problem Problem 1(c)

Internal problem ID [12223]
Internal ﬁle name [OUTPUT/10876_Thursday_September_28_2023_01_05_42_AM_88747455/index.tex]

Book: APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section: Chapter 4, Second and Higher Order Linear Diﬀerential Equations. Problems page 221
Problem number: Problem 1(c).
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_ode_missing_x", "exact nonlinear second order ode"

2.3.1 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }+y y^{\prime }\right )d x &= \int 1d x\\ \frac {y^{2}}{2}+y^{\prime } = x + c_{1} \end {align*}

Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {y^{2}}{2}+x +c_{1} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {y^{2}}{2}+x +c_{1} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x +c_{1}\), \(f_1(x)=0\) and \(f_2(x)=-{\frac {1}{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {u}{2}} \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 &=\frac {x}{4}+\frac {c_{1}}{4} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} -\frac {u^{\prime \prime }\left (x \right )}{2}+\left (\frac {x}{4}+\frac {c_{1}}{4}\right ) u \left (x \right ) &=0 \end {align*}

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

\[ y = \frac {2^{\frac {2}{3}} \left (c_{4} \operatorname {AiryAi}\left (1, \frac {2^{\frac {2}{3}} \left (x +c_{1} \right )}{2}\right )+\operatorname {AiryBi}\left (1, \frac {2^{\frac {2}{3}} \left (x +c_{1} \right )}{2}\right )\right )}{c_{4} \operatorname {AiryAi}\left (\frac {2^{\frac {2}{3}} \left (x +c_{1} \right )}{2}\right )+\operatorname {AiryBi}\left (\frac {2^{\frac {2}{3}} \left (x +c_{1} \right )}{2}\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {2^{\frac {2}{3}} \left (c_{4} \operatorname {AiryAi}\left (1, \frac {2^{\frac {2}{3}} \left (x +c_{1} \right )}{2}\right )+\operatorname {AiryBi}\left (1, \frac {2^{\frac {2}{3}} \left (x +c_{1} \right )}{2}\right )\right )}{c_{4} \operatorname {AiryAi}\left (\frac {2^{\frac {2}{3}} \left (x +c_{1} \right )}{2}\right )+\operatorname {AiryBi}\left (\frac {2^{\frac {2}{3}} \left (x +c_{1} \right )}{2}\right )} \\ \end{align*}

Veriﬁcation of solutions

2.3.2 Solving as second order ode missing x ode

This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}

Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}

Hence the ode becomes \begin {align*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+y p \left (y \right ) = 1 \end {align*}

Which is now solved as ﬁrst order ode for \(p(y)\). Unable to determine ODE type.

2.3.3 Solving as type second_order_integrable_as_is (not using ABC version)

Writing the ode as \[ y^{\prime \prime }+y y^{\prime } = 1 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime \prime }+y y^{\prime }\right )d x &= \int 1d x\\ \frac {y^{2}}{2}+y^{\prime } = x +c_{1} \end {align*}

Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {y^{2}}{2}+x +c_{1} \end {align*}

But \begin {align*} f_2' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\frac {x}{4}+\frac {c_{1}}{4} \end {align*}

Summary

Veriﬁcation of solutions

2.3.4 Solving as exact nonlinear second order ode ode

An exact non-linear second order ode has the form \begin {align*} a_{2} \left (x , y, y^{\prime }\right ) y^{\prime \prime }+a_{1} \left (x , y, y^{\prime }\right ) y^{\prime }+a_{0} \left (x , y, y^{\prime }\right )&=0 \end {align*}

Where the following conditions are satisﬁed \begin {align*} \frac {\partial a_2}{\partial y} &= \frac {\partial a_1}{\partial y'}\\ \frac {\partial a_2}{\partial x} &= \frac {\partial a_0}{\partial y'}\\ \frac {\partial a_1}{\partial x} &= \frac {\partial a_0}{\partial y} \end {align*}

Looking at the the ode given we see that \begin {align*} a_2 &= 1\\ a_1 &= y\\ a_0 &= -1 \end {align*}

Applying the conditions to the above shows this is a nonlinear exact second order ode. Therefore it can be reduced to ﬁrst order ode given by \begin {align*} \int {a_2\,d y'} + \int {a_1\,d y} + \int {a_0\,d x} &= c_{1}\\ \int {1\,d y'} + \int {y\,d y} + \int {-1\,d x} &= c_{1} \end {align*}

Which results in \begin {align*} \frac {y^{2}}{2}-x +y^{\prime } = c_{1} \end {align*}

Which is now solved In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {y^{2}}{2}+x +c_{1} \end {align*}

But \begin {align*} f_2' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\frac {x}{4}+\frac {c_{1}}{4} \end {align*}

Summary

Veriﬁcation of solutions

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
trying 2nd order WeierstrassP 
trying 2nd order JacobiSN 
differential order: 2; trying a linearization to 3rd order 
trying 2nd order ODE linearizable_by_differentiation 
trying 2nd order, 2 integrating factors of the form mu(x,y) 
trying differential order: 2; missing variables 
`, `-> Computing symmetries using: way = 3 
`, `-> Computing symmetries using: way = exp_sym 
-> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+_b(_a)*_a-1 = 0, _b(_a)`   *** Sublevel 2 *** 
   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 
   trying Abel 
   <- Abel successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful`

\[ -2 \,2^{\frac {2}{3}} \left (\int _{}^{y \left (x \right )}\frac {1}{2^{\frac {2}{3}} \textit {\_a}^{2}-4 \operatorname {RootOf}\left (\operatorname {AiryBi}\left (\textit {\_Z} \right ) 2^{\frac {1}{3}} c_{1} \textit {\_a} +2^{\frac {1}{3}} \textit {\_a} \operatorname {AiryAi}\left (\textit {\_Z} \right )-2 \operatorname {AiryBi}\left (1, \textit {\_Z}\right ) c_{1} -2 \operatorname {AiryAi}\left (1, \textit {\_Z}\right )\right )}d \textit {\_a} \right )-x -c_{2} = 0 \]

\[ y(x)\to \frac {2^{2/3} \left (c_2 \operatorname {AiryAiPrime}\left (\frac {x-c_1}{\sqrt [3]{2}}\right )+\operatorname {AiryBiPrime}\left (\frac {x-c_1}{\sqrt [3]{2}}\right )\right )}{c_2 \operatorname {AiryAi}\left (\frac {x-c_1}{\sqrt [3]{2}}\right )+\operatorname {AiryBi}\left (\frac {x-c_1}{\sqrt [3]{2}}\right )} \]

2.3 problem Problem 1(c)

2.3.1 Solving as second order integrable as is ode

2.3.2 Solving as second order ode missing x ode

2.3.3 Solving as type second_order_integrable_as_is (not using ABC version)

2.3.4 Solving as exact nonlinear second order ode ode