18.41 problem 41

Internal problem ID [2313]
Internal file name [OUTPUT/2313_Tuesday_February_27_2024_08_25_56_AM_14896452/index.tex]

Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 35, page 157
Problem number: 41.
ODE order: 2.
ODE degree: 1.

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

Maple gives the following as the ode type

[[_2nd_order, _missing_y]]

\[ \boxed {\left (1-{\mathrm e}^{x}\right ) y^{\prime \prime }-y^{\prime } {\mathrm e}^{x}=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 0, y^{\prime }\left (1\right ) = 1] \end {align*}

18.41.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime \prime } + p(x)y^{\prime } + q(x) y &= F \end {align*}

Where here \begin {align*} p(x) &=-\frac {{\mathrm e}^{x}}{1-{\mathrm e}^{x}}\\ q(x) &=0\\ F &=0 \end {align*}

Hence the ode is \begin {align*} y^{\prime \prime }-\frac {y^{\prime } {\mathrm e}^{x}}{1-{\mathrm e}^{x}} = 0 \end {align*}

The domain of \(p(x)=-\frac {{\mathrm e}^{x}}{1-{\mathrm e}^{x}}\) is \[ \{2 i \pi \_Z256

18.41.2 Solving as second order integrable as is ode

Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (1-{\mathrm e}^{x}\right ) y^{\prime \prime }-y^{\prime } {\mathrm e}^{x}\right )d x &= 0 \\ -\left ({\mathrm e}^{x}-1\right ) y^{\prime } = c_{1} \end {align*}

Which is now solved for \(y\). Integrating both sides gives \begin {align*} y &= \int { -\frac {c_{1}}{{\mathrm e}^{x}-1}\,\mathop {\mathrm {d}x}}\\ &= -c_{1} \left (-x +\ln \left ({\mathrm e}^{x}-1\right )\right )+c_{2} \end {align*}

Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = -c_{1} \left (-x +\ln \left ({\mathrm e}^{x}-1\right )\right )+c_{2} \tag {1} \end {align*}

Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = c_{1} -c_{1} \ln \left (-1+{\mathrm e}\right )+c_{2}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = -c_{1} \left (-1+\frac {{\mathrm e}^{x}}{{\mathrm e}^{x}-1}\right ) \end {align*}

substituting \(y^{\prime } = 1\) and \(x = 1\) in the above gives \begin {align*} 1 = -\frac {c_{1}}{-1+{\mathrm e}}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=1-{\mathrm e}\\ c_{2}&=-\left (-1+\ln \left (-1+{\mathrm e}\right )\right ) \left (-1+{\mathrm e}\right ) \end {align*}

Substituting these values back in above solution results in \begin {align*} y = {\mathrm e} \ln \left ({\mathrm e}^{x}-1\right )-\ln \left (-1+{\mathrm e}\right ) {\mathrm e}-{\mathrm e} x -\ln \left ({\mathrm e}^{x}-1\right )+{\mathrm e}+\ln \left (-1+{\mathrm e}\right )+x -1 \end {align*}

Which simplifies to \[ y = -\left (-1+{\mathrm e}\right ) \left (x +\ln \left (-1+{\mathrm e}\right )-\ln \left ({\mathrm e}^{x}-1\right )-1\right ) \]

Summary

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

Figure 611: Solution plot

Verification of solutions

\[ y = -\left (-1+{\mathrm e}\right ) \left (x +\ln \left (-1+{\mathrm e}\right )-\ln \left ({\mathrm e}^{x}-1\right )-1\right ) \] Verified OK.

18.41.3 Solving as second order ode missing y ode

This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}

Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}

Hence the ode becomes \begin {align*} \left (1-{\mathrm e}^{x}\right ) p^{\prime }\left (x \right )-{\mathrm e}^{x} p \left (x \right ) = 0 \end {align*}

Which is now solve for \(p(x)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= -\frac {{\mathrm e}^{x} p}{{\mathrm e}^{x}-1} \end {align*}

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

Initial conditions are used to solve for \(c_{1}\). Substituting \(x=1\) and \(p=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = \frac {c_{1}}{-1+{\mathrm e}} \end {align*}

The solutions are \begin {align*} c_{1} = -1+{\mathrm e} \end {align*}

Trying the constant \begin {align*} c_{1} = -1+{\mathrm e} \end {align*}

Substituting this in the general solution gives \begin {align*} p \left (x \right )&=\frac {-1+{\mathrm e}}{{\mathrm e}^{x}-1} \end {align*}

The constant \(c_{1} = -1+{\mathrm e}\) gives valid solution.

Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = \frac {-1+{\mathrm e}}{{\mathrm e}^{x}-1} \end {align*}

Integrating both sides gives \begin {align*} y &= \int { \frac {-1+{\mathrm e}}{{\mathrm e}^{x}-1}\,\mathop {\mathrm {d}x}}\\ &= \left (-1+{\mathrm e}\right ) \left (-x +\ln \left ({\mathrm e}^{x}-1\right )\right )+c_{2} \end {align*}

Initial conditions are used to solve for \(c_{2}\). 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+{\mathrm e}\right ) {\mathrm e}-{\mathrm e}-\ln \left (-1+{\mathrm e}\right )+1+c_{2} \end {align*}

The solutions are \begin {align*} c_{2} = -\ln \left (-1+{\mathrm e}\right ) {\mathrm e}+{\mathrm e}+\ln \left (-1+{\mathrm e}\right )-1 \end {align*}

Trying the constant \begin {align*} c_{2} = -\ln \left (-1+{\mathrm e}\right ) {\mathrm e}+{\mathrm e}+\ln \left (-1+{\mathrm e}\right )-1 \end {align*}

Substituting this in the general solution gives \begin {align*} y&={\mathrm e} \ln \left ({\mathrm e}^{x}-1\right )-\ln \left (-1+{\mathrm e}\right ) {\mathrm e}-{\mathrm e} x -\ln \left ({\mathrm e}^{x}-1\right )+{\mathrm e}+\ln \left (-1+{\mathrm e}\right )+x -1 \end {align*}

The constant \(c_{2} = -\ln \left (-1+{\mathrm e}\right ) {\mathrm e}+{\mathrm e}+\ln \left (-1+{\mathrm e}\right )-1\) gives valid solution.

Initial conditions are used to solve for the constants of integration.

Summary

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

Figure 612: Solution plot

Verification of solutions

\[ y = {\mathrm e} \ln \left ({\mathrm e}^{x}-1\right )-\ln \left (-1+{\mathrm e}\right ) {\mathrm e}-{\mathrm e} x -\ln \left ({\mathrm e}^{x}-1\right )+{\mathrm e}+\ln \left (-1+{\mathrm e}\right )+x -1 \] Verified OK.

18.41.4 Solving as type second_order_integrable_as_is (not using ABC version)

Writing the ode as \[ \left (1-{\mathrm e}^{x}\right ) y^{\prime \prime }-y^{\prime } {\mathrm e}^{x} = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (1-{\mathrm e}^{x}\right ) y^{\prime \prime }-y^{\prime } {\mathrm e}^{x}\right )d x &= 0 \\ -y^{\prime } {\mathrm e}^{x}+y^{\prime } = c_{1} \end {align*}

Which is now solved for \(y\). Integrating both sides gives \begin {align*} y &= \int { -\frac {c_{1}}{{\mathrm e}^{x}-1}\,\mathop {\mathrm {d}x}}\\ &= -c_{1} \left (-x +\ln \left ({\mathrm e}^{x}-1\right )\right )+c_{2} \end {align*}

Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = -c_{1} \left (-x +\ln \left ({\mathrm e}^{x}-1\right )\right )+c_{2} \tag {1} \end {align*}

Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = c_{1} -c_{1} \ln \left (-1+{\mathrm e}\right )+c_{2}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = -c_{1} \left (-1+\frac {{\mathrm e}^{x}}{{\mathrm e}^{x}-1}\right ) \end {align*}

substituting \(y^{\prime } = 1\) and \(x = 1\) in the above gives \begin {align*} 1 = -\frac {c_{1}}{-1+{\mathrm e}}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=1-{\mathrm e}\\ c_{2}&=-\left (-1+\ln \left (-1+{\mathrm e}\right )\right ) \left (-1+{\mathrm e}\right ) \end {align*}

