2.1.12 Internal
problem
ID
[8495]Book
:
Second
order
enumerated
odes Section
:
section
1 Problem
number
:
12Date
solved
:
Sunday, November 10, 2024 at 03:55:05 AMCAS
classification
:
[[_2nd_order, _missing_x]]
Solve
\begin{align*} y^{\prime \prime }+y^{\prime }&=0 \end{align*}
Time used: 0.063 (sec)
This is second order with constant coefficients homogeneous ODE. In standard form the
ODE is
\[ A y''(x) + B y'(x) + C y(x) = 0 \]
Where in the above \(A=1, B=1, C=0\) . Let the solution be \(y=e^{\lambda x}\) . Substituting this into the ODE gives
\[ \lambda ^{2} {\mathrm e}^{x \lambda }+\lambda \,{\mathrm e}^{x \lambda } = 0 \tag {1} \]
Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda x}\) gives
\[ \lambda ^{2}+\lambda = 0 \tag {2} \]
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 \(A=1, B=1, C=0\) into the above gives
\begin{align*} \lambda _{1,2} &= \frac {-1}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {1^2 - (4) \left (1\right )\left (0\right )}\\ &= -{\frac {1}{2}} \pm {\frac {1}{2}} \end{align*}
Hence
\begin{align*}
\lambda _1 &= -{\frac {1}{2}} + {\frac {1}{2}} \\
\lambda _2 &= -{\frac {1}{2}} - {\frac {1}{2}} \\
\end{align*}
Which simplifies to
\begin{align*}
\lambda _1 &= 0 \\
\lambda _2 &= -1 \\
\end{align*}
Since roots are real and distinct, then the solution is
\begin{align*}
y &= c_1 e^{\lambda _1 x} + c_2 e^{\lambda _2 x} \\
y &= c_1 e^{\left (0\right )x} +c_2 e^{\left (-1\right )x} \\
\end{align*}
Or
\[
y =c_1 +c_2 \,{\mathrm e}^{-x}
\]
Will
add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_1 +c_2 \,{\mathrm e}^{-x} \\
\end{align*}
Figure 2.36: Slope field plot\(y^{\prime \prime }+y^{\prime } = 0\) Time used: 0.132 (sec)
An ode of the form
\begin{align*} p \left (x \right ) y^{\prime \prime }+q \left (x \right ) y^{\prime }+r \left (x \right ) y&=s \left (x \right ) \end{align*}
is exact if
\begin{align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end{align*}
For the given ode we have
\begin{align*} p(x) &= 1\\ q(x) &= 1\\ 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 (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end{align*}
Integrating gives
\begin{align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end{align*}
Substituting the above values for \(p,q,r,s\) gives
\begin{align*} y^{\prime }+y&=c_1 \end{align*}
We now have a first order ode to solve which is
\begin{align*} y^{\prime }+y = c_1 \end{align*}
Integrating gives
\begin{align*} \int \frac {1}{-y +c_1}d y &= dx\\ -\ln \left (-y +c_1 \right )&= x +c_2 \end{align*}
Singular solutions are found by solving
\begin{align*} -y +c_1&= 0 \end{align*}
for \(y\) . 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*} y = c_1 \end{align*}
Solving for \(y\) gives
\begin{align*}
y &= -{\mathrm e}^{-x -c_2}+c_1 \\
\end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_1 \\
y &= -{\mathrm e}^{-x -c_2}+c_1 \\
\end{align*}
Figure 2.37: Slope field plot\(y^{\prime \prime }+y^{\prime } = 0\) Time used: 0.099 (sec)
This is second order ode with missing dependent variable \(y\) . Let
\begin{align*} p(x) &= y^{\prime } \end{align*}
Then
\begin{align*} p'(x) &= y^{\prime \prime } \end{align*}
Hence the ode becomes
\begin{align*} p^{\prime }\left (x \right )+p \left (x \right ) = 0 \end{align*}
Which is now solve for \(p(x)\) as first order ode. Integrating gives
\begin{align*} \int -\frac {1}{p}d p &= dx\\ -\ln \left (p \right )&= x +c_1 \end{align*}
Singular solutions are found by solving
\begin{align*} -p&= 0 \end{align*}
for \(p \left (x \right )\) . 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*} p \left (x \right ) = 0 \end{align*}
Solving for \(p \left (x \right )\) gives
\begin{align*}
p \left (x \right ) &= {\mathrm e}^{-x -c_1} \\
\end{align*}
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order
ode to solve which is
\begin{align*} y^{\prime } = 0 \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\) , then we only need to integrate \(f(x)\) .
