Internal problem ID [11424]
Internal file name [OUTPUT/10407_Thursday_May_18_2023_04_18_34_AM_84307621/index.tex]
Book: A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag,
NY. 2015.
Section: Chapter 1, First order differential equations. Section 1.4.1. Integrating factors. Exercises
page 41
Problem number: 15(b).
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati", "bernoulli", "first_order_ode_lie_symmetry_lookup"
Maple gives the following as the ode type
[[_1st_order, _with_linear_symmetries], _Bernoulli]
\[ \boxed {x^{\prime }-x \left (1+x \,{\mathrm e}^{t}\right )=0} \]
Writing the ode as \begin {align*} x^{\prime }&=x \left (1+{\mathrm e}^{t} x \right )\\ x^{\prime }&= \omega \left ( t,x\right ) \end {align*}
The condition of Lie symmetry is the linearized PDE given by \begin {align*} \eta _{t}+\omega \left ( \eta _{x}-\xi _{t}\right ) -\omega ^{2}\xi _{x}-\omega _{t}\xi -\omega _{x}\eta =0\tag {A} \end {align*}
The type of this ode is known. It is of type Bernoulli. Therefore we do not need
to solve the PDE (A), and can just use the lookup table shown below to find \(\xi ,\eta \)
|
ODE class |
Form |
\(\xi \) |
\(\eta \) |
|
linear ode |
\(y'=f(x) y(x) +g(x)\) |
\(0\) |
\(e^{\int fdx}\) |
|
separable ode |
\(y^{\prime }=f\left ( x\right ) g\left ( y\right ) \) |
\(\frac {1}{f}\) |
\(0\) |
|
quadrature ode |
\(y^{\prime }=f\left ( x\right ) \) |
\(0\) |
\(1\) |
|
quadrature ode |
\(y^{\prime }=g\left ( y\right ) \) |
\(1\) |
\(0\) |
|
homogeneous ODEs of Class A |
\(y^{\prime }=f\left ( \frac {y}{x}\right ) \) |
\(x\) |
\(y\) |
|
homogeneous ODEs of Class C |
\(y^{\prime }=\left ( a+bx+cy\right ) ^{\frac {n}{m}}\) |
\(1\) |
\(-\frac {b}{c}\) |
|
homogeneous class D |
\(y^{\prime }=\frac {y}{x}+g\left ( x\right ) F\left (\frac {y}{x}\right ) \) |
\(x^{2}\) |
\(xy\) |
|
First order special form ID 1 |
\(y^{\prime }=g\left ( x\right ) e^{h\left (x\right ) +by}+f\left ( x\right ) \) |
\(\frac {e^{-\int bf\left ( x\right )dx-h\left ( x\right ) }}{g\left ( x\right ) }\) |
\(\frac {f\left ( x\right )e^{-\int bf\left ( x\right ) dx-h\left ( x\right ) }}{g\left ( x\right ) }\) |
|
polynomial type ode |
\(y^{\prime }=\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}\) |
\(\frac {a_{1}b_{2}x-a_{2}b_{1}x-b_{1}c_{2}+b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) |
\(\frac {a_{1}b_{2}y-a_{2}b_{1}y-a_{1}c_{2}-a_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}\) |
|
Bernoulli ode |
\(y^{\prime }=f\left ( x\right ) y+g\left ( x\right ) y^{n}\) |
\(0\) |
\(e^{-\int \left ( n-1\right ) f\left ( x\right ) dx}y^{n}\) |
|
Reduced Riccati |
\(y^{\prime }=f_{1}\left ( x\right ) y+f_{2}\left ( x\right )y^{2}\) |
\(0\) |
\(e^{-\int f_{1}dx}\) |
|
|
|||
|
|
|||
The above table shows that \begin {align*} \xi \left (t,x\right ) &=0\\ \tag {A1} \eta \left (t,x\right ) &=x^{2} {\mathrm e}^{-t} \end {align*}
The next step is to determine the canonical coordinates \(R,S\). The canonical coordinates map \(\left ( t,x\right ) \to \left ( R,S \right )\) where \(\left ( R,S \right )\) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.
