1.11 problem 11
Internal
problem
ID
[7975]
Book
:
First
order
enumerated
odes
Section
:
section
1
Problem
number
:
11
Date
solved
:
Monday, October 21, 2024 at 04:40:08 PM
CAS
classification
:
[[_Riccati, _special]]
Solve
\begin{align*} y^{\prime }&=a x +b y^{2} \end{align*}
1.11.1 Solved as first order ode of type Riccati
Time used: 0.095 (sec)
In canonical form the ODE is
\begin{align*} y' &= F(x,y)\\ &= b \,y^{2}+a x \end{align*}
This is a Riccati ODE. Comparing the ODE to solve
\[ y' = b \,y^{2}+a x \]
With Riccati ODE standard form
\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]
Shows that \(f_0(x)=a x\), \(f_1(x)=0\) and \(f_2(x)=b\). Let
\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{u b} \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 &=b^{2} a x \end{align*}
Substituting the above terms back in equation (2) gives
\begin{align*} b u^{\prime \prime }\left (x \right )+b^{2} a x u \left (x \right ) = 0 \end{align*}
This is Airy ODE. It has the general form
\[ a \frac {d^{2}u}{d x^{2}} + b \frac {d u}{d x} + c u x = F(x) \]
Where in this case
\begin{align*} a &= b\\ b &= 0\\ c &= b^{2} a\\ F &= 0 \end{align*}
Therefore the solution to the homogeneous Airy ODE becomes
\[
u = c_1 \operatorname {AiryAi}\left (-\left (a b \right )^{{1}/{3}} x \right )+c_2 \operatorname {AiryBi}\left (-\left (a b \right )^{{1}/{3}} x \right )
\]
Will add steps showing
solving for IC soon.
Taking derivative gives
\[
u^{\prime }\left (x \right ) = -c_1 \left (a b \right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\left (a b \right )^{{1}/{3}} x \right )-c_2 \left (a b \right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\left (a b \right )^{{1}/{3}} x \right )
\]
Doing change of constants, the solution becomes
\[
y = -\frac {-c_3 \left (a b \right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\left (a b \right )^{{1}/{3}} x \right )-\left (a b \right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\left (a b \right )^{{1}/{3}} x \right )}{b \left (c_3 \operatorname {AiryAi}\left (-\left (a b \right )^{{1}/{3}} x \right )+\operatorname {AiryBi}\left (-\left (a b \right )^{{1}/{3}} x \right )\right )}
\]
1.11.2 Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }=a x +b y^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a x +b y^{2} \end {array} \]
1.11.3 Maple trace
Methods for first order ODEs:
1.11.4 Maple dsolve solution
Solving time : 0.001
(sec)
Leaf size : 59
dsolve(diff(y(x),x) = a*x+b*y(x)^2,
y(x),singsol=all)
\[
y = \frac {\left (a b \right )^{{1}/{3}} \left (\operatorname {AiryAi}\left (1, -\left (a b \right )^{{1}/{3}} x \right ) c_1 +\operatorname {AiryBi}\left (1, -\left (a b \right )^{{1}/{3}} x \right )\right )}{b \left (c_1 \operatorname {AiryAi}\left (-\left (a b \right )^{{1}/{3}} x \right )+\operatorname {AiryBi}\left (-\left (a b \right )^{{1}/{3}} x \right )\right )}
\]
1.11.5 Mathematica DSolve solution
Solving time : 0.156
(sec)
Leaf size : 331
DSolve[{D[y[x],x]==a*x+b*y[x]^2,{}},
y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to \frac {\sqrt {a} \sqrt {b} x^{3/2} \left (-2 \operatorname {BesselJ}\left (-\frac {2}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )+c_1 \left (\operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )-\operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )}{2 b x \left (\operatorname {BesselJ}\left (\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )\right )} \\
y(x)\to -\frac {\sqrt {a} \sqrt {b} x^{3/2} \operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )-\sqrt {a} \sqrt {b} x^{3/2} \operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )+\operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )}{2 b x \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )} \\
\end{align*}