Internal problem ID [5625]
Internal file name [OUTPUT/4873_Sunday_June_05_2022_03_08_52_PM_75155354/index.tex]
Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.1. page
174
Problem number: 8.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "first order ode series method. Regular singular point"
Maple gives the following as the ode type
[_separable]
\[ \boxed {y^{\prime } x -3 y=k} \] With the expansion point for the power series method at \(x = 0\).
Writing the ODE as \begin {align*} y^{\prime } + q(x)y &= p(x) \\ y^{\prime }-\frac {3 y}{x} &= \frac {k}{x} \end {align*}
Where \begin {align*} q(x) &= -\frac {3}{x}\\ p(x) &= \frac {k}{x} \end {align*}
Next, the type of the expansion point \(x = 0\) 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 \(x = 0\) 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. \(x = 0\) is called an ordinary point \(q(x)\) has a Taylor series expansion around the point \(x = 0\). \(x = 0\) is called a regular singular point if \(q(x)\) is not not analytic at \(x = 0\) but \(x q(x)\) has Taylor series expansion. And finally, \(x = 0\) 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 \(x = 0\) is checked to see if it is an ordinary point or not.
Since \(x = 0\) is not an ordinary point, we now check to see if it is a regular singular point. \(xq(x)=-3\) has a Taylor series around \(x = 0\). Since \(x = 0\) is regular singular point, then Frobenius power series is used. Since this is an inhomogeneous, then let the solution be \[ y = y_h + y_p \] Where \(y_h\) is the solution to the homogeneous ode \(y^{\prime }-\frac {3 y}{x} = 0\),and \(y_p\) is a particular solution to the inhomogeneous ode. First, we solve for \(y_h\) Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin {align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \end {align*}
Substituting the above back into the ode gives \begin {align*} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-\frac {3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )}{x} = 0\tag {1} \end {align*}
Hence the ODE in Eq (1) becomes \[ \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-\frac {3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )}{x} = 0\tag {1} \] Expanding the second term in (1) gives \[ \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-3\eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\frac {1}{x}\eslowast \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0\tag {1} \] Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r -1} a_{n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r -1} a_{n}\right ) = 0 \end{equation} The indicial equation is obtained from \(n=0\). From Eq (2) this gives \[ \left (n +r \right ) a_{n} x^{n +r -1}-3 x^{n +r -1} a_{n} = 0 \] When \(n=0\) the above becomes \[ r a_{0} x^{-1+r}-3 x^{-1+r} a_{0} = 0 \] The corresponding balance equation is found by replacing \(r\) by \(m\) and \(a\) by \(c\) to avoid confusing terms between particular solution and the homogeneous solution. Hence the balance equation is \begin {align*} \left (x^{-1+m} m -3 x^{-1+m}\right ) c_{0} = \frac {k}{x} \end {align*}
This equation will used later to find the particular solution.
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r -3\right ) x^{-1+r} = 0 \] Since the above is true for all \(x\) then the indicial equation simplifies to \[ r -3 = 0 \] Solving for \(r\) gives the root of the indicial equation as \[ r=3 \]
We start by finding \(y_h\). From the above we see that there is no recurrence relation since there is only one summation term. Therefore all \(a_{n}\) terms are zero except for \(a_{0}\). Hence \begin {align*} y_h &= a_{0} x^r \end {align*}
Therefore the homogeneous solution is \begin {align*} y_h(x) &= a_{0} \left (x^{3}+O\left (x^{6}\right )\right ) \end {align*}
Now we determine the particular solution \(y_p\) by solving the balance equation \[ \left (x^{-1+m} m -3 x^{-1+m}\right ) c_{0} = \frac {k}{x} \] For \(c_{0}\) and \(x\). This results in \begin {align*} c_{0}&=-\frac {k}{3}\\ m&=0 \end {align*}
The particular solution is therefore \begin {align*} y_p &= \sum _{n=0}^{\infty } c_n x^{n+m}\\ &= \sum _{n=0}^{\infty } c_n x^{n+0} \end {align*}
Where in the above \(c_0 = -\frac {k}{3}\). The remaining \(c_n\) values are found using the same recurrence relation used to find the homogeneous solution but using \(c_0\) in place of \(a_{0}\) and using \(m=0\) in place of the root of the indicial equation used to find the homogeneous solution. The following are the values of \(a_n\) found in terms of the indicial root \(r\). These will be now used to find find \(c_n\) by replacing \(a_{0} = -\frac {k}{3}\) and \(r=0\). The following table gives the \(a_n\) values found and the corresponding \(c_n\) values which will be used to find the particular solution
| \(n\) | \(a_n\) | \(c_n\) |
| \(0\) | \(a_{0} = 1\) | \(c_{0} = -\frac {k}{3}\) |
The particular solution is now found using \begin {align*} y_p &= x^{m} \sum _{n=0}^{\infty } c_n x^n\\ &= 1 \sum _{n=0}^{\infty } c_n x^n \end {align*}
Using the values found above for \(c_n\) into the above sum gives \begin {align*} y_p &= 1\left (-\frac {k}{3}\right ) \end {align*}
At \(x = 0\) the solution above becomes \begin {gather*} y = -\frac {k}{3}+O\left (x^{6}\right )+c_{1} \left (x^{3}+O\left (x^{6}\right )\right ) \end {gather*}
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {k}{3}+O\left (x^{6}\right )+c_{1} \left (x^{3}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = -\frac {k}{3}+O\left (x^{6}\right )+c_{1} \left (x^{3}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\frac {3 y}{x}=\frac {k}{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {3 y}{x}+\frac {k}{x} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{3 y+k}=\frac {1}{x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{3 y+k}d x =\int \frac {1}{x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (3 y+k \right )}{3}=\ln \left (x \right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {{\mathrm e}^{3 c_{1}} x^{3}}{3}-\frac {k}{3} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 24
Order:=6; dsolve(x*diff(y(x),x)-3*y(x)=k,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{3} \left (1+\operatorname {O}\left (x^{6}\right )\right )+\left (-\frac {k}{3}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 15
AsymptoticDSolveValue[x*y'[x]-3*y[x]==k,y[x],{x,0,5}]
\[ y(x)\to -\frac {k}{3}+c_1 x^3 \]