\begin{align*} \int {dy} &= \int {0\, dx} + c_2 \\ y &= c_2 \end{align*}
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is
\begin{align*} y^{\prime } = {\mathrm e}^{-x -c_1} \end{align*}
Since the ode has the form \(y^{\prime }=f(x)\) , then we only need to integrate \(f(x)\) .
\begin{align*} \int {dy} &= \int {{\mathrm e}^{-x -c_1}\, dx}\\ y &= -{\mathrm e}^{-x -c_1} + c_3 \end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_2 \\
y &= -{\mathrm e}^{-x -c_1}+c_3 \\
\end{align*}
Figure 2.38: Slope field plot\(y^{\prime \prime }+y^{\prime } = 0\) Time used: 0.056 (sec)
Integrating both sides of the ODE w.r.t \(x\) gives
\begin{align*} \int \left (y^{\prime \prime }+y^{\prime }\right )d x &= 0 \\ y^{\prime }+y = c_1 \end{align*}
Which is now solved for \(y\) . Integrating gives
\begin{align*} \int \frac {1}{-y +c_1}d y &= dx\\ -\ln \left (-y +c_1 \right )&= x +c_2 \end{align*}
Singular solutions are found by solving
\begin{align*} -y +c_1&= 0 \end{align*}
for \(y\) . 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*} y = c_1 \end{align*}
Solving for \(y\) gives
\begin{align*}
y &= -{\mathrm e}^{-x -c_2}+c_1 \\
\end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_1 \\
y &= -{\mathrm e}^{-x -c_2}+c_1 \\
\end{align*}
Figure 2.39: Slope field plot\(y^{\prime \prime }+y^{\prime } = 0\) Time used: 0.049 (sec)
Writing the ode as
\[
y^{\prime \prime }+y^{\prime } = 0
\]
Integrating both sides of the ODE w.r.t \(x\) gives
\begin{align*} \int \left (y^{\prime \prime }+y^{\prime }\right )d x &= 0 \\ y^{\prime }+y = c_1 \end{align*}
Which is now solved for \(y\) . Integrating gives
\begin{align*} \int \frac {1}{-y +c_1}d y &= dx\\ -\ln \left (-y +c_1 \right )&= x +c_2 \end{align*}
Singular solutions are found by solving
\begin{align*} -y +c_1&= 0 \end{align*}
for \(y\) . 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*} y = c_1 \end{align*}
Solving for \(y\) gives
\begin{align*}
y &= -{\mathrm e}^{-x -c_2}+c_1 \\
\end{align*}
Will add steps showing solving for IC soon.
