3.6 problem 6

Internal problem ID [14129]
Internal file name [OUTPUT/13810_Saturday_March_02_2024_02_50_51_PM_27457795/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: 6.
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]

\[ \boxed {y^{\prime }+t y^{2}=4 t^{2}} \] With initial conditions \begin {align*} [y \left (2\right ) = 1] \end {align*}

3.6.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= -t \,y^{2}+4 t^{2} \end {align*}

The \(t\) domain of \(f(t,y)\) when \(y=1\) is \[ \{-\infty

The \(t\) domain of \(\frac {\partial f}{\partial y}\) when \(y=1\) is \[ \{-\infty

3.6.2 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(t,y)\\ &= -t \,y^{2}+4 t^{2} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = -t \,y^{2}+4 t^{2} \] With Riccati ODE standard form \[ y' = f_0(t)+ f_1(t)y+f_2(t)y^{2} \] Shows that \(f_0(t)=4 t^{2}\), \(f_1(t)=0\) and \(f_2(t)=-t\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-t 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' &=-1\\ f_1 f_2 &=0\\ f_2^2 f_0 &=4 t^{4} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} -t u^{\prime \prime }\left (t \right )+u^{\prime }\left (t \right )+4 t^{4} u \left (t \right ) &=0 \end {align*}

Solving the above ODE (this ode solved using Maple, not this program), gives

\[ u \left (t \right ) = t \left (c_{1} \operatorname {BesselI}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )+c_{2} \operatorname {BesselK}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )\right ) \] The above shows that \[ u^{\prime }\left (t \right ) = -2 t^{\frac {5}{2}} \left (\operatorname {BesselK}\left (\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) c_{2} -\operatorname {BesselI}\left (-\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) c_{1} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {2 \sqrt {t}\, \left (\operatorname {BesselK}\left (\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) c_{2} -\operatorname {BesselI}\left (-\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) c_{1} \right )}{c_{1} \operatorname {BesselI}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )+c_{2} \operatorname {BesselK}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = -\frac {2 \sqrt {t}\, \left (\operatorname {BesselK}\left (\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )-\operatorname {BesselI}\left (-\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) c_{3} \right )}{c_{3} \operatorname {BesselI}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )+\operatorname {BesselK}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )} \] Initial conditions are used to solve for \(c_{3}\). Substituting \(t=2\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = \frac {2 \sqrt {2}\, \operatorname {BesselI}\left (-\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) c_{3} -2 \sqrt {2}\, \operatorname {BesselK}\left (\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right )}{c_{3} \operatorname {BesselI}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right )+\operatorname {BesselK}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right )} \end {align*}

The solutions are \begin {align*} c_{3} = -\frac {2 \sqrt {2}\, \operatorname {BesselK}\left (\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right )+\operatorname {BesselK}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right )}{-2 \operatorname {BesselI}\left (-\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \sqrt {2}+\operatorname {BesselI}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right )} \end {align*}

Trying the constant \begin {align*} c_{3} = -\frac {2 \sqrt {2}\, \operatorname {BesselK}\left (\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right )+\operatorname {BesselK}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right )}{-2 \operatorname {BesselI}\left (-\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \sqrt {2}+\operatorname {BesselI}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right )} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=\frac {-4 \sqrt {t}\, \operatorname {BesselK}\left (\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (-\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) \sqrt {2}+4 \sqrt {t}\, \operatorname {BesselI}\left (-\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselK}\left (\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) \sqrt {2}-2 \sqrt {t}\, \operatorname {BesselI}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselK}\left (\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )-2 \sqrt {t}\, \operatorname {BesselK}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (-\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )}{-2 \operatorname {BesselK}\left (\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \sqrt {2}\, \operatorname {BesselI}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )-2 \operatorname {BesselI}\left (-\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \sqrt {2}\, \operatorname {BesselK}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )+\operatorname {BesselI}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselK}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )-\operatorname {BesselK}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )} \end {align*}

