Internal problem ID [2367]
Internal file name [OUTPUT/2367_Tuesday_February_27_2024_08_36_12_AM_97136872/index.tex
]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 40, page 186
Problem number: 4.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "first order ode series method. Ordinary point", "first order ode series method. Taylor series method"
Maple gives the following as the ode type
[_linear]
\[ \boxed {y^{\prime }-\frac {y}{x}=3 x} \] With initial conditions \begin {align*} [y \left (1\right ) = 3] \end {align*}
With the expansion point for the power series method at \(x = 1\).
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {1}{x}\\ q(x) &=3 x \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {y}{x} = 3 x \end {align*}
The domain of \(p(x)=-\frac {1}{x}\) is \[
\{x <0\boldsymbol {\lor }0
The ode does not have its expansion point at \(x = 0\), therefore to simplify the computation of
power series expansion, change of variable is made on the independent variable to shift the
initial conditions and the expasion point back to zero. The new ode is then solved more
easily since the expansion point is now at zero. The solution converted back to the original
independent variable. Let \[ t = x -1 \] The ode is converted to be in terms of the new independent
variable \(t\). This results in \[
\frac {d}{d t}y \left (t \right )-\frac {y \left (t \right )}{t +1} = 3 t +3
\] With its expansion point and initial conditions now at \(t = 0\). With
initial conditions now becoming \begin {align*} y(0) &= 3\\ \end {align*}
The transformed ODE is now solved.
Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving first order ode. Let \[ y^{\prime }=f\left ( x,y\right ) \] Where \(f\left ( x,y\right ) \) is analytic at expansion point \(x_{0}\). We can always
shift to \(x_{0}=0\) if \(x_{0}\) is not zero. So from now we assume \(x_{0}=0\,\). Assume also that \(y\left ( x_{0}\right ) =y_{0}\). Using Taylor
series\begin {align*} y\left ( x\right ) & =y\left ( x_{0}\right ) +\left ( x-x_{0}\right ) y^{\prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{2}}{2}y^{\prime \prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{3}}{3!}y^{\prime \prime }\left ( x_{0}\right ) +\cdots \\ & =y_{0}+xf+\frac {x^{2}}{2}\left . \frac {df}{dx}\right \vert _{x_{0},y_{0}}+\frac {x^{3}}{3!}\left . \frac {d^{2}f}{dx^{2}}\right \vert _{x_{0},y_{0}}+\cdots \\ & =y_{0}+\sum _{n=0}^{\infty }\frac {x^{n+1}}{\left ( n+1\right ) !}\left . \frac {d^{n}f}{dx^{n}}\right \vert _{x_{0},y_{0}} \end {align*}
But \begin {align} \frac {df}{dx} & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}f\tag {1}\\ \frac {d^{2}f}{dx^{2}} & =\frac {d}{dx}\left ( \frac {df}{dx}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {df}{dx}\right ) +\frac {\partial }{\partial y}\left ( \frac {df}{dx}\right ) f\tag {2}\\ \frac {d^{3}f}{dx^{3}} & =\frac {d}{dx}\left ( \frac {d^{2}f}{dx^{2}}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {d^{2}f}{dx^{2}}\right ) +\left ( \frac {\partial }{\partial y}\frac {d^{2}f}{dx^{2}}\right ) f\tag {3}\\ & \vdots \nonumber \end {align}
And so on. Hence if we name \(F_{0}=f\left ( x,y\right ) \) then the above can be written as \begin {align} F_{0} & =f\left ( x,y\right ) \tag {4}\\ F_{n} & =\frac {d}{dx}\left ( F_{n-1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) F_{0} \tag {5} \end {align}
For example, for \(n=1\,\) we see that \begin {align*} F_{1} & =\frac {d}{dx}\left ( F_{0}\right ) \\ & =\frac {\partial }{\partial x}F_{0}+\left ( \frac {\partial F_{0}}{\partial y}\right ) F_{0}\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}f \end {align*}
Which is (1). And when \(n=2\)\begin {align*} F_{2} & =\frac {d}{dx}\left ( F_{1}\right ) \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) F_{0}\\ & =\frac {\partial }{\partial x}\left ( \frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}f\right ) +\frac {\partial }{\partial y}\left ( \frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}f\right ) f\\ & =\frac {\partial }{\partial x}\left ( \frac {df}{dx}\right ) +\frac {\partial }{\partial y}\left ( \frac {df}{dx}\right ) f \end {align*}
Which is (2) and so on. Therefore (4,5) can be used from now on along with \begin {equation} y\left ( x\right ) =y_{0}+\sum _{n=0}^{\infty }\frac {x^{n+1}}{\left ( n+1\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0}} \tag {6} \end {equation} Hence
