2.2.7 Problem 5 (iv)

Internal problem ID [19691]
Book : Elementary Differential Equations. By R.L.E. Schwarzenberger. Chapman and Hall. London. First Edition (1969)
Section : Chapter 4. Autonomous systems. Exercises at page 69
Problem number : 5 (iv)
Date solved : Thursday, December 11, 2025 at 01:42:38 PM
CAS classification : [[_2nd_order, _missing_x]]

2.2.7.1 second order linear constant coeff

0.052 (sec)

\begin{align*} x^{\prime \prime }+3 x^{\prime }&=0 \\ \end{align*}

Entering second order linear constant coefficient ode solver

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

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

Where in the above \(A=1, B=3, C=0\). Let the solution be \(x=e^{\lambda t}\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{t \lambda }+3 \lambda \,{\mathrm e}^{t \lambda } = 0 \tag {1} \]

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

\[ \lambda ^{2}+3 \lambda = 0 \tag {2} \]

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

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

Substituting \(A=1, B=3, C=0\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {-3}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {3^2 - (4) \left (1\right )\left (0\right )}\\ &= -{\frac {3}{2}} \pm {\frac {3}{2}} \end{align*}

Hence

\begin{align*} \lambda _1 &= -{\frac {3}{2}} + {\frac {3}{2}} \\ \lambda _2 &= -{\frac {3}{2}} - {\frac {3}{2}} \\ \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= 0 \\ \lambda _2 &= -3 \\ \end{align*}

Since roots are distinct, then the solution is

\begin{align*} x &= c_1 e^{\lambda _1 t} + c_2 e^{\lambda _2 t} \\ x &= c_1 e^{\left (0\right )t} +c_2 e^{\left (-3\right )t} \\ \end{align*}

Or

\[ x =c_1 +c_2 \,{\mathrm e}^{-3 t} \]

Summary of solutions found

\begin{align*} x &= c_1 +c_2 \,{\mathrm e}^{-3 t} \\ \end{align*}

Figure 2.59: Slope field \(x^{\prime \prime }+3 x^{\prime } = 0\)

2.2.7.2 second order linear exact ode

0.210 (sec)

\begin{align*} x^{\prime \prime }+3 x^{\prime }&=0 \\ \end{align*}

Entering second order linear exact ode solverAn ode of the form

\begin{align*} p \left (t \right ) x^{\prime \prime }+q \left (t \right ) x^{\prime }+r \left (t \right ) x&=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 we have

\begin{align*} p(x) &= 1\\ q(x) &= 3\\ r(x) &= 0\\ s(x) &= 0 \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*}

Hence the ode is exact. Since we now know the ode is exact, it can be written as

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

Integrating gives

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

Substituting the above values for \(p,q,r,s\) gives

\begin{align*} x^{\prime }+3 x&=c_1 \end{align*}

We now have a first order ode to solve which is

\begin{align*} x^{\prime }+3 x = c_1 \end{align*}

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int \frac {1}{-3 x +c_1}d x &= dt\\ -\frac {\ln \left (-3 x +c_1 \right )}{3}&= t +c_2 \end{align*}

Singular solutions are found by solving

\begin{align*} -3 x +c_1&= 0 \end{align*}

for \(x\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} x = \frac {c_1}{3} \end{align*}

Solving for \(x\) gives

\begin{align*} x &= \frac {c_1}{3} \\ x &= -\frac {{\mathrm e}^{-3 t -3 c_2}}{3}+\frac {c_1}{3} \\ \end{align*}

Summary of solutions found

\begin{align*} x &= \frac {c_1}{3} \\ x &= -\frac {{\mathrm e}^{-3 t -3 c_2}}{3}+\frac {c_1}{3} \\ \end{align*}

Figure 2.60: Slope field \(x^{\prime \prime }+3 x^{\prime } = 0\)

2.2.7.3 second order ode missing y

0.315 (sec)

\begin{align*} x^{\prime \prime }+3 x^{\prime }&=0 \\ \end{align*}

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

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

Then

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

Hence the ode becomes

\begin{align*} u^{\prime }\left (t \right )+3 u \left (t \right ) = 0 \end{align*}

