Internal problem ID [2608]
Internal file name [OUTPUT/2100_Sunday_June_05_2022_02_48_29_AM_15930513/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 1, First-Order Differential Equations. Section 1.2, Basic Ideas and Terminology.
page 21
Problem number: Problem 30.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "exact"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(y)]`]]
\[ \boxed {y^{\prime }-\frac {\left (1-y \,{\mathrm e}^{x y}\right ) {\mathrm e}^{-x y}}{x}=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 0] \end {align*}
This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= -\frac {\left (y \,{\mathrm e}^{x y}-1\right ) {\mathrm e}^{-x y}}{x} \end {align*}
The \(x\) domain of \(f(x,y)\) when \(y=0\) is \[
\{x <0\boldsymbol {\lor }0 The \(x\) domain of \(\frac {\partial f}{\partial y}\) when \(y=0\) is \[
\{x <0\boldsymbol {\lor }0
Entering Exact first order ODE solver. (Form one type)
To solve an ode of the form\begin {equation} M\left ( x,y\right ) +N\left ( x,y\right ) \frac {dy}{dx}=0\tag {A} \end {equation} We assume there exists a function \(\phi \left ( x,y\right ) =c\) where \(c\) is constant, that
satisfies the ode. Taking derivative of \(\phi \) w.r.t. \(x\) gives\[ \frac {d}{dx}\phi \left ( x,y\right ) =0 \] Hence\begin {equation} \frac {\partial \phi }{\partial x}+\frac {\partial \phi }{\partial y}\frac {dy}{dx}=0\tag {B} \end {equation} Comparing (A,B) shows
that\begin {align*} \frac {\partial \phi }{\partial x} & =M\\ \frac {\partial \phi }{\partial y} & =N \end {align*}
But since \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) then for the above to be valid, we require that\[ \frac {\partial M}{\partial y}=\frac {\partial N}{\partial x}\] If the above condition is satisfied,
then the original ode is called exact. We still need to determine \(\phi \left ( x,y\right ) \) but at least we know
now that we can do that since the condition \(\frac {\partial ^{2}\phi }{\partial x\partial y}=\frac {\partial ^{2}\phi }{\partial y\partial x}\) is satisfied. If this condition is not
satisfied then this method will not work and we have to now look for an integrating
factor to force this condition, which might or might not exist. The first step is
to write the ODE in standard form to check for exactness, which is \[ M(x,y) \mathop {\mathrm {d}x}+ N(x,y) \mathop {\mathrm {d}y}=0 \tag {1A} \] Therefore
\begin {align*} \left (x \,{\mathrm e}^{x y}\right )\mathop {\mathrm {d}y} &= \left (1-y \,{\mathrm e}^{x y}\right )\mathop {\mathrm {d}x}\\ \left (y \,{\mathrm e}^{x y}-1\right )\mathop {\mathrm {d}x} + \left (x \,{\mathrm e}^{x y}\right )\mathop {\mathrm {d}y} &= 0 \tag {2A} \end {align*}
Comparing (1A) and (2A) shows that \begin {align*} M(x,y) &= y \,{\mathrm e}^{x y}-1\\ N(x,y) &= x \,{\mathrm e}^{x y} \end {align*}
The next step is to determine if the ODE is is exact or not. The ODE is exact when the
following condition is satisfied \[ \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x} \] Using result found above gives \begin {align*} \frac {\partial M}{\partial y} &= \frac {\partial }{\partial y} \left (y \,{\mathrm e}^{x y}-1\right )\\ &= {\mathrm e}^{x y} \left (x y +1\right ) \end {align*}
And \begin {align*} \frac {\partial N}{\partial x} &= \frac {\partial }{\partial x} \left (x \,{\mathrm e}^{x y}\right )\\ &= {\mathrm e}^{x y} \left (x y +1\right ) \end {align*}