\begin {align*} F_0 &= \frac {3 t^{2}+6 t +y \left (t \right )+3}{t +1}\\ F_1 &= \frac {d F_0}{dt} \\ &= \frac {\partial F_0}{\partial t}+ \frac {\partial F_0}{\partial y} F_0 \\ &= 6\\ F_2 &= \frac {d F_1}{dt} \\ &= \frac {\partial F_1}{\partial t}+ \frac {\partial F_1}{\partial y} F_1 \\ &= 0\\ F_3 &= \frac {d F_2}{dt} \\ &= \frac {\partial F_2}{\partial t}+ \frac {\partial F_2}{\partial y} F_2 \\ &= 0 \end {align*}
And so on. Evaluating all the above at initial conditions \(t \left (0\right ) = 0\) and \(y \left (0\right ) = 3\) gives \begin {align*} F_0 &= 6\\ F_1 &= 6\\ F_2 &= 0\\ F_3 &= 0 \end {align*}
Substituting all the above in (6) and simplifying gives the solution as \[
y \left (t \right ) = 3 t^{2}+6 t +3+O\left (t^{5}\right )
\] Now we substitute the
given initial conditions in the above to solve for \(y \left (0\right )\). Solving for \(y \left (0\right )\) from initial conditions gives
\begin {align*} y \left (0\right ) = y \left (0\right ) \end {align*}
Therefore the solution becomes \begin {align*} y \left (t \right ) = 3 t^{2}+6 t +3 \end {align*}
Hence the solution can be written as \begin {gather*} y \left (t \right ) = 3 t^{2}+6 t +3+O\left (t^{5}\right ) \end {gather*} which simplifies to \begin {gather*} y \left (t \right ) = 3 t^{2}+6 t +3+O\left (t^{5}\right ) \end {gather*} Since \(t = 0\) is also an ordinary point,
then standard power series can also be used. Writing the ODE as \begin {align*} \frac {d}{d t}y \left (t \right ) + q(t)y \left (t \right ) &= p(t) \\ \frac {d}{d t}y \left (t \right )-\frac {y \left (t \right )}{t +1} &= \frac {3 t^{2}+6 t +3}{t +1} \end {align*}
Where \begin {align*} q(t) &= -\frac {1}{t +1}\\ p(t) &= \frac {3 t^{2}+6 t +3}{t +1} \end {align*}
Next, the type of the expansion point \(t = 0\) is determined. This point can be an ordinary point, a
regular singular point (also called removable singularity), or irregular singular point (also
called non-removable singularity or essential singularity). When \(t = 0\) is an ordinary point, then
the standard power series is used. If the point is a regular singular point, Frobenius series is
used instead. Irregular singular point requires more advanced methods (asymptotic methods)
and is not supported now. Hopefully this will be added in the future. \(t = 0\) is called
an ordinary point \(q(t)\) has a Taylor series expansion around the point \(t = 0\). \(t = 0\) is called a
regular singular point if \(q(t)\) is not not analytic at \(t = 0\) but \(t q(t)\) has Taylor series expansion.
And finally, \(t = 0\) is an irregular singular point if the point is not ordinary and not
regular singular. This is the most complicated case. Now the expansion point \(t = 0\)
is checked to see if it is an ordinary point or not. Now the ode is normalized by
writing it as \[ \left (t +1\right ) \left (\frac {d}{d t}y \left (t \right )\right )-y \left (t \right ) = 3 \left (t +1\right )^{2} \] Let the solution be represented as power series of the form \[ y \left (t \right ) = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n} \] Then
\begin {align*} \frac {d}{d t}y \left (t \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1} \end {align*}
Substituting the above back into the ode gives \begin {align*} \left (t +1\right ) \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = 3 \left (t +1\right )^{2}\tag {1} \end {align*}
Expanding \(3 \left (t +1\right )^{2}\) as Taylor series around \(t=0\) and keeping only the first \(5\) terms gives \begin {align*} 3 \left (t +1\right )^{2} &= 3 t^{2}+6 t +3 + \dots \\ &= 3 t^{2}+6 t +3 \end {align*}
Hence the ODE in Eq (1) becomes \[ \left (t +1\right ) \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right ) = 3 t^{2}+6 t +3\tag {1} \] Which simplifies to \begin{equation}
\tag{2} \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} t^{n}\right ) = 3 t^{2}+6 t +3
\end{equation} The next step is to make all powers
of \(t\) be \(n\) in each summation term. Going over each summation term above with power of \(t\) in it
which is not already \(t^{n}\) and adjusting the power and the corresponding index gives \begin{align*}
\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n -1} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +1\right ) a_{n +1} t^{n} \\
\end{align*}
Substituting all the above in Eq (2) gives the following equation where now all powers of \(t\) are
the same and equal to \(n\). \begin{equation}
\tag{3} \left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} t^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +1\right ) a_{n +1} t^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} t^{n}\right ) = 3 t^{2}+6 t +3
\end{equation} \(n=0\) gives \begin {align*} \left (a_{1}-a_{0}\right ) t^0 &= 3\\ a_{1}-a_{0} &= 3 \\ a_{1} &= a_{0}+3 \end {align*}