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

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int -\frac {1}{3 u}d u &= dt\\ -\frac {\ln \left (u \right )}{3}&= t +c_1 \end{align*}

Singular solutions are found by solving

\begin{align*} -3 u&= 0 \end{align*}

for \(u \left (t \right )\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} u \left (t \right ) = 0 \end{align*}

Solving for \(u \left (t \right )\) gives

\begin{align*} u \left (t \right ) &= 0 \\ u \left (t \right ) &= {\mathrm e}^{-3 t -3 c_1} \\ \end{align*}

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

\begin{align*} u \left (t \right ) &= 0 \\ u \left (t \right ) &= {\mathrm e}^{-3 t -3 c_1} \\ \end{align*}

For solution \(u \left (t \right ) = 0\), since \(u=x^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} x^{\prime } = 0 \end{align*}

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

\begin{align*} \int {dx} &= \int {0\, dt} + c_2 \\ x &= c_2 \end{align*}

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

\begin{align*} x &= c_2 \\ \end{align*}

For solution \(u \left (t \right ) = {\mathrm e}^{-3 t -3 c_1}\), since \(u=x^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} x^{\prime } = {\mathrm e}^{-3 t -3 c_1} \end{align*}

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

\begin{align*} \int {dx} &= \int {{\mathrm e}^{-3 t -3 c_1}\, dt}\\ x &= -\frac {{\mathrm e}^{-3 t -3 c_1}}{3} + c_3 \end{align*}

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

\begin{align*} x &= -\frac {{\mathrm e}^{-3 t -3 c_1}}{3}+c_3 \\ \end{align*}

Summary of solutions found

\begin{align*} x &= c_2 \\ x &= -\frac {{\mathrm e}^{-3 t -3 c_1}}{3}+c_3 \\ \end{align*}

Figure 2.61: Slope field \(x^{\prime \prime }+3 x^{\prime } = 0\)

2.2.7.4 second order integrable as is

0.054 (sec)

\begin{align*} x^{\prime \prime }+3 x^{\prime }&=0 \\ \end{align*}

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

\begin{align*} \int \left (x^{\prime \prime }+3 x^{\prime }\right )d t &= 0 \\ x^{\prime }+3 x = c_1 \end{align*}

Which is now solved for \(x\). Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int \frac {1}{-3 x +c_1}d x &= dt\\ -\frac {\ln \left (-3 x +c_1 \right )}{3}&= t +c_2 \end{align*}

Singular solutions are found by solving

\begin{align*} -3 x +c_1&= 0 \end{align*}

for \(x\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} x = \frac {c_1}{3} \end{align*}

Solving for \(x\) gives

\begin{align*} x &= \frac {c_1}{3} \\ x &= -\frac {{\mathrm e}^{-3 t -3 c_2}}{3}+\frac {c_1}{3} \\ \end{align*}

Summary of solutions found

\begin{align*} x &= \frac {c_1}{3} \\ x &= -\frac {{\mathrm e}^{-3 t -3 c_2}}{3}+\frac {c_1}{3} \\ \end{align*}

Figure 2.62: Slope field \(x^{\prime \prime }+3 x^{\prime } = 0\)

2.2.7.5 second order kovacic

0.065 (sec)

\begin{align*} x^{\prime \prime }+3 x^{\prime }&=0 \\ \end{align*}

Entering kovacic solverWriting the ode as

\begin{align*} x^{\prime \prime }+3 x^{\prime } &= 0 \tag {1} \\ A x^{\prime \prime } + B x^{\prime } + C x &= 0 \tag {2} \end{align*}

Comparing (1) and (2) shows that

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

Applying the Liouville transformation on the dependent variable gives

\begin{align*} z(t) &= x 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 {9}{4}\tag {6} \end{align*}

Comparing the above to (5) shows that

\begin{align*} s &= 9\\ t &= 4 \end{align*}

Therefore eq. (4) becomes

\begin{align*} z''(t) &= \frac {9 z \left (t \right )}{4} \tag {7} \end{align*}

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

\begin{align*} x &= 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 2.16: 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 - 0 \\ &= 0 \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 \(0\) then the necessary conditions for case one are met. Therefore

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

Since \(r = {\frac {9}{4}}\) 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

\[ z_1(t) = {\mathrm e}^{-\frac {3 t}{2}} \]

