problem 119

Internal problem ID [3241]
Internal ﬁle name [OUTPUT/2733_Sunday_June_05_2022_08_39_52_AM_49446695/index.tex]

Book: Diﬀerential equations for engineers by Wei-Chau XIE, Cambridge Press 2010
Section: Chapter 2. First-Order and Simple Higher-Order Diﬀerential Equations. Page 78
Problem number: 119.
ODE order: 1.
ODE degree: 0.

1.96.1 Solving as homogeneousTypeD2 ode

Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} u^{\prime }\left (x \right ) x +u \left (x \right ) = {\mathrm e}^{\frac {\frac {d}{d x}\left (u \left (x \right ) x \right )}{u \left (x \right )}} \end {align*}

In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {-\operatorname {LambertW}\left (-\frac {1}{u}\right ) u -u}{x} \end {align*}

Where \(f(x)=\frac {1}{x}\) and \(g(u)=-\operatorname {LambertW}\left (-\frac {1}{u}\right ) u -u\). Integrating both sides gives \begin{align*} \frac {1}{-\operatorname {LambertW}\left (-\frac {1}{u}\right ) u -u} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{-\operatorname {LambertW}\left (-\frac {1}{u}\right ) u -u} \,du} &= \int {\frac {1}{x} \,d x} \\ \ln \left (\operatorname {LambertW}\left (-\frac {1}{u}\right )\right )&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \operatorname {LambertW}\left (-\frac {1}{u}\right ) &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}

Which simpliﬁes to \begin {align*} \operatorname {LambertW}\left (-\frac {1}{u}\right ) &= c_{3} x \end {align*}

Therefore the solution \(y\) is \begin {align*} y&=x u\\ &=-\frac {{\mathrm e}^{-c_{2}} {\mathrm e}^{-c_{3} {\mathrm e}^{c_{2}} x}}{c_{3}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {{\mathrm e}^{-c_{2}} {\mathrm e}^{-c_{3} {\mathrm e}^{c_{2}} x}}{c_{3}} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -\frac {{\mathrm e}^{-c_{2}} {\mathrm e}^{-c_{3} {\mathrm e}^{c_{2}} x}}{c_{3}} \] Veriﬁed OK.

1.96.2 Solving as dAlembert ode

Let \(p=y^{\prime }\) the ode becomes \begin {align*} p = {\mathrm e}^{\frac {x p}{y}} \end {align*}

Solving for \(y\) from the above results in \begin {align*} y &= \frac {x p}{\ln \left (p \right )}\tag {1A} \end {align*}

Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved. Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}

Comparing the form \(y=x f + g\) to (1A) shows that \begin {align*} f &= \frac {p}{\ln \left (p \right )}\\ g &= 0 \end {align*}

Hence (2) becomes \begin {align*} p -\frac {p}{\ln \left (p \right )} = x \left (\frac {1}{\ln \left (p \right )}-\frac {1}{\ln \left (p \right )^{2}}\right ) p^{\prime }\left (x \right )\tag {2A} \end {align*}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} p -\frac {p}{\ln \left (p \right )} = 0 \end {align*}

Solving for \(p\) from the above gives \begin {align*} p&={\mathrm e} \end {align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = \frac {p \left (x \right )-\frac {p \left (x \right )}{\ln \left (p \left (x \right )\right )}}{x \left (\frac {1}{\ln \left (p \left (x \right )\right )}-\frac {1}{\ln \left (p \left (x \right )\right )^{2}}\right )}\tag {3} \end {align*}

This ODE is now solved for \(p \left (x \right )\). In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= \frac {\ln \left (p \right ) p}{x} \end {align*}

Where \(f(x)=\frac {1}{x}\) and \(g(p)=\ln \left (p \right ) p\). Integrating both sides gives \begin{align*} \frac {1}{\ln \left (p \right ) p} \,dp &= \frac {1}{x} \,d x \\ \int { \frac {1}{\ln \left (p \right ) p} \,dp} &= \int {\frac {1}{x} \,d x} \\ \ln \left (\ln \left (p \right )\right )&=\ln \left (x \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \ln \left (p \right ) &= {\mathrm e}^{\ln \left (x \right )+c_{1}} \end {align*}

Which simpliﬁes to \begin {align*} \ln \left (p \right ) &= c_{2} x \end {align*}

Substituing the above solution for \(p\) in (2A) gives \begin {align*} y = \frac {x \,{\mathrm e}^{c_{2} x \,{\mathrm e}^{c_{1}}}}{\ln \left ({\mathrm e}^{c_{2} x \,{\mathrm e}^{c_{1}}}\right )}\\ \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e} x \\ \tag{2} y &= \frac {x \,{\mathrm e}^{c_{2} x \,{\mathrm e}^{c_{1}}}}{\ln \left ({\mathrm e}^{c_{2} x \,{\mathrm e}^{c_{1}}}\right )} \\ \end{align*}

Veriﬁcation of solutions

\[ y = {\mathrm e} x \] Veriﬁed OK.

\[ y = \frac {x \,{\mathrm e}^{c_{2} x \,{\mathrm e}^{c_{1}}}}{\ln \left ({\mathrm e}^{c_{2} x \,{\mathrm e}^{c_{1}}}\right )} \] Veriﬁed OK.

1.96.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{\frac {x y^{\prime }}{y}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y \mathit {LambertW}\left (-\frac {x}{y}\right )}{x} \end {array} \]

Maple trace

`Methods for first order ODEs: 
-> Solving 1st order ODE of high degree, 1st attempt 
trying 1st order WeierstrassP solution for high degree ODE 
trying 1st order WeierstrassPPrime solution for high degree ODE 
trying 1st order JacobiSN solution for high degree ODE 
trying 1st order ODE linearizable_by_differentiation 
trying differential order: 1; missing variables 
trying homogeneous B 
trying homogeneous types: 
trying homogeneous D 
<- homogeneous successful`

1.96 problem 119

1.96.1 Solving as homogeneousTypeD2 ode

1.96.2 Solving as dAlembert ode

1.96.3 Maple step by step solution