For \(1\le n\), the recurrence equation is \begin{equation}
\tag{4} \left (n a_{n}+\left (n +1\right ) a_{n +1}-a_{n}\right ) t^{n} = 3 t^{2}+6 t +3
\end{equation} For \(n = 1\) the recurrence equation gives \begin{align*}
\left (2 a_{2}\right ) t&=6 t \\
2 a_{2} &= 6 \\
\end{align*} Which after substituting
the earlier terms found becomes \[
a_{2} = 3
\] For \(n = 2\) the recurrence equation gives \begin{align*}
\left (a_{2}+3 a_{3}\right ) t^{2}&=3 t^{2} \\
a_{2}+3 a_{3} &= 3 \\
\end{align*} Which after substituting
the earlier terms found becomes \[
a_{3} = 0
\] For \(n = 3\) the recurrence equation gives \begin{align*}
\left (2 a_{3}+4 a_{4}\right ) t^{3}&=0 \\
2 a_{3}+4 a_{4} &= 0 \\
\end{align*} Which after substituting
the earlier terms found becomes \[
a_{4} = 0
\] For \(n = 4\) the recurrence equation gives \begin{align*}
\left (3 a_{4}+5 a_{5}\right ) t^{4}&=0 \\
3 a_{4}+5 a_{5} &= 0 \\
\end{align*} Which after
substituting the earlier terms found becomes \[
a_{5} = 0
\] And so on. Therefore the solution is
\begin {align*} y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\\ &= a_{3} t^{3}+a_{2} t^{2}+a_{1} t +a_{0} + \dots \end {align*}
Substituting the values for \(a_{n}\) found above, the solution becomes \[
y \left (t \right ) = a_{0}+\left (a_{0}+3\right ) t +3 t^{2}+\dots
\] Collecting terms, the solution
becomes \begin{equation}
\tag{3} y \left (t \right ) = \left (t +1\right ) a_{0}+3 t^{2}+O\left (t^{5}\right )+3 t
\end{equation} At \(t = 0\) the solution above becomes \[ y(0) = a_{0} \] Therefore the solution in Eq(3) now can be
written as \[
y \left (t \right ) = \left (t +1\right ) y \left (0\right )+3 t^{2}+O\left (t^{5}\right )+3 t
\] Now we substitute the given initial conditions in the above to solve for \(y \left (0\right )\). Solving
for \(y \left (0\right )\) from initial conditions gives \begin {align*} y \left (0\right ) = 3 \end {align*}
Therefore the solution becomes \begin {align*} y \left (t \right ) = 3 t^{2}+6 t +3 \end {align*}
Hence the solution can be written as \begin {gather*} y \left (t \right ) = 3 t^{2}+6 t +3+O\left (t^{5}\right ) \end {gather*} which simplifies to \begin {gather*} y \left (t \right ) = 3 t^{2}+6 t +3+O\left (t^{5}\right ) \end {gather*} Replacing \(t\)
in the above with the original independent variable \(xs\) using \(t = x -1\) results in \begin {gather*} y = 3 \left (x -1\right )^{2}+6 x -3+O\left (\left (x -1\right )^{5}\right ) \end {gather*}
The solution(s) found are the following \begin{align*}
\tag{1} y &= 3 \left (x -1\right )^{2}+6 x -3+O\left (\left (x -1\right )^{5}\right ) \\
\end{align*} Verification of solutions
\[
y = 3 \left (x -1\right )^{2}+6 x -3+O\left (\left (x -1\right )^{5}\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {y}{x}=3 x , y \left (1\right )=3\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=3 x +\frac {y}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {y}{x}=3 x \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }-\frac {y}{x}\right )=3 \mu \left (x \right ) x \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (y \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (y^{\prime }-\frac {y}{x}\right )=y^{\prime } \mu \left (x \right )+y \mu ^{\prime }\left (x \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (x \right ) \\ {} & {} & \mu ^{\prime }\left (x \right )=-\frac {\mu \left (x \right )}{x} \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )=\frac {1}{x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (y \mu \left (x \right )\right )\right )d x =\int 3 \mu \left (x \right ) x d x +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (x \right )=\int 3 \mu \left (x \right ) x d x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int 3 \mu \left (x \right ) x d x +c_{1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )=\frac {1}{x} \\ {} & {} & y=x \left (\int 3d x +c_{1} \right ) \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=x \left (3 x +c_{1} \right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=3 \\ {} & {} & 3=3+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y=3 x^{2} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=3 x^{2} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 16
\[
y \left (x \right ) = 3 x^{2}
\]
✓ Solution by Mathematica
Time used: 0.028 (sec). Leaf size: 17
\[
y(x)\to 3 (x-1)^2+6 (x-1)+3
\]
22.4.2 Solving as series ode
22.4.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful`
Order:=5;
dsolve([diff(y(x),x)=3*x+y(x)/x,y(1) = 3],y(x),type='series',x=1);
AsymptoticDSolveValue[{y'[x]==3*x+y[x]/x,{y[1]==3}},y[x],{x,1,4}]