Internal problem ID [3325]
Internal file name [OUTPUT/2817_Sunday_June_05_2022_08_41_04_AM_83860826/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 3
Problem number: 61.
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]
\[ \boxed {y^{\prime }-a \left (x -y\right ) y=1} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a x y -a \,y^{2}+1 \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a x y -a \,y^{2}+1 \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=1\), \(f_1(x)=a x\) and \(f_2(x)=-a\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-a 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 &=-a^{2} x\\ f_2^2 f_0 &=a^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -a u^{\prime \prime }\left (x \right )+a^{2} x u^{\prime }\left (x \right )+a^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {a \,x^{2}}{2}} \left (c_{1} \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right )+c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {a x \sqrt {\pi }\, \left (c_{1} \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right )+c_{2} \right ) {\mathrm e}^{\frac {a \,x^{2}}{2}}+\sqrt {a}\, \sqrt {2}\, c_{1}}{\sqrt {\pi }} \] Using the above in (1) gives the solution \[ y = \frac {\left (a x \sqrt {\pi }\, \left (c_{1} \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right )+c_{2} \right ) {\mathrm e}^{\frac {a \,x^{2}}{2}}+\sqrt {a}\, \sqrt {2}\, c_{1} \right ) {\mathrm e}^{-\frac {a \,x^{2}}{2}}}{\sqrt {\pi }\, a \left (c_{1} \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right )+c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\sqrt {a}\, \sqrt {2}\, c_{3} {\mathrm e}^{-\frac {a \,x^{2}}{2}}+\sqrt {\pi }\, a x \left (c_{3} \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right )+1\right )}{\sqrt {\pi }\, a \left (c_{3} \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right )+1\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sqrt {a}\, \sqrt {2}\, c_{3} {\mathrm e}^{-\frac {a \,x^{2}}{2}}+\sqrt {\pi }\, a x \left (c_{3} \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right )+1\right )}{\sqrt {\pi }\, a \left (c_{3} \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right )+1\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\sqrt {a}\, \sqrt {2}\, c_{3} {\mathrm e}^{-\frac {a \,x^{2}}{2}}+\sqrt {\pi }\, a x \left (c_{3} \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right )+1\right )}{\sqrt {\pi }\, a \left (c_{3} \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right )+1\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \left (x -y\right ) y=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=1+a \left (x -y\right ) y \end {array} \]
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 found: 2 potential symmetries. Proceeding with integration step`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 73
dsolve(diff(y(x),x) = 1+a*(x-y(x))*y(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {2 a^{\frac {3}{2}} c_{1} x +\sqrt {2}\, \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right ) \sqrt {\pi }\, a x +2 \sqrt {a}\, {\mathrm e}^{-\frac {a \,x^{2}}{2}}}{a \left (\sqrt {\pi }\, \sqrt {2}\, \operatorname {erf}\left (\frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right )+2 c_{1} \sqrt {a}\right )} \]
✓ Solution by Mathematica
Time used: 2.067 (sec). Leaf size: 93
DSolve[y'[x]==1+a(x-y[x])y[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sqrt {2 \pi } c_1 x \text {erf}\left (\frac {\sqrt {a} x}{\sqrt {2}}\right )+\frac {2 \left (a x+c_1 e^{-\frac {a x^2}{2}}\right )}{\sqrt {a}}}{2 \sqrt {a}+\sqrt {2 \pi } c_1 \text {erf}\left (\frac {\sqrt {a} x}{\sqrt {2}}\right )} \\ y(x)\to x \\ \end{align*}