Internal problem ID [1701]
Internal file name [OUTPUT/1702_Sunday_June_05_2022_02_28_06_AM_82219360/index.tex]
Book: Differential equations and their applications, 3rd ed., M. Braun
Section: Section 1.10. Page 80
Problem number: 8.
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]
\[ \boxed {y^{\prime }-y^{2}={\mathrm e}^{-t^{2}}} \]
In canonical form the ODE is \begin {align*} y' &= F(t,y)\\ &= {\mathrm e}^{-t^{2}}+y^{2} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = {\mathrm e}^{-t^{2}}+y^{2} \] With Riccati ODE standard form \[ y' = f_0(t)+ f_1(t)y+f_2(t)y^{2} \] Shows that \(f_0(t)={\mathrm e}^{-t^{2}}\), \(f_1(t)=0\) and \(f_2(t)=1\). 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''(t) -\left ( f_2' + f_1 f_2 \right ) u'(t) + f_2^2 f_0 u(t) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &={\mathrm e}^{-t^{2}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (t \right )+{\mathrm e}^{-t^{2}} u \left (t \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (t \right ) = \operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (t \right )+{\mathrm e}^{-t^{2}} \textit {\_Y} \left (t \right )\right \}, \left \{\textit {\_Y} \left (t \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (t \right ) = \frac {d}{d t}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (t \right )+{\mathrm e}^{-t^{2}} \textit {\_Y} \left (t \right )\right \}, \left \{\textit {\_Y} \left (t \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\frac {d}{d t}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (t \right )+{\mathrm e}^{-t^{2}} \textit {\_Y} \left (t \right )\right \}, \left \{\textit {\_Y} \left (t \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (t \right )+{\mathrm e}^{-t^{2}} \textit {\_Y} \left (t \right )\right \}, \left \{\textit {\_Y} \left (t \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\frac {d}{d t}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (t \right )+{\mathrm e}^{-t^{2}} \textit {\_Y} \left (t \right )\right \}, \left \{\textit {\_Y} \left (t \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (t \right )+{\mathrm e}^{-t^{2}} \textit {\_Y} \left (t \right )\right \}, \left \{\textit {\_Y} \left (t \right )\right \}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\frac {d}{d t}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (t \right )+{\mathrm e}^{-t^{2}} \textit {\_Y} \left (t \right )\right \}, \left \{\textit {\_Y} \left (t \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (t \right )+{\mathrm e}^{-t^{2}} \textit {\_Y} \left (t \right )\right \}, \left \{\textit {\_Y} \left (t \right )\right \}\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\frac {d}{d t}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (t \right )+{\mathrm e}^{-t^{2}} \textit {\_Y} \left (t \right )\right \}, \left \{\textit {\_Y} \left (t \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (t \right )+{\mathrm e}^{-t^{2}} \textit {\_Y} \left (t \right )\right \}, \left \{\textit {\_Y} \left (t \right )\right \}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}={\mathrm e}^{-t^{2}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{-t^{2}}+y^{2} \end {array} \]
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 Special trying inverse_Riccati trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 4 `, `-> Computing symmetries using: way = 2 `, `-> Computing symmetries using: way = 6 trying symmetry patterns for 1st order ODEs -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying symmetry patterns of the forms [F(x),G(y)] and [G(y),F(x)] -> trying a symmetry pattern of the form [F(x),G(x)] -> trying a symmetry pattern of the form [F(y),G(y)] -> trying a symmetry pattern of the form [F(x)+G(y), 0] -> trying a symmetry pattern of the form [0, F(x)+G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] -> trying a symmetry pattern of conformal type`
✗ Solution by Maple
dsolve(diff(y(t),t)= exp(-t^2)+y(t)^2,y(t), singsol=all)
\[ \text {No solution found} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[t]== Exp[-t^2]+y[t]^2,y[t],t,IncludeSingularSolutions -> True]
Not solved