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

\begin{align*} x_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dt} \\ &= z_1 e^{ -\int \frac {1}{2} \frac {3}{1} \,dt} \\ &= z_1 e^{-\frac {3 t}{2}} \\ &= z_1 \left ({\mathrm e}^{-\frac {3 t}{2}}\right ) \\ \end{align*}

Which simplifies to

\[ x_1 = {\mathrm e}^{-3 t} \]

The second solution \(x_2\) to the original ode is found using reduction of order

\[ x_2 = x_1 \int \frac { e^{\int -\frac {B}{A} \,dt}}{x_1^2} \,dt \]

Substituting gives

\begin{align*} x_2 &= x_1 \int \frac { e^{\int -\frac {3}{1} \,dt}}{\left (x_1\right )^2} \,dt \\ &= x_1 \int \frac { e^{-3 t}}{\left (x_1\right )^2} \,dt \\ &= x_1 \left (\frac {{\mathrm e}^{3 t}}{3}\right ) \\ \end{align*}

Therefore the solution is

\begin{align*} x &= c_1 x_1 + c_2 x_2 \\ &= c_1 \left ({\mathrm e}^{-3 t}\right ) + c_2 \left ({\mathrm e}^{-3 t}\left (\frac {{\mathrm e}^{3 t}}{3}\right )\right ) \\ \end{align*}

Summary of solutions found

\begin{align*} x &= c_1 \,{\mathrm e}^{-3 t}+\frac {c_2}{3} \\ \end{align*}

Figure 2.63: Slope field \(x^{\prime \prime }+3 x^{\prime } = 0\)

2.2.7.6 SSolved using second order ode arccos transformation

0.701 (sec)

\begin{align*} x^{\prime \prime }+3 x^{\prime }&=0 \\ \end{align*}

Applying change of variable \(t = \arccos \left (\tau \right )\) to the above ode results in the following new ode

\[ -\left (\frac {d^{2}}{d \tau ^{2}}x \left (\tau \right )\right ) \tau ^{2}-3 \left (\frac {d}{d \tau }x \left (\tau \right )\right ) \sqrt {-\tau ^{2}+1}-\left (\frac {d}{d \tau }x \left (\tau \right )\right ) \tau +\frac {d^{2}}{d \tau ^{2}}x \left (\tau \right ) = 0 \]

Which is now solved for \(x \left (\tau \right )\). Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(x \left (\tau \right )\). Let

\begin{align*} u(\tau ) &= \frac {d}{d \tau }x \left (\tau \right ) \end{align*}

Then

\begin{align*} u'(\tau ) &= \frac {d^{2}}{d \tau ^{2}}x \left (\tau \right ) \end{align*}

Hence the ode becomes

\begin{align*} -\left (\frac {d}{d \tau }u \left (\tau \right )\right ) \tau ^{2}-3 u \left (\tau \right ) \sqrt {-\tau ^{2}+1}-u \left (\tau \right ) \tau +\frac {d}{d \tau }u \left (\tau \right ) = 0 \end{align*}

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

Entering first order ode linear solverIn canonical form a linear first order is

\begin{align*} \frac {d}{d \tau }u \left (\tau \right ) + q(\tau )u \left (\tau \right ) &= p(\tau ) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(\tau ) &=-\frac {-3 \sqrt {-\tau ^{2}+1}-\tau }{\tau ^{2}-1}\\ p(\tau ) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,d\tau }}\\ &= {\mathrm e}^{\int -\frac {-3 \sqrt {-\tau ^{2}+1}-\tau }{\tau ^{2}-1}d \tau }\\ &= \sqrt {\tau -1}\, \sqrt {\tau +1}\, {\mathrm e}^{-3 \arcsin \left (\tau \right )} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }} \left (u \sqrt {\tau -1}\, \sqrt {\tau +1}\, {\mathrm e}^{-3 \arcsin \left (\tau \right )}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \sqrt {\tau -1}\, \sqrt {\tau +1}\, {\mathrm e}^{-3 \arcsin \left (\tau \right )}&= \int {0 \,d\tau } + c_1 \\ &=c_1 \end{align*}

