Internal
problem
ID
[7893]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
4.0
Problem
number
:
24
Date
solved
:
Monday, October 21, 2024 at 04:31:32 PM
CAS
classification
:
[[_2nd_order, _linear, _nonhomogeneous]]
Solve
Using series expansion around \(x=0\)
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE.
The following is summary of singularities for the above ode. Writing the ode as
Where
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be
Since this is an inhomogeneous, then let the solution be
Where \(y_h\) is the solution to the homogeneous ode \(2 x^{2} y^{\prime \prime }-x y^{\prime }+\left (-x^{2}+1\right ) y = 0\), and \(y_p\) is a particular solution to the inhomogeneous ode.which is found using the balance equation generated from indicial equation
First, we solve for \(y_h\) Let the solution be represented as Frobenius power series of the form
Then
Substituting the above back into the ode gives
Which simplifies to
The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) 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 \(n +r\).
The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives
When \(n = 0\) the above becomes
Or
Since \(a_{0}\neq 0\) then the above simplifies to
Since the above is true for all \(x\) then the indicial equation becomes
Solving for \(r\) gives the roots of the indicial equation as
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
This equation will used later to find the particular solution.
Since \(a_{0}\neq 0\) then the indicial equation becomes
Solving for \(r\) gives the roots of the indicial equation as \(\left [1, {\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = {\frac {1}{2}}\) is not an integer, then we can construct two linearly independent solutions
Or
We start by finding \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives
For \(2\le n\) the recursive equation is
Solving for \(a_{n}\) from recursive equation (4) gives
Which for the root \(r = 1\) becomes
At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) and after as more terms are found using the above recursive equation.
\(n\) | \(a_{n ,r}\) | \(a_{n}\) |
\(a_{0}\) | \(1\) | \(1\) |
\(a_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives
Which for the root \(r = 1\) becomes
And the table now becomes
\(n\) | \(a_{n ,r}\) | \(a_{n}\) |
\(a_{0}\) | \(1\) | \(1\) |
\(a_{1}\) | \(0\) | \(0\) |
\(a_{2}\) | \(\frac {1}{2 r^{2}+5 r +3}\) | \(\frac {1}{10}\) |
For \(n = 3\), using the above recursive equation gives
And the table now becomes
\(n\) | \(a_{n ,r}\) | \(a_{n}\) |
\(a_{0}\) | \(1\) | \(1\) |
\(a_{1}\) | \(0\) | \(0\) |
\(a_{2}\) | \(\frac {1}{2 r^{2}+5 r +3}\) | \(\frac {1}{10}\) |
\(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives
Which for the root \(r = 1\) becomes
And the table now becomes
\(n\) | \(a_{n ,r}\) | \(a_{n}\) |
\(a_{0}\) | \(1\) | \(1\) |
\(a_{1}\) | \(0\) | \(0\) |
\(a_{2}\) | \(\frac {1}{2 r^{2}+5 r +3}\) | \(\frac {1}{10}\) |
\(a_{3}\) | \(0\) | \(0\) |
\(a_{4}\) | \(\frac {1}{4 r^{4}+36 r^{3}+113 r^{2}+144 r +63}\) | \(\frac {1}{360}\) |
For \(n = 5\), using the above recursive equation gives
And the table now becomes
\(n\) | \(a_{n ,r}\) | \(a_{n}\) |
\(a_{0}\) | \(1\) | \(1\) |
\(a_{1}\) | \(0\) | \(0\) |
\(a_{2}\) | \(\frac {1}{2 r^{2}+5 r +3}\) | \(\frac {1}{10}\) |
\(a_{3}\) | \(0\) | \(0\) |
\(a_{4}\) | \(\frac {1}{4 r^{4}+36 r^{3}+113 r^{2}+144 r +63}\) | \(\frac {1}{360}\) |
\(a_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives
For \(2\le n\) the recursive equation is
Solving for \(b_{n}\) from recursive equation (4) gives
Which for the root \(r = {\frac {1}{2}}\) becomes
At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) and after as more terms are found using the above recursive equation.
\(n\) | \(b_{n ,r}\) | \(b_{n}\) |
\(b_{0}\) | \(1\) | \(1\) |
\(b_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives
Which for the root \(r = {\frac {1}{2}}\) becomes
And the table now becomes
\(n\) | \(b_{n ,r}\) | \(b_{n}\) |
\(b_{0}\) | \(1\) | \(1\) |
\(b_{1}\) | \(0\) | \(0\) |
\(b_{2}\) | \(\frac {1}{2 r^{2}+5 r +3}\) | \(\frac {1}{6}\) |
For \(n = 3\), using the above recursive equation gives
And the table now becomes
\(n\) | \(b_{n ,r}\) | \(b_{n}\) |
\(b_{0}\) | \(1\) | \(1\) |
\(b_{1}\) | \(0\) | \(0\) |
\(b_{2}\) | \(\frac {1}{2 r^{2}+5 r +3}\) | \(\frac {1}{6}\) |
\(b_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives
Which for the root \(r = {\frac {1}{2}}\) becomes
And the table now becomes
\(n\) | \(b_{n ,r}\) | \(b_{n}\) |
\(b_{0}\) | \(1\) | \(1\) |
\(b_{1}\) | \(0\) | \(0\) |
\(b_{2}\) | \(\frac {1}{2 r^{2}+5 r +3}\) | \(\frac {1}{6}\) |
\(b_{3}\) | \(0\) | \(0\) |
\(b_{4}\) | \(\frac {1}{4 r^{4}+36 r^{3}+113 r^{2}+144 r +63}\) | \(\frac {1}{168}\) |
For \(n = 5\), using the above recursive equation gives
And the table now becomes
\(n\) | \(b_{n ,r}\) | \(b_{n}\) |
\(b_{0}\) | \(1\) | \(1\) |
\(b_{1}\) | \(0\) | \(0\) |
\(b_{2}\) | \(\frac {1}{2 r^{2}+5 r +3}\) | \(\frac {1}{6}\) |
\(b_{3}\) | \(0\) | \(0\) |
\(b_{4}\) | \(\frac {1}{4 r^{4}+36 r^{3}+113 r^{2}+144 r +63}\) | \(\frac {1}{168}\) |
\(b_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is
Therefore the homogeneous solution is
The particular solution is found by solving for \(c,m\) the balance equation
Where \(F(x)\) is the RHS of the ode. If \(F(x)\) has more than one term, then this is done for each term one at a time and then all the particular solutions are added. The function \(F(x)\) will be converted to series if needed. in order to solve for \(c_n,m\) for each term, the same recursive relation used to find \(y_h(x)\) is used to find \(c_n,m\) which is used to find the particular solution \(\sum _{n=0} c_n x^{n+m}\) by replacing \(a_n\) by \(c_n\) and \(r\) by \(m\).