Figure 2.40: Slope field plot\(y^{\prime \prime }+y^{\prime } = 0\) Time used: 0.040 (sec)
Writing the ode as
\begin{align*} y^{\prime \prime }+y^{\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 &= 1\tag {3} \\ C &= 0 \end{align*}
Applying the Liouville transformation on the dependent variable gives
\begin{align*} z(x) &= y e^{\int \frac {B}{2 A} \,dx} \end{align*}
Then (2) becomes
\begin{align*} z''(x) = r z(x)\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 {1}{4}\tag {6} \end{align*}
Comparing the above to (5) shows that
\begin{align*} s &= 1\\ t &= 4 \end{align*}
Therefore eq. (4) becomes
\begin{align*} z''(x) &= \frac {z \left (x \right )}{4} \tag {7} \end{align*}
Equation (7) is now solved. After finding \(z(x)\) then \(y\) is found using the inverse transformation
\begin{align*} y &= z \left (x \right ) e^{-\int \frac {B}{2 A} \,dx} \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 2.5: 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 \(r = {\frac {1}{4}}\) is not a function of \(x\) , then there is no need run Kovacic algorithm to obtain a solution
for transformed ode \(z''=r z\) as one solution is
\[ z_1(x) = {\mathrm e}^{-\frac {x}{2}} \]
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
\begin{align*}
y_1 &= z_1 e^{ \int -\frac {1}{2} \frac {B}{A} \,dx} \\
&= z_1 e^{ -\int \frac {1}{2} \frac {1}{1} \,dx} \\
&= z_1 e^{-\frac {x}{2}} \\
&= z_1 \left ({\mathrm e}^{-\frac {x}{2}}\right ) \\
\end{align*}
Which simplifies to
\[
y_1 = {\mathrm e}^{-x}
\]
The second solution \(y_2\) to the original ode is found using reduction of order
\[ y_2 = y_1 \int \frac { e^{\int -\frac {B}{A} \,dx}}{y_1^2} \,dx \]
Substituting gives
\begin{align*}
y_2 &= y_1 \int \frac { e^{\int -\frac {1}{1} \,dx}}{\left (y_1\right )^2} \,dx \\
&= y_1 \int \frac { e^{-x}}{\left (y_1\right )^2} \,dx \\
&= y_1 \left ({\mathrm e}^{x}\right ) \\
\end{align*}
Therefore the solution is
\begin{align*}
y &= c_1 y_1 + c_2 y_2 \\
&= c_1 \left ({\mathrm e}^{-x}\right ) + c_2 \left ({\mathrm e}^{-x}\left ({\mathrm e}^{x}\right )\right ) \\
\end{align*}
Will add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_1 \,{\mathrm e}^{-x}+c_2 \\
\end{align*}
Figure 2.41: Slope field plot\(y^{\prime \prime }+y^{\prime } = 0\) Time used: 0.406 (sec)
In normal form the ode
\begin{align*} y^{\prime \prime }+y^{\prime } = 0 \tag {1} \end{align*}
Becomes
\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=r \left (x \right ) \tag {2} \end{align*}
Where
\begin{align*} p \left (x \right )&=1\\ q \left (x \right )&=0\\ r \left (x \right )&=0 \end{align*}
The Lagrange adjoint ode is given by
\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\xi \left (x \right )\right )' + \left (0\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )-\xi ^{\prime }\left (x \right )&= 0 \end{align*}
Which is solved for \(\xi (x)\) . This is second order with constant coefficients homogeneous ODE. In
standard form the ODE is
\[ A \xi ''(x) + B \xi '(x) + C \xi (x) = 0 \]
Where in the above \(A=1, B=-1, C=0\) . Let the solution be \(\xi =e^{\lambda x}\) . Substituting this into
the ODE gives
\[ \lambda ^{2} {\mathrm e}^{x \lambda }-\lambda \,{\mathrm e}^{x \lambda } = 0 \tag {1} \]
Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda x}\)
gives
\[ \lambda ^{2}-\lambda = 0 \tag {2} \]
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 \(A=1, B=-1, C=0\) into the above gives
\begin{align*} \lambda _{1,2} &= \frac {1}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {-1^2 - (4) \left (1\right )\left (0\right )}\\ &= {\frac {1}{2}} \pm {\frac {1}{2}} \end{align*}
Hence
\begin{align*}
\lambda _1 &= {\frac {1}{2}} + {\frac {1}{2}} \\
\lambda _2 &= {\frac {1}{2}} - {\frac {1}{2}} \\
\end{align*}
Which simplifies to
\begin{align*}
\lambda _1 &= 1 \\
\lambda _2 &= 0 \\
\end{align*}
Since roots are real and distinct, then the solution is
\begin{align*}
\xi &= c_1 e^{\lambda _1 x} + c_2 e^{\lambda _2 x} \\
\xi &= c_1 e^{\left (1\right )x} +c_2 e^{\left (0\right )x} \\
\end{align*}
Or
\[
\xi =c_1 \,{\mathrm e}^{x}+c_2
\]
Will
add steps showing solving for IC soon.