Dividing throughout by the integrating factor \(\sqrt {\tau -1}\, \sqrt {\tau +1}\, {\mathrm e}^{-3 \arcsin \left (\tau \right )}\) gives the final solution

\[ u \left (\tau \right ) = \frac {{\mathrm e}^{3 \arcsin \left (\tau \right )} c_1}{\sqrt {\tau -1}\, \sqrt {\tau +1}} \]

In summary, these are the solution found for \(x \left (\tau \right )\)

\begin{align*} u \left (\tau \right ) &= \frac {{\mathrm e}^{3 \arcsin \left (\tau \right )} c_1}{\sqrt {\tau -1}\, \sqrt {\tau +1}} \\ \end{align*}

For solution \(u \left (\tau \right ) = \frac {{\mathrm e}^{3 \arcsin \left (\tau \right )} c_1}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\), since \(u=\frac {d}{d \tau }x \left (\tau \right )\) then we now have a new first order ode to solve which is

\begin{align*} \frac {d}{d \tau }x \left (\tau \right ) = \frac {{\mathrm e}^{3 \arcsin \left (\tau \right )} c_1}{\sqrt {\tau -1}\, \sqrt {\tau +1}} \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(\frac {d}{d \tau }x \left (\tau \right )=f(\tau )\), then we only need to integrate \(f(\tau )\).

\begin{align*} \int {dx} &= \int {\frac {{\mathrm e}^{3 \arcsin \left (\tau \right )} c_1}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\, d\tau }\\ x \left (\tau \right ) &= \int \frac {{\mathrm e}^{3 \arcsin \left (\tau \right )} c_1}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau + c_2 \end{align*}

\begin{align*} x \left (\tau \right )&= \int \frac {{\mathrm e}^{3 \arcsin \left (\tau \right )} c_1}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau +c_2 \end{align*}

In summary, these are the solution found for \((x \left (\tau \right ))\)

\begin{align*} x \left (\tau \right ) &= \int \frac {{\mathrm e}^{3 \arcsin \left (\tau \right )} c_1}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau +c_2 \\ \end{align*}

Applying change of variable \(\tau = \cos \left (t \right )\) to the solutions above gives

\begin{align*} x &= \int -\frac {{\mathrm e}^{3 \arcsin \left (\cos \left (t \right )\right )} c_1 \sin \left (t \right )}{\sqrt {\cos \left (t \right )-1}\, \sqrt {\cos \left (t \right )+1}}d t +c_2 \\ \end{align*}

Figure 2.64: Slope field \(x^{\prime \prime }+3 x^{\prime } = 0\)

2.2.7.7 ✓ Maple. Time used: 0.001 (sec). Leaf size: 12

ode:=diff(diff(x(t),t),t)+3*diff(x(t),t) = 0; 
dsolve(ode,x(t), singsol=all);
 

\[ x = c_1 +c_2 \,{\mathrm e}^{-3 t} \]

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d t^{2}}x \left (t \right )+3 \frac {d}{d t}x \left (t \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d t^{2}}x \left (t \right ) \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+3 r =0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & r \left (r +3\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-3, 0\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & x_{1}\left (t \right )={\mathrm e}^{-3 t} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & x_{2}\left (t \right )=1 \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & x \left (t \right )=\mathit {C1} x_{1}\left (t \right )+\mathit {C2} x_{2}\left (t \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & x \left (t \right )=\mathit {C1} \,{\mathrm e}^{-3 t}+\mathit {C2} \end {array} \]

2.2.7.8 ✓ Mathematica. Time used: 0.012 (sec). Leaf size: 19

ode=D[x[t],{t,2}]+3*D[x[t],t]==0; 
ic={}; 
DSolve[{ode,ic},x[t],t,IncludeSingularSolutions->True]
 

\begin{align*} x(t)&\to c_2-\frac {1}{3} c_1 e^{-3 t} \end{align*}

2.2.7.9 ✓ Sympy. Time used: 0.077 (sec). Leaf size: 10

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq(3*Derivative(x(t), t) + Derivative(x(t), (t, 2)),0) 
ics = {} 
dsolve(ode,func=x(t),ics=ics)
 

\[ x{\left (t \right )} = C_{1} + C_{2} e^{- 3 t} \]