Internal problem ID [2371]
Internal file name [OUTPUT/2371_Tuesday_February_27_2024_08_36_19_AM_27802779/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: 8.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "first order ode series method. Taylor series method"
Maple gives the following as the ode type
[`y=_G(x,y')`]
\[ \boxed {y^{\prime }-\sqrt {1+y x}=0} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}
With the expansion point for the power series method at \(x = 0\).
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 &= \sqrt {1+y x}\\ F_1 &= \frac {d F_0}{dx} \\ &= \frac {\partial F_0}{\partial x}+ \frac {\partial F_0}{\partial y} F_0 \\ &= \frac {x \sqrt {1+y x}+y}{2 \sqrt {1+y x}}\\ F_2 &= \frac {d F_1}{dx} \\ &= \frac {\partial F_1}{\partial x}+ \frac {\partial F_1}{\partial y} F_1 \\ &= \frac {3 y x \sqrt {1+y x}-y^{2}+4 \sqrt {1+y x}}{4 \left (1+y x \right )^{\frac {3}{2}}} \end {align*}
And so on. Evaluating all the above at initial conditions \(x \left (0\right ) = 0\) and \(y \left (0\right ) = 1\) gives \begin {align*} F_0 &= 1\\ F_1 &= {\frac {1}{2}}\\ F_2 &= {\frac {3}{4}} \end {align*}
Substituting all the above in (6) and simplifying gives the solution as \[ y = x +1+\frac {x^{2}}{4}+\frac {x^{3}}{8}+O\left (x^{4}\right ) \] 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 = x +1+\frac {1}{4} x^{2}+\frac {1}{8} x^{3} \end {align*}
Hence the solution can be written as \begin {gather*} y = x +1+\frac {x^{2}}{4}+\frac {x^{3}}{8}+O\left (x^{4}\right ) \end {gather*} which simplifies to \begin {gather*} y = x +1+\frac {x^{2}}{4}+\frac {x^{3}}{8}+O\left (x^{4}\right ) \end {gather*} Unable to also solve using normal power series since not linear ode. Not currently supported.
The solution(s) found are the following \begin{align*} \tag{1} y &= x +1+\frac {x^{2}}{4}+\frac {x^{3}}{8}+O\left (x^{4}\right ) \\ \end{align*}
Verification of solutions
\[ y = x +1+\frac {x^{2}}{4}+\frac {x^{3}}{8}+O\left (x^{4}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\sqrt {1+y x}, y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sqrt {1+y x} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying an equivalence to an Abel ODE trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = 5 trying symmetry patterns for 1st order ODEs -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying symmetry patterns of the forms [F(x),G(y)] and [G(y),F(x)] -> trying a symmetry pattern of the form [F(x),G(x)] -> trying a symmetry pattern of the form [F(y),G(y)] -> trying a symmetry pattern of the form [F(x)+G(y), 0] -> trying a symmetry pattern of the form [0, F(x)+G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] -> trying a symmetry pattern of conformal type`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 16
Order:=4; dsolve([diff(y(x),x)=sqrt(1+x*y(x)),y(0) = 1],y(x),type='series',x=0);
\[ y \left (x \right ) = 1+x +\frac {1}{4} x^{2}+\frac {1}{8} x^{3}+\operatorname {O}\left (x^{4}\right ) \]
✓ Solution by Mathematica
Time used: 0.032 (sec). Leaf size: 20
AsymptoticDSolveValue[{y'[x]==Sqrt[1+x*y[x]],{y[0]==1}},y[x],{x,0,3}]
\[ y(x)\to \frac {x^3}{8}+\frac {x^2}{4}+x+1 \]