Since \(\frac {\partial M}{\partial y}= \frac {\partial N}{\partial x}\), then the ODE is exact The following equations are now set up to solve for the
function \(\phi \left (x,y\right )\) \begin {align*} \frac {\partial \phi }{\partial x } &= M\tag {1} \\ \frac {\partial \phi }{\partial y } &= N\tag {2} \end {align*}
Integrating (1) w.r.t. \(x\) gives \begin{align*}
\int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int M\mathop {\mathrm {d}x} \\
\int \frac {\partial \phi }{\partial x} \mathop {\mathrm {d}x} &= \int y \,{\mathrm e}^{x y}-1\mathop {\mathrm {d}x} \\
\tag{3} \phi &= -x +{\mathrm e}^{x y}+ f(y) \\
\end{align*} Where \(f(y)\) is used for the constant of integration since \(\phi \) is a
function of both \(x\) and \(y\). Taking derivative of equation (3) w.r.t \(y\) gives \begin{equation}
\tag{4} \frac {\partial \phi }{\partial y} = x \,{\mathrm e}^{x y}+f'(y)
\end{equation} But equation
(2) says that \(\frac {\partial \phi }{\partial y} = x \,{\mathrm e}^{x y}\). Therefore equation (4) becomes \begin{equation}
\tag{5} x \,{\mathrm e}^{x y} = x \,{\mathrm e}^{x y}+f'(y)
\end{equation} Solving equation (5) for \( f'(y)\) gives \[ f'(y) = 0 \]
Therefore \[ f(y) = c_{1} \] Where \(c_{1}\) is constant of integration. Substituting this result for \(f(y)\) into
equation (3) gives \(\phi \) \[
\phi = -x +{\mathrm e}^{x y}+ c_{1}
\] But since \(\phi \) itself is a constant function, then let \(\phi =c_{2}\) where \(c_{2}\) is new
constant and combining \(c_{1}\) and \(c_{2}\) constants into new constant \(c_{1}\) gives the solution as \[
c_{1} = -x +{\mathrm e}^{x y}
\]
The solution becomes\[
y = \frac {\ln \left (x +c_{1} \right )}{x}
\] Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(y=0\)
in the above solution gives an equation to solve for the constant of integration.
\begin {align*} 0 = \ln \left (1+c_{1} \right ) \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting this in the general solution gives \begin {align*} y&=\frac {\ln \left (x \right )}{x} \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
Summary
The solution(s) found are the following \begin{align*}
\tag{1} y &= \frac {\ln \left (x \right )}{x} \\
\end{align*} Verification of solutions
\[
y = \frac {\ln \left (x \right )}{x}
\] Verified OK. \[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-\frac {1-y \,{\mathrm e}^{x y}}{x \,{\mathrm e}^{x y}}=0, y \left (1\right )=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {1-y \,{\mathrm e}^{x y}}{x \,{\mathrm e}^{x y}} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]
Maple trace
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 10
\[
y \left (x \right ) = \frac {\ln \left (x \right )}{x}
\]
✓ Solution by Mathematica
Time used: 0.403 (sec). Leaf size: 11
\[
y(x)\to \frac {\log (x)}{x}
\]
1.22.2 Solving as exact ode
1.22.3 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 inverse_Riccati
trying an equivalence to an Abel ODE
differential order: 1; trying a linearization to 2nd order
--- trying a change of variables {x -> y(x), y(x) -> x}
differential order: 1; trying a linearization to 2nd order
trying 1st order ODE linearizable_by_differentiation
--- Trying Lie symmetry methods, 1st order ---
`, `-> Computing symmetries using: way = 3
`, `-> Computing symmetries using: way = 4`[1, -y/x], [-x, (x*y-1)/x]
dsolve([diff(y(x),x)=(1-y(x)*exp(x*y(x)))/(x*exp(x*y(x))),y(1) = 0],y(x), singsol=all)
DSolve[{y'[x]==(1-y[x]*Exp[x*y[x]])/(x*Exp[x*y[x]]),{y[1]==0}},y[x],x,IncludeSingularSolutions -> True]