The characteristic pde which is used to find the canonical coordinates is \begin {align*} \frac {d t}{\xi } &= \frac {d x}{\eta } = dS \tag {1} \end {align*}
The above comes from the requirements that \(\left ( \xi \frac {\partial }{\partial t} + \eta \frac {\partial }{\partial x}\right ) S(t,x) = 1\). Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable \(R\) in the canonical coordinates, where \(S(R)\). Since \(\xi =0\) then in this special case \begin {align*} R = t \end {align*}
\(S\) is found from \begin {align*} S &= \int { \frac {1}{\eta }} dy\\ &= \int { \frac {1}{x^{2} {\mathrm e}^{-t}}} dy \end {align*}
Which results in \begin {align*} S&= -\frac {{\mathrm e}^{t}}{x} \end {align*}
Now that \(R,S\) are found, we need to setup the ode in these coordinates. This is done by evaluating \begin {align*} \frac {dS}{dR} &= \frac { S_{t} + \omega (t,x) S_{x} }{ R_{t} + \omega (t,x) R_{x} }\tag {2} \end {align*}
Where in the above \(R_{t},R_{x},S_{t},S_{x}\) are all partial derivatives and \(\omega (t,x)\) is the right hand side of the original ode given by \begin {align*} \omega (t,x) &= x \left (1+{\mathrm e}^{t} x \right ) \end {align*}
Evaluating all the partial derivatives gives \begin {align*} R_{t} &= 1\\ R_{x} &= 0\\ S_{t} &= -\frac {{\mathrm e}^{t}}{x}\\ S_{x} &= \frac {{\mathrm e}^{t}}{x^{2}} \end {align*}
Substituting all the above in (2) and simplifying gives the ode in canonical coordinates. \begin {align*} \frac {dS}{dR} &= {\mathrm e}^{2 t}\tag {2A} \end {align*}
We now need to express the RHS as function of \(R\) only. This is done by solving for \(t,x\) in terms of \(R,S\) from the result obtained earlier and simplifying. This gives \begin {align*} \frac {dS}{dR} &= {\mathrm e}^{2 R} \end {align*}
The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates \(R,S\). Integrating the above gives \begin {align*} S \left (R \right ) = \frac {{\mathrm e}^{2 R}}{2}+c_{1}\tag {4} \end {align*}
To complete the solution, we just need to transform (4) back to \(t,x\) coordinates. This results in \begin {align*} -\frac {{\mathrm e}^{t}}{x} = \frac {{\mathrm e}^{2 t}}{2}+c_{1} \end {align*}
Which simplifies to \begin {align*} -\frac {{\mathrm e}^{t}}{x} = \frac {{\mathrm e}^{2 t}}{2}+c_{1} \end {align*}
Which gives \begin {align*} x = -\frac {2 \,{\mathrm e}^{t}}{{\mathrm e}^{2 t}+2 c_{1}} \end {align*}
The following diagram shows solution curves of the original ode and how they transform in the canonical coordinates space using the mapping shown.
|
Original ode in \(t,x\) coordinates |
Canonical coordinates transformation |
ODE in canonical coordinates \((R,S)\) |
| \( \frac {dx}{dt} = x \left (1+{\mathrm e}^{t} x \right )\) |
|
\( \frac {d S}{d R} = {\mathrm e}^{2 R}\) |
| |
\(\!\begin {aligned} R&= t\\ S&= -\frac {{\mathrm e}^{t}}{x} \end {aligned} \) |
|
Summary
The solution(s) found are the following \begin{align*} \tag{1} x &= -\frac {2 \,{\mathrm e}^{t}}{{\mathrm e}^{2 t}+2 c_{1}} \\ \end{align*}
Verification of solutions
\[ x = -\frac {2 \,{\mathrm e}^{t}}{{\mathrm e}^{2 t}+2 c_{1}} \] Verified OK.
In canonical form, the ODE is \begin {align*} x' &= F(t,x)\\ &= x \left (1+{\mathrm e}^{t} x \right ) \end {align*}
This is a Bernoulli ODE. \[ x' = x +{\mathrm e}^{t} x^{2} \tag {1} \] The standard Bernoulli ODE has the form \[ x' = f_0(t)x+f_1(t)x^n \tag {2} \] The first step is to divide the above equation by \(x^n \) which gives \[ \frac {x'}{x^n} = f_0(t) x^{1-n} +f_1(t) \tag {3} \] The next step is use the substitution \(w = x^{1-n}\) in equation (3) which generates a new ODE in \(w \left (t \right )\) which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution \(x(t)\) which is what we want.
This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that \begin {align*} f_0(t)&=1\\ f_1(t)&={\mathrm e}^{t}\\ n &=2 \end {align*}
Dividing both sides of ODE (1) by \(x^n=x^{2}\) gives \begin {align*} x'\frac {1}{x^{2}} &= \frac {1}{x} +{\mathrm e}^{t} \tag {4} \end {align*}
Let \begin {align*} w &= x^{1-n} \\ &= \frac {1}{x} \tag {5} \end {align*}
Taking derivative of equation (5) w.r.t \(t\) gives \begin {align*} w' &= -\frac {1}{x^{2}}x' \tag {6} \end {align*}
Substituting equations (5) and (6) into equation (4) gives \begin {align*} -w^{\prime }\left (t \right )&= w \left (t \right )+{\mathrm e}^{t}\\ w' &= -w -{\mathrm e}^{t} \tag {7} \end {align*}
The above now is a linear ODE in \(w \left (t \right )\) which is now solved.