The constant \(c_{3} = -\frac {2 \sqrt {2}\, \operatorname {BesselK}\left (\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right )+\operatorname {BesselK}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right )}{-2 \operatorname {BesselI}\left (-\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \sqrt {2}+\operatorname {BesselI}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right )}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-4 \sqrt {t}\, \operatorname {BesselK}\left (\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (-\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) \sqrt {2}+4 \sqrt {t}\, \operatorname {BesselI}\left (-\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselK}\left (\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) \sqrt {2}-2 \sqrt {t}\, \operatorname {BesselI}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselK}\left (\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )-2 \sqrt {t}\, \operatorname {BesselK}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (-\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )}{-2 \operatorname {BesselK}\left (\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \sqrt {2}\, \operatorname {BesselI}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )-2 \operatorname {BesselI}\left (-\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \sqrt {2}\, \operatorname {BesselK}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )+\operatorname {BesselI}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselK}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )-\operatorname {BesselK}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {-4 \sqrt {t}\, \operatorname {BesselK}\left (\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (-\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) \sqrt {2}+4 \sqrt {t}\, \operatorname {BesselI}\left (-\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselK}\left (\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) \sqrt {2}-2 \sqrt {t}\, \operatorname {BesselI}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselK}\left (\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )-2 \sqrt {t}\, \operatorname {BesselK}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (-\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )}{-2 \operatorname {BesselK}\left (\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \sqrt {2}\, \operatorname {BesselI}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )-2 \operatorname {BesselI}\left (-\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \sqrt {2}\, \operatorname {BesselK}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )+\operatorname {BesselI}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselK}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )-\operatorname {BesselK}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )} \] Verified OK.

3.6.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+t y^{2}=4 t^{2}, y \left (2\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=4 t^{2}-t y^{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (2\right )=1 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]

Maple trace

`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: 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (diff(y(x), x))/x+4*y(x)*x^3, y(x)`      *** Sublevel 2 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a quadrature 
      checking if the LODE has constant coefficients 
      checking if the LODE is of Euler type 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      checking if the LODE is missing y 
      -> Trying a Liouvillian solution using Kovacics algorithm 
      <- No Liouvillian solutions exists 
      -> Trying a solution in terms of special functions: 
         -> Bessel 
         <- Bessel successful 
      <- special function solution successful 
   <- Riccati to 2nd Order successful`
 

✓ Solution by Maple

Time used: 0.219 (sec). Leaf size: 127

dsolve([diff(y(t),t)=4*t^2-t*y(t)^2,y(2) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = -\frac {2 \left (\left (2 \sqrt {2}\, \operatorname {BesselK}\left (\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right )+\operatorname {BesselK}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right )\right ) \operatorname {BesselI}\left (-\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )+\operatorname {BesselK}\left (\frac {3}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) \left (-2 \operatorname {BesselI}\left (-\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \sqrt {2}+\operatorname {BesselI}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right )\right )\right ) \sqrt {t}}{\left (-2 \sqrt {2}\, \operatorname {BesselK}\left (\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right )-\operatorname {BesselK}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right )\right ) \operatorname {BesselI}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right )+\operatorname {BesselK}\left (\frac {2}{5}, \frac {4 t^{\frac {5}{2}}}{5}\right ) \left (-2 \operatorname {BesselI}\left (-\frac {3}{5}, \frac {16 \sqrt {2}}{5}\right ) \sqrt {2}+\operatorname {BesselI}\left (\frac {2}{5}, \frac {16 \sqrt {2}}{5}\right )\right )} \]

