Internal problem ID [14792]
Internal file name [OUTPUT/14472_Monday_April_08_2024_06_25_36_AM_50735726/index.tex
]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.8, page 203
Problem number: 19.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_y", "second order series method. Ordinary point", "second order series method. Taylor series method"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {y^{\prime \prime }+y^{\prime } x=\sin \left (x \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 1, y^{\prime }\left (0\right ) = 0] \end {align*}
With the expansion point for the power series method at \(x = 0\).
This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(x)y^{\prime } + q(x) y &= F \end {align*}
Where here \begin {align*} p(x) &=x\\ q(x) &=0\\ F &=\sin \left (x \right ) \end {align*}
Hence the ode is \begin {align*} y^{\prime \prime }+y^{\prime } x = \sin \left (x \right ) \end {align*}
The domain of \(p(x)=x\) is \[
\{-\infty Solving ode using Taylor series method. This gives review on how the Taylor series method
works for solving second order ode.
Let \[ y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \] Assuming expansion is at \(x_{0}=0\) (we can always shift the actual expansion point to \(0\) by change
of variables) and assuming \(f\left ( x,y,y^{\prime }\right ) \) is analytic at \(x_{0}\) which must be the case for an ordinary point. Let
initial conditions be \(y\left ( x_{0}\right ) =y_{0}\) and \(y^{\prime }\left ( x_{0}\right ) =y_{0}^{\prime }\). Using Taylor series gives\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 \prime }\left ( x_{0}\right ) +\cdots \\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}\left . f\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\frac {x^{3}}{3!}\left . f^{\prime }\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\cdots \\ & =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . \frac {d^{n}f}{dx^{n}}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \end {align*}
But \begin {align} \frac {df}{dx} & =\frac {\partial f}{\partial x}\frac {dx}{dx}+\frac {\partial f}{\partial y}\frac {dy}{dx}+\frac {\partial f}{\partial y^{\prime }}\frac {dy^{\prime }}{dx}\tag {1}\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\\ \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 ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\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 ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {d^{2}f}{dx^{2}}\right ) f\tag {3}\\ & \vdots \nonumber \end {align}
And so on. Hence if we name \(F_{0}=f\left ( x,y,y^{\prime }\right ) \) then the above can be written as \begin {align} F_{0} & =f\left ( x,y,y^{\prime }\right ) \tag {4}\\ F_{1} & =\frac {df}{dx}\nonumber \\ & =\frac {dF_{0}}{dx}\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\tag {5}\\ & =\frac {\partial F_{0}}{\partial x}+\frac {\partial F_{0}}{\partial y}y^{\prime }+\frac {\partial F_{0}}{\partial y^{\prime }}F_{0}\nonumber \\ F_{2} & =\frac {d}{dx}\left ( \frac {d}{dx}f\right ) \nonumber \\ & =\frac {d}{dx}\left ( F_{1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) F_{0}\nonumber \\ & \vdots \nonumber \\ 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 ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) F_{0} \tag {6} \end {align}
Therefore (6) can be used from now on along with \begin {equation} y\left ( x\right ) =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \tag {7} \end {equation} To find \(y\left ( x\right ) \) series solution around \(x=0\). Hence