Substituting these values back in above solution results in \begin {align*} y = {\mathrm e} \ln \left ({\mathrm e}^{x}-1\right )-\ln \left (-1+{\mathrm e}\right ) {\mathrm e}-{\mathrm e} x -\ln \left ({\mathrm e}^{x}-1\right )+{\mathrm e}+\ln \left (-1+{\mathrm e}\right )+x -1 \end {align*}

Which simplifies to \[ y = -\left (-1+{\mathrm e}\right ) \left (x +\ln \left (-1+{\mathrm e}\right )-\ln \left ({\mathrm e}^{x}-1\right )-1\right ) \]

Summary

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

Figure 613: Solution plot

Verification of solutions

\[ y = -\left (-1+{\mathrm e}\right ) \left (x +\ln \left (-1+{\mathrm e}\right )-\ln \left ({\mathrm e}^{x}-1\right )-1\right ) \] Verified OK.

18.41.5 Solving as exact linear second order ode ode

An ode of the form \begin {align*} p \left (x \right ) y^{\prime \prime }+q \left (x \right ) y^{\prime }+r \left (x \right ) y&=s \left (x \right ) \end {align*}

is exact if \begin {align*} p''(x) - q'(x) + r(x) &= 0 \tag {1} \end {align*}

For the given ode we have \begin {align*} p(x) &= 1-{\mathrm e}^{x}\\ q(x) &= -{\mathrm e}^{x}\\ r(x) &= 0\\ s(x) &= 0 \end {align*}

Hence \begin {align*} p''(x) &= -{\mathrm e}^{x}\\ q'(x) &= -{\mathrm e}^{x} \end {align*}

Therefore (1) becomes \begin {align*} -{\mathrm e}^{x}- \left (-{\mathrm e}^{x}\right ) + \left (0\right )&=0 \end {align*}

Hence the ode is exact. Since we now know the ode is exact, it can be written as \begin {align*} \left (p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y\right )' &= s(x) \end {align*}

Integrating gives \begin {align*} p \left (x \right ) y^{\prime }+\left (q \left (x \right )-p^{\prime }\left (x \right )\right ) y&=\int {s \left (x \right )\, dx} \end {align*}

Substituting the above values for \(p,q,r,s\) gives \begin {align*} \left (1-{\mathrm e}^{x}\right ) y^{\prime }&=c_{1} \end {align*}

We now have a first order ode to solve which is \begin {align*} \left (1-{\mathrm e}^{x}\right ) y^{\prime } = c_{1} \end {align*}

Integrating both sides gives \begin {align*} y &= \int { -\frac {c_{1}}{{\mathrm e}^{x}-1}\,\mathop {\mathrm {d}x}}\\ &= -c_{1} \left (-x +\ln \left ({\mathrm e}^{x}-1\right )\right )+c_{2} \end {align*}

Initial conditions are used to solve for the constants of integration.

Looking at the above solution \begin {align*} y = -c_{1} \left (-x +\ln \left ({\mathrm e}^{x}-1\right )\right )+c_{2} \tag {1} \end {align*}

Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 0\) and \(x = 1\) in the above gives \begin {align*} 0 = c_{1} -c_{1} \ln \left (-1+{\mathrm e}\right )+c_{2}\tag {1A} \end {align*}

Taking derivative of the solution gives \begin {align*} y^{\prime } = -c_{1} \left (-1+\frac {{\mathrm e}^{x}}{{\mathrm e}^{x}-1}\right ) \end {align*}

substituting \(y^{\prime } = 1\) and \(x = 1\) in the above gives \begin {align*} 1 = -\frac {c_{1}}{-1+{\mathrm e}}\tag {2A} \end {align*}

Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). Solving for the constants gives \begin {align*} c_{1}&=1-{\mathrm e}\\ c_{2}&=-\left (-1+\ln \left (-1+{\mathrm e}\right )\right ) \left (-1+{\mathrm e}\right ) \end {align*}

Substituting these values back in above solution results in \begin {align*} y = {\mathrm e} \ln \left ({\mathrm e}^{x}-1\right )-\ln \left (-1+{\mathrm e}\right ) {\mathrm e}-{\mathrm e} x -\ln \left ({\mathrm e}^{x}-1\right )+{\mathrm e}+\ln \left (-1+{\mathrm e}\right )+x -1 \end {align*}

Which simplifies to \[ y = -\left (-1+{\mathrm e}\right ) \left (x +\ln \left (-1+{\mathrm e}\right )-\ln \left ({\mathrm e}^{x}-1\right )-1\right ) \]

Summary

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

Figure 614: Solution plot

Verification of solutions

\[ y = -\left (-1+{\mathrm e}\right ) \left (x +\ln \left (-1+{\mathrm e}\right )-\ln \left ({\mathrm e}^{x}-1\right )-1\right ) \] Verified OK.