The following are the values of \(a_n\) found in terms of the indicial root \(r\).
\(a_{1} = 0\) |
\(a_{2} = \frac {a_{0}}{2 r^{2}+5 r +3}\) |
\(a_{3} = 0\) |
\(a_{4} = \frac {a_{0}}{\left (2 r^{2}+5 r +3\right ) \left (2 r^{2}+13 r +21\right )}\) |
\(a_{5} = 0\) |
Expanding the rhs of the ode \(\cos \left (x \right )\) in series gives
Since the \(F=1-\frac {1}{2} x^{2}+\frac {1}{24} x^{4}\) has more than one term then we find a particular solution for each term and add the result to find the particular solution to the ode.
Now we determine the particular solution \(y_p\) associated with \(F=1\) by solving the balance equation
For \(c_{0}\) and \(x\). This results in
The particular solution is therefore
Where in the above \(c_0 = 1\).
The remaining \(c_n\) values are found using the same recurrence relation given in the earlier table which was 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. By letting \(a_{0} = c_{0}\) or \(a_{0} = 1\) and \(r = m\) or \(r = 0\). The following table gives the resulting \(c_n\) values. These values will be used to find the particular solution. Values of \(c_n\) found not defined when doing the substitution will be discarded and not used
\(c_{0} = 1\) |
\(c_{1} = 0\) |
\(c_{2} = {\frac {1}{3}}\) |
\(c_{3} = 0\) |
\(c_{4} = {\frac {1}{63}}\) |
\(c_{5} = 0\) |
The particular solution is now found using
Using the values found above for \(c_n\) into the above sum gives
Now we determine the particular solution \(y_p\) associated with \(F=-\frac {x^{2}}{2}\) by solving the balance equation
For \(c_{0}\) and \(x\). This results in
The particular solution is therefore
Where in the above \(c_0 = -{\frac {1}{6}}\).
The remaining \(c_n\) values are found using the same recurrence relation given in the earlier table which was used to find the homogeneous solution but using \(c_0\) in place of \(a_{0}\) and using \(m=2\) in place of the root of the indicial equation used to find the homogeneous solution. By letting \(a_{0} = c_{0}\) or \(a_{0} = -{\frac {1}{6}}\) and \(r = m\) or \(r = 2\). The following table gives the resulting \(c_n\) values. These values will be used to find the particular solution. Values of \(c_n\) found not defined when doing the substitution will be discarded and not used
\(c_{0} = -{\frac {1}{6}}\) |
\(c_{1} = 0\) |
\(c_{2} = -{\frac {1}{126}}\) |
\(c_{3} = 0\) |
\(c_{4} = -{\frac {1}{6930}}\) |
\(c_{5} = 0\) |
The particular solution is now found using
Using the values found above for \(c_n\) into the above sum gives
Now we determine the particular solution \(y_p\) associated with \(F=\frac {x^{4}}{24}\) by solving the balance equation
For \(c_{0}\) and \(x\). This results in
The particular solution is therefore
Where in the above \(c_0 = {\frac {1}{504}}\).
The remaining \(c_n\) values are found using the same recurrence relation given in the earlier table which was used to find the homogeneous solution but using \(c_0\) in place of \(a_{0}\) and using \(m=4\) in place of the root of the indicial equation used to find the homogeneous solution. By letting \(a_{0} = c_{0}\) or \(a_{0} = {\frac {1}{504}}\) and \(r = m\) or \(r = 4\). The following table gives the resulting \(c_n\) values. These values will be used to find the particular solution. Values of \(c_n\) found not defined when doing the substitution will be discarded and not used
\(c_{0} = {\frac {1}{504}}\) |
\(c_{1} = 0\) |
\(c_{2} = {\frac {1}{27720}}\) |
\(c_{3} = 0\) |
\(c_{4} = {\frac {1}{2910600}}\) |
\(c_{5} = 0\) |
The particular solution is now found using
Using the values found above for \(c_n\) into the above sum gives
Adding all the above particular solution(s) gives
Truncating the particular solution to the order of series requested gives
Hence the final solution is
Methods for second order ODEs:
Solving time : 0.013
(sec)
Leaf size : 43
dsolve(2*x^2*diff(diff(y(x),x),x)-x*diff(y(x),x)+(-x^2+1)*y(x) = cos(x),y(x), series,x=0)
Solving time : 0.049
(sec)
Leaf size : 176
AsymptoticDSolveValue[{2*x^2*D[y[x],{x,2}]-x*D[y[x],x]+(1-x^2)*y[x]==Cos[x],{}}, y[x],{x,0,5}]