\begin {align*} F_0 &= -y^{\prime } x +\sin \left (x \right )\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_{0}}{\partial x}+ \frac {\partial F_{0}}{\partial y} y^{\prime }+ \frac {\partial F_{0}}{\partial y^{\prime }} F_0 \\ &= x^{2} y^{\prime }-\sin \left (x \right ) x -y^{\prime }+\cos \left (x \right )\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_{1}}{\partial x}+ \frac {\partial F_{1}}{\partial y} y^{\prime }+ \frac {\partial F_{1}}{\partial y^{\prime }} F_1 \\ &= \left (-x^{3}+3 x \right ) y^{\prime }+x^{2} \sin \left (x \right )-\cos \left (x \right ) x -3 \sin \left (x \right )\\ F_3 &= \frac {d F_2}{dx} \\ &= \frac {\partial F_{2}}{\partial x}+ \frac {\partial F_{2}}{\partial y} y^{\prime }+ \frac {\partial F_{2}}{\partial y^{\prime }} F_2 \\ &= \left (x^{4}-6 x^{2}+3\right ) y^{\prime }-x^{3} \sin \left (x \right )+\cos \left (x \right ) x^{2}+6 \sin \left (x \right ) x -4 \cos \left (x \right )\\ F_4 &= \frac {d F_3}{dx} \\ &= \frac {\partial F_{3}}{\partial x}+ \frac {\partial F_{3}}{\partial y} y^{\prime }+ \frac {\partial F_{3}}{\partial y^{\prime }} F_3 \\ &= \left (-x^{5}+10 x^{3}-15 x \right ) y^{\prime }+\left (x^{4}-10 x^{2}+13\right ) \sin \left (x \right )-\cos \left (x \right ) x \left (x^{2}-8\right ) \end {align*}
And so on. Evaluating all the above at initial conditions \(x = 0\) and \(y \left (0\right ) = 1\) and \(y^{\prime }\left (0\right ) = 0\) gives \begin {align*} F_0 &= 0\\ F_1 &= 1\\ F_2 &= 0\\ F_3 &= -4\\ F_4 &= 0 \end {align*}
Substituting all the above in (7) and simplifying gives the solution as \[
y = 1+\frac {x^{3}}{6}-\frac {x^{5}}{30}+O\left (x^{6}\right )
\] \[
y = 1+\frac {x^{3}}{6}-\frac {x^{5}}{30}+O\left (x^{6}\right )
\] Since the expansion
point \(x = 0\) is an ordinary, we can also solve this using standard power series Let the solution be
represented as power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n} \] Then \begin {align*} y^{\prime } &= \moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} \end {align*}
Substituting the above back into the ode gives \begin {align*} \moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} = -\left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right ) x +\sin \left (x \right )\tag {1} \end {align*}
Expanding \(\sin \left (x \right )\) as Taylor series around \(x=0\) and keeping only the first \(6\) terms gives \begin {align*} \sin \left (x \right ) &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5} + \dots \\ &= x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5} \end {align*}
Hence the ODE in Eq (1) becomes \[
\left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n a_{n} x^{n -1}\right ) x = x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}
\] Which simplifies to \begin{equation}
\tag{2} \left (\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n \,x^{n} a_{n}\right ) = x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}
\end{equation} The next step is to make all powers
of \(x\) be \(n\) in each summation term. Going over each summation term above with power of \(x\) in it
which is not already \(x^{n}\) and adjusting the power and the corresponding index gives \begin{align*}
\moverset {\infty }{\munderset {n =2}{\sum }}n \left (n -1\right ) a_{n} x^{n -2} &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n} \\
\end{align*}
Substituting all the above in Eq (2) gives the following equation where now all powers of \(x\)
are the same and equal to \(n\). \begin{equation}
\tag{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +2\right ) a_{n +2} \left (n +1\right ) x^{n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}n \,x^{n} a_{n}\right ) = x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}
\end{equation} For \(1\le n\), the recurrence equation is \begin{equation}
\tag{4} \left (\left (n +2\right ) a_{n +2} \left (n +1\right )+n a_{n}\right ) x^{n} = x -\frac {1}{6} x^{3}+\frac {1}{120} x^{5}
\end{equation} For \(n = 1\) the recurrence
equation gives \begin{align*}
\left (6 a_{3}+a_{1}\right ) x&=x \\
6 a_{3}+a_{1} &= 1 \\