18.41.6 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [\left (1-{\mathrm e}^{x}\right ) y^{\prime \prime }-y^{\prime } {\mathrm e}^{x}=0, y \left (1\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (1-{\mathrm e}^{x}\right ) u^{\prime }\left (x \right )-u \left (x \right ) {\mathrm e}^{x}=0 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{u \left (x \right )}=\frac {{\mathrm e}^{x}}{1-{\mathrm e}^{x}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{u \left (x \right )}d x =\int \frac {{\mathrm e}^{x}}{1-{\mathrm e}^{x}}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (x \right )\right )=-\ln \left (1-{\mathrm e}^{x}\right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {{\mathrm e}^{c_{1}}}{{\mathrm e}^{x}-1} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {{\mathrm e}^{c_{1}}}{{\mathrm e}^{x}-1} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {{\mathrm e}^{c_{1}}}{{\mathrm e}^{x}-1} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {{\mathrm e}^{c_{1}}}{{\mathrm e}^{x}-1}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-{\mathrm e}^{c_{1}} \left (-\ln \left ({\mathrm e}^{x}\right )+\ln \left ({\mathrm e}^{x}-1\right )\right )+c_{2} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} y=-{\mathrm e}^{c_{1}} \left (-\ln \left ({\mathrm e}^{x}\right )+\ln \left ({\mathrm e}^{x}-1\right )\right )+c_{2} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=-{\mathrm e}^{c_{1}} \left (-1+\ln \left (-1+{\mathrm e}\right )\right )+c_{2} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & y^{\prime }=-{\mathrm e}^{c_{1}} \left (-1+\frac {{\mathrm e}^{x}}{{\mathrm e}^{x}-1}\right ) \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1 \\ {} & {} & 1=-{\mathrm e}^{c_{1}} \left (-1+\frac {{\mathrm e}}{-1+{\mathrm e}}\right ) \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & {} & \left \{c_{1} =\ln \left (1-{\mathrm e}\right ), c_{2} =-\ln \left (-1+{\mathrm e}\right ) {\mathrm e}+{\mathrm e}+\ln \left (-1+{\mathrm e}\right )-1\right \} \\ {} & \circ & \textrm {Substitute constant values into general solution and simplify}\hspace {3pt} \\ {} & {} & y=-\left (\ln \left ({\mathrm e}^{x}\right )+\ln \left (-1+{\mathrm e}\right )-\ln \left ({\mathrm e}^{x}-1\right )-1\right ) \left (-1+{\mathrm e}\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y=-\left (\ln \left ({\mathrm e}^{x}\right )+\ln \left (-1+{\mathrm e}\right )-\ln \left ({\mathrm e}^{x}-1\right )-1\right ) \left (-1+{\mathrm e}\right ) \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
<- LODE missing y successful`
 

✓ Solution by Maple

Time used: 0.047 (sec). Leaf size: 27

dsolve([(1-exp(x))*diff(y(x),x$2)=exp(x)*diff(y(x),x),y(1) = 0, D(y)(1) = 1],y(x), singsol=all)
 

\[ y \left (x \right ) = -\left (\ln \left ({\mathrm e}^{x}\right )+\ln \left (-1+{\mathrm e}\right )-\ln \left ({\mathrm e}^{x}-1\right )-1\right ) \left (-1+{\mathrm e}\right ) \]

✓ Solution by Mathematica

Time used: 0.041 (sec). Leaf size: 27

DSolve[{(1-Exp[x])*y''[x]==Exp[x]*y'[x],{y[1]==0,y'[1]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to -2 (e-1) \left (\text {arctanh}(1-2 e)-\text {arctanh}\left (1-2 e^x\right )\right ) \]