*} y^{\prime \prime }&=1 \\ \end{align*}

Entering second order integrable as is solverIntegrating both sides of the ODE w.r.t \(t\) gives

\begin{align*} \int y^{\prime \prime }d t &= \int 1d t\\ y^{\prime } = t + c_1 \end{align*}

Which is now solved for \(y\). Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(t)\), then we only need to integrate \(f(t)\).

\begin{align*} \int {dy} &= \int {t +c_1\, dt}\\ y &= \frac {1}{2} t^{2}+c_1 t + c_2 \end{align*}

\begin{align*} y&= \frac {1}{2} t^{2}+c_1 t +c_2 \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {1}{2} t^{2}+c_1 t +c_2 \\ \end{align*}

Figure 2.44: Phase plot for \(y^{\prime \prime } = 1\)

2.1.55.6 second order kovacic

0.105 (sec)

\begin{align*} y^{\prime \prime }&=1 \\ \end{align*}

Entering kovacic solverWriting the ode as

\begin{align*} y^{\prime \prime } &= 0 \tag {1} \\ A y^{\prime \prime } + B y^{\prime } + C y &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

\begin{align*} A &= 1 \\ B &= 0\tag {3} \\ C &= 0 \end{align*}

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(t) &= y e^{\int \frac {B}{2 A} \,dt} \end{align*}

Then (2) becomes

\begin{align*} z''(t) = r z(t)\tag {4} \end{align*}

Where \(r\) is given by

\begin{align*} r &= \frac {s}{t}\tag {5} \\ &= \frac {2 A B' - 2 B A' + B^2 - 4 A C}{4 A^2} \end{align*}

Substituting the values of \(A,B,C\) from (3) in the above and simplifying gives

\begin{align*} r &= \frac {0}{1}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= 0\\ t &= 1 \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(t) &= 0 \tag {7} \end{align*}

Equation (7) is now solved. After finding \(z(t)\) then \(y\) is found using the inverse transformation

\begin{align*} y &= z \left (t \right ) e^{-\int \frac {B}{2 A} \,dt} \end{align*}

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of \(r\) and the order of \(r\) at \(\infty \). The following table summarizes these cases.

Case	Allowed pole order for \(r\)	Allowed value for \(\mathcal {O}(\infty )\)
1	\(\left \{ 0,1,2,4,6,8,\cdots \right \} \)	\(\left \{ \cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \right \} \)
2	Need to have at least one pole that is either order \(2\) or odd order greater than \(2\). Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. \(\{1,2\}\),\(\{1,3\}\),\(\{2\}\),\(\{3\}\),\(\{3,4\}\),\(\{1,2,5\}\).	no condition
3	\(\left \{ 1,2\right \} \)	\(\left \{ 2,3,4,5,6,7,\cdots \right \} \)

Table 2.21: Necessary conditions for each Kovacic case

The order of \(r\) at \(\infty \) is the degree of \(t\) minus the degree of \(s\). Therefore

\begin{align*} O\left (\infty \right ) &= \text {deg}(t) - \text {deg}(s) \\ &= 0 - -\infty \\ &= \infty \end{align*}

There are no poles in \(r\). Therefore the set of poles \(\Gamma \) is empty. Since there is no odd order pole larger than \(2\) and the order at \(\infty \) is \(infinity\) then the necessary conditions for case one are met. Therefore

\begin{align*} L &= [1] \end{align*}

Since \(r = 0\) is not a function of \(t\), then there is no need run Kovacic algorithm to obtain a solution for transformed ode \(z''=r z\) as one solution is

\[ z_1(t) = 1 \]

Using the above, the solution for the original ode can now be found. The first solution to the original ode in \(y\) is found from

\[ y_1 = z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dt} \]

Since \(B=0\) then the above reduces to

\begin{align*} y_1 &= z_1 \\ &= 1 \\ \end{align*}

Which simplifies to

\[ y_1 = 1 \]

The second solution \(y_2\) to the original ode is found using reduction of order

\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dt}}{y_1^2} \,dt \]

Since \(B=0\) then the above becomes

\begin{align*} y_2 &= y_1 \int \frac {1}{y_1^2} \,dt \\ &= 1\int \frac {1}{1} \,dt \\ &= 1\left (t\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} y &= c_1 y_1 + c_2 y_2 \\ &= c_1 \left (1\right ) + c_2 \left (1\left (t\right )\right ) \\ \end{align*}

This is second order nonhomogeneous ODE. Let the solution be

\[ y = y_h + y_p \]

Where \(y_h\) is the solution to the homogeneous ODE

\[ A y''(t) + B y'(t) + C y(t) = 0 \]

And \(y_p\) is a particular solution to the nonhomogeneous ODE

\[ A y''(t) + B y'(t) + C y(t) = f(t) \]

\(y_h\) is the solution to

\[ y^{\prime \prime } = 0 \]