✓ Solution by Mathematica

Time used: 0.268 (sec). Leaf size: 709

DSolve[{y'[t]==4*t^2-t*y[t]^2,{y[2]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {-3 t^{5/2} \operatorname {BesselI}\left (\frac {2}{5},\frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (\frac {3}{5},\frac {4 t^{5/2}}{5}\right )+4 \sqrt {2} t^{5/2} \operatorname {BesselI}\left (-\frac {3}{5},\frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (\frac {3}{5},\frac {4 t^{5/2}}{5}\right )-4 \sqrt {2} t^{5/2} \operatorname {BesselI}\left (\frac {3}{5},\frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (\frac {7}{5},\frac {4 t^{5/2}}{5}\right )+3 t^{5/2} \operatorname {BesselI}\left (-\frac {2}{5},\frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (\frac {7}{5},\frac {4 t^{5/2}}{5}\right )-4 \sqrt {2} t^{5/2} \operatorname {BesselI}\left (-\frac {7}{5},\frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (\frac {7}{5},\frac {4 t^{5/2}}{5}\right )+t^{5/2} \left (4 \sqrt {2} \operatorname {BesselI}\left (-\frac {3}{5},\frac {16 \sqrt {2}}{5}\right )-3 \operatorname {BesselI}\left (\frac {2}{5},\frac {16 \sqrt {2}}{5}\right )+4 \sqrt {2} \operatorname {BesselI}\left (\frac {7}{5},\frac {16 \sqrt {2}}{5}\right )\right ) \operatorname {BesselI}\left (-\frac {7}{5},\frac {4 t^{5/2}}{5}\right )+4 \sqrt {2} t^{5/2} \operatorname {BesselI}\left (\frac {7}{5},\frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (\frac {3}{5},\frac {4 t^{5/2}}{5}\right )+t^{5/2} \left (-4 \sqrt {2} \operatorname {BesselI}\left (-\frac {7}{5},\frac {16 \sqrt {2}}{5}\right )+3 \operatorname {BesselI}\left (-\frac {2}{5},\frac {16 \sqrt {2}}{5}\right )-4 \sqrt {2} \operatorname {BesselI}\left (\frac {3}{5},\frac {16 \sqrt {2}}{5}\right )\right ) \operatorname {BesselI}\left (-\frac {3}{5},\frac {4 t^{5/2}}{5}\right )+4 \sqrt {2} \operatorname {BesselI}\left (-\frac {3}{5},\frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (-\frac {2}{5},\frac {4 t^{5/2}}{5}\right )+3 \operatorname {BesselI}\left (-\frac {2}{5},\frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (\frac {2}{5},\frac {4 t^{5/2}}{5}\right )-4 \sqrt {2} \operatorname {BesselI}\left (-\frac {7}{5},\frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (\frac {2}{5},\frac {4 t^{5/2}}{5}\right )+4 \sqrt {2} \operatorname {BesselI}\left (\frac {7}{5},\frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (-\frac {2}{5},\frac {4 t^{5/2}}{5}\right )-4 \sqrt {2} \operatorname {BesselI}\left (\frac {3}{5},\frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (\frac {2}{5},\frac {4 t^{5/2}}{5}\right )-3 \operatorname {BesselI}\left (\frac {2}{5},\frac {16 \sqrt {2}}{5}\right ) \operatorname {BesselI}\left (-\frac {2}{5},\frac {4 t^{5/2}}{5}\right )}{t^2 \left (\left (4 \sqrt {2} \operatorname {BesselI}\left (-\frac {3}{5},\frac {16 \sqrt {2}}{5}\right )-3 \operatorname {BesselI}\left (\frac {2}{5},\frac {16 \sqrt {2}}{5}\right )+4 \sqrt {2} \operatorname {BesselI}\left (\frac {7}{5},\frac {16 \sqrt {2}}{5}\right )\right ) \operatorname {BesselI}\left (-\frac {2}{5},\frac {4 t^{5/2}}{5}\right )+\left (-4 \sqrt {2} \operatorname {BesselI}\left (-\frac {7}{5},\frac {16 \sqrt {2}}{5}\right )+3 \operatorname {BesselI}\left (-\frac {2}{5},\frac {16 \sqrt {2}}{5}\right )-4 \sqrt {2} \operatorname {BesselI}\left (\frac {3}{5},\frac {16 \sqrt {2}}{5}\right )\right ) \operatorname {BesselI}\left (\frac {2}{5},\frac {4 t^{5/2}}{5}\right )\right )} \]