Internal problem ID [7346]
Internal file name [OUTPUT/6327_Sunday_June_05_2022_04_40_15_PM_24901680/index.tex]
Book: First order enumerated odes
Section: section 1
Problem number: 30.
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]
\[ \boxed {y^{\prime }-y-b y^{2}=x} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= b \,y^{2}+x +y \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = b \,y^{2}+x +y \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x\), \(f_1(x)=1\) and \(f_2(x)=b\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{b 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 &=b\\ f_2^2 f_0 &=x \,b^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} b u^{\prime \prime }\left (x \right )-b u^{\prime }\left (x \right )+x \,b^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {x}{2}} \left (\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{1} +\operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -{\mathrm e}^{\frac {x}{2}} \left (b^{\frac {1}{3}} \operatorname {AiryAi}\left (1, -\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{1} +b^{\frac {1}{3}} \operatorname {AiryBi}\left (1, -\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{2} -\frac {\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{1}}{2}-\frac {\operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{2}}{2}\right ) \] Using the above in (1) gives the solution \[ y = \frac {b^{\frac {1}{3}} \operatorname {AiryAi}\left (1, -\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{1} +b^{\frac {1}{3}} \operatorname {AiryBi}\left (1, -\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{2} -\frac {\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{1}}{2}-\frac {\operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{2}}{2}}{b \left (\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{1} +\operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {2 b^{\frac {1}{3}} \operatorname {AiryAi}\left (1, -\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{3} -\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{3} +2 b^{\frac {1}{3}} \operatorname {AiryBi}\left (1, -\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right )-\operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right )}{2 b \left (\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{3} +\operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {2 b^{\frac {1}{3}} \operatorname {AiryAi}\left (1, -\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{3} -\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{3} +2 b^{\frac {1}{3}} \operatorname {AiryBi}\left (1, -\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right )-\operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right )}{2 b \left (\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{3} +\operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {2 b^{\frac {1}{3}} \operatorname {AiryAi}\left (1, -\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{3} -\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{3} +2 b^{\frac {1}{3}} \operatorname {AiryBi}\left (1, -\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right )-\operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right )}{2 b \left (\operatorname {AiryAi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right ) c_{3} +\operatorname {AiryBi}\left (-\frac {4 x b -1}{4 b^{\frac {2}{3}}}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y-b y^{2}=x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=x +y+b y^{2} \end {array} \]
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 sub-methods: <- Abel AIR successful: ODE belongs to the 0F1 0-parameter (Airy type) class`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 105
dsolve(diff(y(x),x)=x+y(x)+b*y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {2 b^{\frac {1}{3}} \operatorname {AiryAi}\left (1, -\frac {4 b x -1}{4 b^{\frac {2}{3}}}\right ) c_{1} +2 \operatorname {AiryBi}\left (1, -\frac {4 b x -1}{4 b^{\frac {2}{3}}}\right ) b^{\frac {1}{3}}-\operatorname {AiryAi}\left (-\frac {4 b x -1}{4 b^{\frac {2}{3}}}\right ) c_{1} -\operatorname {AiryBi}\left (-\frac {4 b x -1}{4 b^{\frac {2}{3}}}\right )}{2 b \left (\operatorname {AiryAi}\left (-\frac {4 b x -1}{4 b^{\frac {2}{3}}}\right ) c_{1} +\operatorname {AiryBi}\left (-\frac {4 b x -1}{4 b^{\frac {2}{3}}}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.222 (sec). Leaf size: 211
DSolve[y'[x]==x+y[x]+b*y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {-(-b)^{2/3} \operatorname {AiryBi}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )+2 b \operatorname {AiryBiPrime}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )+c_1 \left (2 b \operatorname {AiryAiPrime}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )-(-b)^{2/3} \operatorname {AiryAi}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )\right )}{2 (-b)^{5/3} \left (\operatorname {AiryBi}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )+c_1 \operatorname {AiryAi}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )\right )} \\ y(x)\to -\frac {\frac {2 \sqrt [3]{-b} \operatorname {AiryAiPrime}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )}{\operatorname {AiryAi}\left (\frac {\frac {1}{4}-b x}{(-b)^{2/3}}\right )}+1}{2 b} \\ \end{align*}