The homogeneous solution is found using the Kovacic algorithm which results in

\[ y_h = c_2 t +c_1 \]

The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

\[ 1 \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{1\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \{1, t\} \]

Since \(1\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes

\[ [\{t\}] \]

Since \(t\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes

\[ [\{t^{2}\}] \]

Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set.

\[ y_p = A_{1} t^{2} \]

The unknowns \(\{A_{1}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

\[ 2 A_{1} = 1 \]

Solving for the unknowns by comparing coefficients results in

\[ \left [A_{1} = {\frac {1}{2}}\right ] \]

Substituting the above back in the above trial solution \(y_p\), gives the particular solution

\[ y_p = \frac {t^{2}}{2} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_2 t +c_1\right ) + \left (\frac {t^{2}}{2}\right ) \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {1}{2} t^{2}+c_2 t +c_1 \\ \end{align*}

Figure 2.45: Phase plot for \(y^{\prime \prime } = 1\)

2.1.55.7 ✓ Maple. Time used: 0.000 (sec). Leaf size: 14

ode:=diff(diff(y(t),t),t) = 1; 
dsolve(ode,y(t), singsol=all);
 

\[ y = \frac {1}{2} t^{2}+c_1 t +c_2 \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
<- quadrature successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right )=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d t^{2}}y \left (t \right ) \\ \bullet & {} & \textrm {Characteristic polynomial of homogeneous ODE}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Use quadratic formula to solve for}\hspace {3pt} r \\ {} & {} & r =\frac {0\pm \left (\sqrt {0}\right )}{2} \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {1st solution of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (t \right )=1 \\ \bullet & {} & \textrm {Repeated root, multiply}\hspace {3pt} y_{1}\left (t \right )\hspace {3pt}\textrm {by}\hspace {3pt} t \hspace {3pt}\textrm {to ensure linear independence}\hspace {3pt} \\ {} & {} & y_{2}\left (t \right )=t \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} y_{1}\left (t \right )+\mathit {C2} y_{2}\left (t \right )+y_{p}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions of the homogeneous ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} +\mathit {C2} t +y_{p}\left (t \right ) \\ \square & {} & \textrm {Find a particular solution}\hspace {3pt} y_{p}\left (t \right )\hspace {3pt}\textrm {of the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Use variation of parameters to find}\hspace {3pt} y_{p}\hspace {3pt}\textrm {here}\hspace {3pt} f \left (t \right )\hspace {3pt}\textrm {is the forcing function}\hspace {3pt} \\ {} & {} & \left [y_{p}\left (t \right )=-y_{1}\left (t \right ) \int \frac {y_{2}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t +y_{2}\left (t \right ) \int \frac {y_{1}\left (t \right ) f \left (t \right )}{W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )}d t , f \left (t \right )=1\right ] \\ {} & \circ & \textrm {Wronskian of solutions of the homogeneous equation}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=\left [\begin {array}{cc} 1 & t \\ 0 & 1 \end {array}\right ] \\ {} & \circ & \textrm {Compute Wronskian}\hspace {3pt} \\ {} & {} & W \left (y_{1}\left (t \right ), y_{2}\left (t \right )\right )=1 \\ {} & \circ & \textrm {Substitute functions into equation for}\hspace {3pt} y_{p}\left (t \right ) \\ {} & {} & y_{p}\left (t \right )=-\int t d t +t \int 1d t \\ {} & \circ & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y_{p}\left (t \right )=\frac {t^{2}}{2} \\ \bullet & {} & \textrm {Substitute particular solution into general solution to ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} +\mathit {C2} t +\frac {1}{2} t^{2} \end {array} \]

2.1.55.8 ✓ Mathematica. Time used: 0.002 (sec). Leaf size: 19

ode=D[y[t],{t,2}]==1; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

\begin{align*} y(t)&\to \frac {t^2}{2}+c_2 t+c_1 \end{align*}

2.1.55.9 ✓ Sympy. Time used: 0.155 (sec). Leaf size: 12

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(Derivative(y(t), (t, 2)) - 1,0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 

\[ y{\left (t \right )} = C_{1} + C_{2} t + \frac {t^{2}}{2} \]

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(t)) 
 
('nth_algebraic', 'nth_linear_constant_coeff_undetermined_coefficients', 'nth_linear_euler_eq_nonhomogeneous_undetermined_coefficients', 'nth_linear_constant_coeff_variation_of_parameters', 'nth_linear_euler_eq_nonhomogeneous_variation_of_parameters', 'nth_algebraic_Integral', 'nth_linear_constant_coeff_variation_of_parameters_Integral', 'nth_linear_euler_eq_nonhomogeneous_variation_of_parameters_Integral', '2nd_nonlinear_autonomous_conserved', '2nd_nonlinear_autonomous_conserved_Integral')