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} w^{\prime }\left (t \right ) + p(t)w \left (t \right ) &= q(t) \end {align*}
Where here \begin {align*} p(t) &=1\\ q(t) &=-{\mathrm e}^{t} \end {align*}
Hence the ode is \begin {align*} w^{\prime }\left (t \right )+w \left (t \right ) = -{\mathrm e}^{t} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int 1d t} \\ &= {\mathrm e}^{t} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu w\right ) &= \left (\mu \right ) \left (-{\mathrm e}^{t}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left ({\mathrm e}^{t} w\right ) &= \left ({\mathrm e}^{t}\right ) \left (-{\mathrm e}^{t}\right )\\ \mathrm {d} \left ({\mathrm e}^{t} w\right ) &= \left (-{\mathrm e}^{2 t}\right )\, \mathrm {d} t \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{t} w &= \int {-{\mathrm e}^{2 t}\,\mathrm {d} t}\\ {\mathrm e}^{t} w &= -\frac {{\mathrm e}^{2 t}}{2} + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{t}\) results in \begin {align*} w \left (t \right ) &= -\frac {{\mathrm e}^{-t} {\mathrm e}^{2 t}}{2}+c_{1} {\mathrm e}^{-t} \end {align*}
which simplifies to \begin {align*} w \left (t \right ) &= -\frac {{\mathrm e}^{t}}{2}+c_{1} {\mathrm e}^{-t} \end {align*}
Replacing \(w\) in the above by \(\frac {1}{x}\) using equation (5) gives the final solution. \begin {align*} \frac {1}{x} = -\frac {{\mathrm e}^{t}}{2}+c_{1} {\mathrm e}^{-t} \end {align*}
Or \begin {align*} x = \frac {1}{-\frac {{\mathrm e}^{t}}{2}+c_{1} {\mathrm e}^{-t}} \end {align*}
Which is simplified to \[ x = -\frac {2 \,{\mathrm e}^{t}}{{\mathrm e}^{2 t}-2 c_{1}} \]
The solution(s) found are the following \begin{align*} \tag{1} x &= -\frac {2 \,{\mathrm e}^{t}}{{\mathrm e}^{2 t}-2 c_{1}} \\ \end{align*}
Verification of solutions
\[ x = -\frac {2 \,{\mathrm e}^{t}}{{\mathrm e}^{2 t}-2 c_{1}} \] Verified OK.
In canonical form the ODE is \begin {align*} x' &= F(t,x)\\ &= x \left (1+{\mathrm e}^{t} x \right ) \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ x' = {\mathrm e}^{t} x^{2}+x \] With Riccati ODE standard form \[ x' = f_0(t)+ f_1(t)x+f_2(t)x^{2} \] Shows that \(f_0(t)=0\), \(f_1(t)=1\) and \(f_2(t)={\mathrm e}^{t}\). Let \begin {align*} x &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{{\mathrm e}^{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' &={\mathrm e}^{t}\\ f_1 f_2 &={\mathrm e}^{t}\\ f_2^2 f_0 &=0 \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} {\mathrm e}^{t} u^{\prime \prime }\left (t \right )-2 \,{\mathrm e}^{t} u^{\prime }\left (t \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (t \right ) = c_{1} +c_{2} {\mathrm e}^{2 t} \] The above shows that \[ u^{\prime }\left (t \right ) = 2 c_{2} {\mathrm e}^{2 t} \] Using the above in (1) gives the solution \[ x = -\frac {2 c_{2} {\mathrm e}^{2 t} {\mathrm e}^{-t}}{c_{1} +c_{2} {\mathrm e}^{2 t}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ x = -\frac {2 \,{\mathrm e}^{t}}{c_{3} +{\mathrm e}^{2 t}} \]
The solution(s) found are the following \begin{align*} \tag{1} x &= -\frac {2 \,{\mathrm e}^{t}}{c_{3} +{\mathrm e}^{2 t}} \\ \end{align*}
Verification of solutions
\[ x = -\frac {2 \,{\mathrm e}^{t}}{c_{3} +{\mathrm e}^{2 t}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{\prime }-x \left (1+x \,{\mathrm e}^{t}\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }=x \left (1+x \,{\mathrm e}^{t}\right ) \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli <- Bernoulli successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 18
dsolve(diff(x(t),t)=x(t)*(1+x(t)*exp(t)),x(t), singsol=all)
\[ x \left (t \right ) = -\frac {2 \,{\mathrm e}^{t}}{{\mathrm e}^{2 t}-2 c_{1}} \]
✓ Solution by Mathematica
Time used: 0.323 (sec). Leaf size: 27
DSolve[x'[t]==x[t]*(1+x[t]*Exp[t]),x[t],t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to -\frac {2 e^t}{e^{2 t}-2 c_1} \\ x(t)\to 0 \\ \end{align*}