\end{align*} Which after substituting the earlier terms found becomes \[
a_{3} = \frac {1}{6}-\frac {a_{1}}{6}
\] For \(n = 2\)
the recurrence equation gives \begin{align*}
\left (12 a_{4}+2 a_{2}\right ) x^{2}&=0 \\
12 a_{4}+2 a_{2} &= 0 \\
\end{align*} Which after substituting the earlier terms found
becomes \[
a_{4} = 0
\] For \(n = 3\) the recurrence equation gives \begin{align*}
\left (20 a_{5}+3 a_{3}\right ) x^{3}&=-\frac {x^{3}}{6} \\
20 a_{5}+3 a_{3} &= -{\frac {1}{6}} \\
\end{align*} Which after substituting the earlier
terms found becomes \[
a_{5} = -\frac {1}{30}+\frac {a_{1}}{40}
\] For \(n = 4\) the recurrence equation gives \begin{align*}
\left (30 a_{6}+4 a_{4}\right ) x^{4}&=0 \\
30 a_{6}+4 a_{4} &= 0 \\
\end{align*} Which after substituting
the earlier terms found becomes \[
a_{6} = 0
\] For \(n = 5\) the recurrence equation gives \begin{align*}
\left (42 a_{7}+5 a_{5}\right ) x^{5}&=\frac {x^{5}}{120} \\
42 a_{7}+5 a_{5} &= {\frac {1}{120}} \\
\end{align*} Which after
substituting the earlier terms found becomes \[
a_{7} = \frac {1}{240}-\frac {a_{1}}{336}
\] And so on. Therefore the solution is
\begin {align*} y &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ &= a_{3} x^{3}+a_{2} x^{2}+a_{1} x +a_{0} + \dots \end {align*}
Substituting the values for \(a_{n}\) found above, the solution becomes \[
y = a_{0}+a_{1} x +\left (\frac {1}{6}-\frac {a_{1}}{6}\right ) x^{3}+\left (-\frac {1}{30}+\frac {a_{1}}{40}\right ) x^{5}+\dots
\]
Collecting terms, the solution becomes \begin{equation}
\tag{3} y = a_{0}+\left (x -\frac {1}{6} x^{3}+\frac {1}{40} x^{5}\right ) a_{1}+\frac {x^{3}}{6}-\frac {x^{5}}{30}+O\left (x^{6}\right )
\end{equation} At \(x = 0\) the solution above becomes \[
y = c_{1} +\left (x -\frac {1}{6} x^{3}+\frac {1}{40} x^{5}\right ) c_{2} +\frac {x^{3}}{6}-\frac {x^{5}}{30}+O\left (x^{6}\right )
\] \[
y = 1+\frac {x^{3}}{6}-\frac {x^{5}}{30}+O\left (x^{6}\right )
\]
The solution(s) found are the following \begin{align*}
\tag{1} y &= 1+\frac {x^{3}}{6}-\frac {x^{5}}{30}+O\left (x^{6}\right ) \\
\tag{2} y &= 1+\frac {x^{3}}{6}-\frac {x^{5}}{30}+O\left (x^{6}\right ) \\
\end{align*} Verification of solutions
\[
y = 1+\frac {x^{3}}{6}-\frac {x^{5}}{30}+O\left (x^{6}\right )
\] Verified OK.
\[
y = 1+\frac {x^{3}}{6}-\frac {x^{5}}{30}+O\left (x^{6}\right )
\] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }=-y^{\prime } x +\sin \left (x \right ), y \left (0\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}0\right \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-u \left (x \right ) x +\sin \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-u \left (x \right ) x +\sin \left (x \right ) \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} u \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )+u \left (x \right ) x =\sin \left (x \right ) \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (x \right ) \\ {} & {} & \mu \left (x \right ) \left (u^{\prime }\left (x \right )+u \left (x \right ) x \right )=\mu \left (x \right ) \sin \left (x \right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d x}\left (u \left (x \right ) \mu \left (x \right )\right ) \\ {} & {} & \mu \left (x \right ) \left (u^{\prime }\left (x \right )+u \left (x \right ) x \right )=u^{\prime }\left (x \right ) \mu \left (x \right )+u \left (x \right ) \mu ^{\prime }\left (x \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (x \right ) \\ {} & {} & \mu ^{\prime }\left (x \right )=\mu \left (x \right ) x \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (x \right )={\mathrm e}^{\frac {x^{2}}{2}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \left (\frac {d}{d x}\left (u \left (x \right ) \mu \left (x \right )\right )\right )d x =\int \mu \left (x \right ) \sin \left (x \right )d x +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & u \left (x \right ) \mu \left (x \right )=\int \mu \left (x \right ) \sin \left (x \right )d x +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {\int \mu \left (x \right ) \sin \left (x \right )d x +c_{1}}{\mu \left (x \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (x \right )={\mathrm e}^{\frac {x^{2}}{2}} \\ {} & {} & u \left (x \right )=\frac {\int {\mathrm e}^{\frac {x^{2}}{2}} \sin \left (x \right )d x +c_{1}}{{\mathrm e}^{\frac {x^{2}}{2}}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & u \left (x \right )=\frac {-\frac {\sqrt {\pi }\, {\mathrm e}^{\frac {1}{2}} \sqrt {2}\, \mathrm {erf}\left (\frac {\mathrm {I} \sqrt {2}\, x}{2}-\frac {\sqrt {2}}{2}\right )}{4}+\frac {\sqrt {\pi }\, {\mathrm e}^{\frac {1}{2}} \sqrt {2}\, \mathrm {erf}\left (\frac {\mathrm {I} \sqrt {2}\, x}{2}+\frac {\sqrt {2}}{2}\right )}{4}+c_{1}}{{\mathrm e}^{\frac {x^{2}}{2}}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & u \left (x \right )=-\frac {\left (\sqrt {\pi }\, {\mathrm e}^{\frac {1}{2}} \sqrt {2}\, \mathrm {erf}\left (\frac {\sqrt {2}\, \left (\mathrm {I} x -1\right )}{2}\right )-\sqrt {\pi }\, {\mathrm e}^{\frac {1}{2}} \sqrt {2}\, \mathrm {erf}\left (\frac {\sqrt {2}\, \left (1+\mathrm {I} x \right )}{2}\right )-4 c_{1} \right ) {\mathrm e}^{-\frac {x^{2}}{2}}}{4} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {\left (\sqrt {\pi }\, {\mathrm e}^{\frac {1}{2}} \sqrt {2}\, \mathrm {erf}\left (\frac {\sqrt {2}\, \left (\mathrm {I} x -1\right )}{2}\right )-\sqrt {\pi }\, {\mathrm e}^{\frac {1}{2}} \sqrt {2}\, \mathrm {erf}\left (\frac {\sqrt {2}\, \left (1+\mathrm {I} x \right )}{2}\right )-4 c_{1} \right ) {\mathrm e}^{-\frac {x^{2}}{2}}}{4} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {\left (\sqrt {\pi }\, {\mathrm e}^{\frac {1}{2}} \sqrt {2}\, \mathrm {erf}\left (\frac {\sqrt {2}\, \left (\mathrm {I} x -1\right )}{2}\right )-\sqrt {\pi }\, {\mathrm e}^{\frac {1}{2}} \sqrt {2}\, \mathrm {erf}\left (\frac {\sqrt {2}\, \left (1+\mathrm {I} x \right )}{2}\right )-4 c_{1} \right ) {\mathrm e}^{-\frac {x^{2}}{2}}}{4} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {\left (\sqrt {\pi }\, {\mathrm e}^{\frac {1}{2}} \sqrt {2}\, \mathrm {erf}\left (\frac {\sqrt {2}\, \left (\mathrm {I} x -1\right )}{2}\right )-\sqrt {\pi }\, {\mathrm e}^{\frac {1}{2}} \sqrt {2}\, \mathrm {erf}\left (\frac {\sqrt {2}\, \left (1+\mathrm {I} x \right )}{2}\right )-4 c_{1} \right ) {\mathrm e}^{-\frac {x^{2}}{2}}}{4}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=\int -\frac {\left (\sqrt {\pi }\, {\mathrm e}^{\frac {1}{2}} \sqrt {2}\, \mathrm {erf}\left (\frac {\sqrt {2}\, \left (\mathrm {I} x -1\right )}{2}\right )-\sqrt {\pi }\, {\mathrm e}^{\frac {1}{2}} \sqrt {2}\, \mathrm {erf}\left (\frac {\sqrt {2}\, \left (1+\mathrm {I} x \right )}{2}\right )-4 c_{1} \right ) {\mathrm e}^{-\frac {x^{2}}{2}}}{4}d x +c_{2} \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 14
\[
y \left (x \right ) = 1+\frac {1}{6} x^{3}-\frac {1}{30} x^{5}+\operatorname {O}\left (x^{6}\right )
\]
✓ Solution by Mathematica
Time used: 0.03 (sec). Leaf size: 19
\[
y(x)\to -\frac {7 x^5}{120}+\frac {x^3}{6}+1
\]
16.17.2 Maple step by step solution
`Methods for second order ODEs:
--- Trying classification methods ---
trying a quadrature
trying high order exact linear fully integrable
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = -_b(_a)*_a+sin(_a), _b(_a)` *** Sublevel 2 ***
Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
<- 1st order linear successful
<- high order exact linear fully integrable successful`
Order:=6;
dsolve([diff(y(x),x$2)+x*diff(y(x),x)=sin(x),y(0) = 1, D(y)(0) = 0],y(x),type='series',x=0);
AsymptoticDSolveValue[{y''[x]+2*x*y'[x]==Sin[x],{y[0]==1,y'[0]==0}},y[x],{x,0,5}]