The original ode now reduces to first order ode
\begin{align*} \xi \left (x \right ) y^{\prime }-y \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) y&=\int \xi \left (x \right ) r \left (x \right )d x\\ y^{\prime }+y \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )} \end{align*}
Or
\begin{align*} y^{\prime }+y \left (1-\frac {c_1 \,{\mathrm e}^{x}}{c_1 \,{\mathrm e}^{x}+c_2}\right )&=0 \end{align*}
Which is now a first order ode. This is now solved for \(y\) . In canonical form a linear first order
is
\begin{align*} y^{\prime } + q(x)y &= p(x) \end{align*}
Comparing the above to the given ode shows that
\begin{align*} q(x) &=\frac {c_2}{c_1 \,{\mathrm e}^{x}+c_2}\\ p(x) &=0 \end{align*}
The integrating factor \(\mu \) is
\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {c_2}{c_1 \,{\mathrm e}^{x}+c_2}d x}\\ &= \frac {{\mathrm e}^{x}}{c_1 \,{\mathrm e}^{x}+c_2} \end{align*}
The ode becomes
\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y \,{\mathrm e}^{x}}{c_1 \,{\mathrm e}^{x}+c_2}\right ) &= 0 \end{align*}
Integrating gives
\begin{align*} \frac {y \,{\mathrm e}^{x}}{c_1 \,{\mathrm e}^{x}+c_2}&= \int {0 \,dx} + c_3 \\ &=c_3 \end{align*}
Dividing throughout by the integrating factor \(\frac {{\mathrm e}^{x}}{c_1 \,{\mathrm e}^{x}+c_2}\) gives the final solution
\[ y = c_3 \left (c_1 +c_2 \,{\mathrm e}^{-x}\right ) \]
Hence, the solution
found using Lagrange adjoint equation method is
\begin{align*}
y &= c_3 \left (c_1 +c_2 \,{\mathrm e}^{-x}\right ) \\
\end{align*}
The constants can be merged to give
\[
y = c_1 +c_2 \,{\mathrm e}^{-x}
\]
Will
add steps showing solving for IC soon.
Summary of solutions found
\begin{align*}
y &= c_1 +c_2 \,{\mathrm e}^{-x} \\
\end{align*}
Figure 2.42: Slope field plot\(y^{\prime \prime }+y^{\prime } = 0\) \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )+\frac {d}{d x}y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Characteristic polynomial of ODE}\hspace {3pt} \\ {} & {} & r^{2}+r =0 \\ \bullet & {} & \textrm {Factor the characteristic polynomial}\hspace {3pt} \\ {} & {} & r \left (r +1\right )=0 \\ \bullet & {} & \textrm {Roots of the characteristic polynomial}\hspace {3pt} \\ {} & {} & r =\left (-1, 0\right ) \\ \bullet & {} & \textrm {1st solution of the ODE}\hspace {3pt} \\ {} & {} & y_{1}\left (x \right )={\mathrm e}^{-x} \\ \bullet & {} & \textrm {2nd solution of the ODE}\hspace {3pt} \\ {} & {} & y_{2}\left (x \right )=1 \\ \bullet & {} & \textrm {General solution of the ODE}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} y_{1}\left (x \right )+\mathit {C2} y_{2}\left (x \right ) \\ \bullet & {} & \textrm {Substitute in solutions}\hspace {3pt} \\ {} & {} & y \left (x \right )=\mathit {C1} \,{\mathrm e}^{-x}+\mathit {C2} \end {array} \]
` Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
checking if the LODE has constant coefficients
<- constant coefficients successful `
Solving time : 0.004
dsolve ( diff ( y ( x ), x )+ diff ( diff ( y ( x ), x ), x ) = 0,
y(x),singsol=all)
\[
y = c_{1} +{\mathrm e}^{-x} c_{2}
\]
Solving time : 0.01
DSolve [{ D [ y [ x ],{ x ,2}]+ D [ y [ x ], x ]==0,{}}, y[x],x,IncludeSingularSolutions-> True ]
\[
y(x)\to c_2-c_1 e^{-x}
\]