Internal problem ID [14124]
Internal file name [OUTPUT/13805_Saturday_March_02_2024_02_50_47_PM_53832926/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 2. First Order Equations. Exercises 2.1, page 32
Problem number: 1 (a).
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]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-y^{2}=-t^{2}} \] With initial conditions \begin {align*} [y \left (0\right ) = 0] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= -t^{2}+y^{2} \end {align*}
The \(t\) domain of \(f(t,y)\) when \(y=0\) is \[
\{-\infty The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(t=0\) is \[
\{-\infty
In canonical form the ODE is \begin {align*} y' &= F(t,y)\\ &= -t^{2}+y^{2} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -t^{2}+y^{2} \] With Riccati ODE standard form \[ y' = f_0(t)+ f_1(t)y+f_2(t)y^{2} \]
Shows that \(f_0(t)=-t^{2}\), \(f_1(t)=0\) 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 &=-t^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (t \right )-t^{2} u \left (t \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[
u \left (t \right ) = \left (\operatorname {BesselK}\left (\frac {1}{4}, \frac {t^{2}}{2}\right ) c_{2} +\operatorname {BesselI}\left (\frac {1}{4}, \frac {t^{2}}{2}\right ) c_{1} \right ) \sqrt {t}
\] The above shows that \[
u^{\prime }\left (t \right ) = \left (-\operatorname {BesselK}\left (\frac {3}{4}, \frac {t^{2}}{2}\right ) c_{2} +\operatorname {BesselI}\left (-\frac {3}{4}, \frac {t^{2}}{2}\right ) c_{1} \right ) t^{\frac {3}{2}}
\] Using the above in (1) gives the solution \[
y = -\frac {\left (-\operatorname {BesselK}\left (\frac {3}{4}, \frac {t^{2}}{2}\right ) c_{2} +\operatorname {BesselI}\left (-\frac {3}{4}, \frac {t^{2}}{2}\right ) c_{1} \right ) t}{\operatorname {BesselK}\left (\frac {1}{4}, \frac {t^{2}}{2}\right ) c_{2} +\operatorname {BesselI}\left (\frac {1}{4}, \frac {t^{2}}{2}\right ) c_{1}}
\] Dividing both
numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following
solution
\[
y = \frac {\left (-\operatorname {BesselI}\left (-\frac {3}{4}, \frac {t^{2}}{2}\right ) c_{3} +\operatorname {BesselK}\left (\frac {3}{4}, \frac {t^{2}}{2}\right )\right ) t}{\operatorname {BesselK}\left (\frac {1}{4}, \frac {t^{2}}{2}\right )+\operatorname {BesselI}\left (\frac {1}{4}, \frac {t^{2}}{2}\right ) c_{3}}
\] Initial conditions are used to solve for \(c_{3}\). Substituting \(t=0\) and \(y=0\) in the above solution gives an
equation to solve for the constant of integration.
Warning: Failed to find \(c_{3}\) using initial conditions. Solution could be wrong or there is no
solution that satisfies the given initial conditions.
Verification of solutions N/A
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y^{2}=-t^{2}, y \left (0\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-t^{2}+y^{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=0 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.859 (sec). Leaf size: 59
\[
y \left (t \right ) = -\left (\left \{\begin {array}{cc} 0 & t =0 \\ \frac {\left (\operatorname {BesselI}\left (-\frac {3}{4}, \frac {t^{2}}{2}\right ) \pi \sqrt {2}-2 \operatorname {BesselK}\left (\frac {3}{4}, \frac {t^{2}}{2}\right )\right ) t}{\pi \sqrt {2}\, \operatorname {BesselI}\left (\frac {1}{4}, \frac {t^{2}}{2}\right )+2 \operatorname {BesselK}\left (\frac {1}{4}, \frac {t^{2}}{2}\right )} & \operatorname {otherwise} \end {array}\right .\right )
\]
✓ Solution by Mathematica
Time used: 0.663 (sec). Leaf size: 81
\[
y(t)\to -\frac {i t^2 \operatorname {BesselJ}\left (-\frac {5}{4},\frac {i t^2}{2}\right )-i t^2 \operatorname {BesselJ}\left (\frac {3}{4},\frac {i t^2}{2}\right )+\operatorname {BesselJ}\left (-\frac {1}{4},\frac {i t^2}{2}\right )}{2 t \operatorname {BesselJ}\left (-\frac {1}{4},\frac {i t^2}{2}\right )}
\]
3.1.2 Solving as riccati ode
3.1.3 Maple step by step solution
`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`
dsolve([diff(y(t),t)+t^2=y(t)^2,y(0) = 0],y(t), singsol=all)
DSolve[{y'[t]+t^2==y[t]^2,{y[0]==0}},y[t],t,IncludeSingularSolutions -> True]