Internal problem ID [14968]
Internal file name [OUTPUT/14978_Monday_April_15_2024_12_04_35_AM_14496990/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Section 3. The method of successive approximation. Exercises page 31
Problem number: 42.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_Riccati, _special]]
\[ \boxed {y^{\prime }-y^{2}=x} \] With initial conditions \begin {align*} [y \left (0\right ) = 0] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= y^{2}+x \end {align*}
The \(x\) domain of \(f(x,y)\) when \(y=0\) is \[
\{-\infty The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=0\) is \[
\{-\infty
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+x \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+x \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]
Shows that \(f_0(x)=x\), \(f_1(x)=0\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second
order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=x \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+x u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[
u \left (x \right ) = c_{1} \operatorname {AiryAi}\left (-x \right )+c_{2} \operatorname {AiryBi}\left (-x \right )
\] The above shows that \[
u^{\prime }\left (x \right ) = -c_{1} \operatorname {AiryAi}\left (1, -x \right )-c_{2} \operatorname {AiryBi}\left (1, -x \right )
\] Using the above in (1) gives the solution \[
y = -\frac {-c_{1} \operatorname {AiryAi}\left (1, -x \right )-c_{2} \operatorname {AiryBi}\left (1, -x \right )}{c_{1} \operatorname {AiryAi}\left (-x \right )+c_{2} \operatorname {AiryBi}\left (-x \right )}
\] Dividing both
numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following
solution
\[
y = \frac {c_{3} \operatorname {AiryAi}\left (1, -x \right )+\operatorname {AiryBi}\left (1, -x \right )}{c_{3} \operatorname {AiryAi}\left (-x \right )+\operatorname {AiryBi}\left (-x \right )}
\] Initial conditions are used to solve for \(c_{3}\). Substituting \(x=0\) and \(y=0\) in the above solution gives an
equation to solve for the constant of integration. \begin {align*} 0 = \frac {3 \Gamma \left (\frac {2}{3}\right )^{2} 3^{\frac {2}{3}}-3 \Gamma \left (\frac {2}{3}\right )^{2} c_{3} 3^{\frac {1}{6}}}{2 \,3^{\frac {5}{6}} \pi +2 \pi c_{3} 3^{\frac {1}{3}}} \end {align*}
The solutions are \begin {align*} c_{3} = \sqrt {3} \end {align*}
Trying the constant \begin {align*} c_{3} = \sqrt {3} \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {\operatorname {AiryAi}\left (1, -x \right ) \sqrt {3}+\operatorname {AiryBi}\left (1, -x \right )}{\operatorname {AiryAi}\left (-x \right ) \sqrt {3}+\operatorname {AiryBi}\left (-x \right )} \end {align*}
The constant \(c_{3} = \sqrt {3}\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {\operatorname {AiryAi}\left (1, -x \right ) \sqrt {3}+\operatorname {AiryBi}\left (1, -x \right )}{\operatorname {AiryAi}\left (-x \right ) \sqrt {3}+\operatorname {AiryBi}\left (-x \right )} \\
\end{align*} Verification of solutions
\[
y = \frac {\operatorname {AiryAi}\left (1, -x \right ) \sqrt {3}+\operatorname {AiryBi}\left (1, -x \right )}{\operatorname {AiryAi}\left (-x \right ) \sqrt {3}+\operatorname {AiryBi}\left (-x \right )}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y^{2}=x , y \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+x \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 35
\[
y \left (x \right ) = \frac {\sqrt {3}\, \operatorname {AiryAi}\left (1, -x \right )+\operatorname {AiryBi}\left (1, -x \right )}{\sqrt {3}\, \operatorname {AiryAi}\left (-x \right )+\operatorname {AiryBi}\left (-x \right )}
\]
✓ Solution by Mathematica
Time used: 1.269 (sec). Leaf size: 80
\[
y(x)\to -\frac {x^{3/2} \operatorname {BesselJ}\left (-\frac {4}{3},\frac {2 x^{3/2}}{3}\right )-x^{3/2} \operatorname {BesselJ}\left (\frac {2}{3},\frac {2 x^{3/2}}{3}\right )+\operatorname {BesselJ}\left (-\frac {1}{3},\frac {2 x^{3/2}}{3}\right )}{2 x \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2 x^{3/2}}{3}\right )}
\]
3.2.2 Solving as riccati ode
3.2.3 Maple step by step solution
`Methods for first order ODEs:
--- Trying classification methods ---
trying a quadrature
trying 1st order linear
trying Bernoulli
trying separable
trying inverse linear
trying homogeneous types:
trying Chini
differential order: 1; looking for linear symmetries
trying exact
Looking for potential symmetries
trying Riccati
trying Riccati Special
<- Riccati Special successful`
dsolve([diff(y(x),x)=x+y(x)^2,y(0) = 0],y(x), singsol=all)
DSolve[{y'[x]==x+y[x]^2,{y[0]==0}},y[x],x,IncludeSingularSolutions -> True]