Internal problem ID [14953]
Internal file name [OUTPUT/14963_Monday_April_15_2024_12_04_23_AM_63877819/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 2. The method of isoclines. Exercises page 27
Problem number: 27.
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]
\[ \boxed {y^{2}+y^{\prime }=x^{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= x^{2}-y^{2} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = x^{2}-y^{2} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x^{2}\), \(f_1(x)=0\) 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^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -u^{\prime \prime }\left (x \right )+x^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (\operatorname {BesselI}\left (\frac {1}{4}, \frac {x^{2}}{2}\right ) c_{1} +\operatorname {BesselK}\left (\frac {1}{4}, \frac {x^{2}}{2}\right ) c_{2} \right ) \sqrt {x} \] The above shows that \[ u^{\prime }\left (x \right ) = x^{\frac {3}{2}} \left (\operatorname {BesselI}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_{1} -\operatorname {BesselK}\left (\frac {3}{4}, \frac {x^{2}}{2}\right ) c_{2} \right ) \] Using the above in (1) gives the solution \[ y = \frac {x \left (\operatorname {BesselI}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_{1} -\operatorname {BesselK}\left (\frac {3}{4}, \frac {x^{2}}{2}\right ) c_{2} \right )}{\operatorname {BesselI}\left (\frac {1}{4}, \frac {x^{2}}{2}\right ) c_{1} +\operatorname {BesselK}\left (\frac {1}{4}, \frac {x^{2}}{2}\right ) c_{2}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {x \left (\operatorname {BesselI}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_{3} -\operatorname {BesselK}\left (\frac {3}{4}, \frac {x^{2}}{2}\right )\right )}{\operatorname {BesselI}\left (\frac {1}{4}, \frac {x^{2}}{2}\right ) c_{3} +\operatorname {BesselK}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x \left (\operatorname {BesselI}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_{3} -\operatorname {BesselK}\left (\frac {3}{4}, \frac {x^{2}}{2}\right )\right )}{\operatorname {BesselI}\left (\frac {1}{4}, \frac {x^{2}}{2}\right ) c_{3} +\operatorname {BesselK}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {x \left (\operatorname {BesselI}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_{3} -\operatorname {BesselK}\left (\frac {3}{4}, \frac {x^{2}}{2}\right )\right )}{\operatorname {BesselI}\left (\frac {1}{4}, \frac {x^{2}}{2}\right ) c_{3} +\operatorname {BesselK}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{2}+y^{\prime }=x^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-y^{2}+x^{2} \end {array} \]
Maple trace
`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`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 44
dsolve(diff(y(x),x)=x^2-y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x \left (\operatorname {BesselI}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_{1} -\operatorname {BesselK}\left (\frac {3}{4}, \frac {x^{2}}{2}\right )\right )}{c_{1} \operatorname {BesselI}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+\operatorname {BesselK}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )} \]
✓ Solution by Mathematica
Time used: 0.114 (sec). Leaf size: 197
DSolve[y'[x]==x^2-y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {-i x^2 \left (2 \operatorname {BesselJ}\left (-\frac {3}{4},\frac {i x^2}{2}\right )+c_1 \left (\operatorname {BesselJ}\left (-\frac {5}{4},\frac {i x^2}{2}\right )-\operatorname {BesselJ}\left (\frac {3}{4},\frac {i x^2}{2}\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{4},\frac {i x^2}{2}\right )}{2 x \left (\operatorname {BesselJ}\left (\frac {1}{4},\frac {i x^2}{2}\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{4},\frac {i x^2}{2}\right )\right )} \\ y(x)\to \frac {i x^2 \operatorname {BesselJ}\left (-\frac {5}{4},\frac {i x^2}{2}\right )-i x^2 \operatorname {BesselJ}\left (\frac {3}{4},\frac {i x^2}{2}\right )+\operatorname {BesselJ}\left (-\frac {1}{4},\frac {i x^2}{2}\right )}{2 x \operatorname {BesselJ}\left (-\frac {1}{4},\frac {i x^2}{2}\right )} \\ \end{align*}