Internal problem ID [9328]
Internal file name [OUTPUT/8264_Monday_June_06_2022_02_34_58_AM_12507954/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 995.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
\[ \boxed {y^{\prime }-\left (y-{\mathrm e}^{x}\right )^{2}={\mathrm e}^{x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= {\mathrm e}^{2 x}-2 y \,{\mathrm e}^{x}+y^{2}+{\mathrm e}^{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = {\mathrm e}^{2 x}-2 y \,{\mathrm e}^{x}+y^{2}+{\mathrm e}^{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)={\mathrm e}^{2 x}+{\mathrm e}^{x}\), \(f_1(x)=-2 \,{\mathrm e}^{x}\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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''(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 &=-2 \,{\mathrm e}^{x}\\ f_2^2 f_0 &={\mathrm e}^{2 x}+{\mathrm e}^{x} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+2 \,{\mathrm e}^{x} u^{\prime }\left (x \right )+\left ({\mathrm e}^{2 x}+{\mathrm e}^{x}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-{\mathrm e}^{x}} \left (c_{2} x +c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \left (-c_{2} x -c_{1} \right ) {\mathrm e}^{x -{\mathrm e}^{x}}+{\mathrm e}^{-{\mathrm e}^{x}} c_{2} \] Using the above in (1) gives the solution \[ y = -\frac {\left (\left (-c_{2} x -c_{1} \right ) {\mathrm e}^{x -{\mathrm e}^{x}}+{\mathrm e}^{-{\mathrm e}^{x}} c_{2} \right ) {\mathrm e}^{{\mathrm e}^{x}}}{c_{2} x +c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-1+\left (c_{3} +x \right ) {\mathrm e}^{x}}{c_{3} +x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-1+\left (c_{3} +x \right ) {\mathrm e}^{x}}{c_{3} +x} \\ \end{align*}
Verification of solutions
\[ y = \frac {-1+\left (c_{3} +x \right ) {\mathrm e}^{x}}{c_{3} +x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\left (y-{\mathrm e}^{x}\right )^{2}={\mathrm e}^{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (y-{\mathrm e}^{x}\right )^{2}+{\mathrm e}^{x} \end {array} \]
Maple trace
`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 Riccati trying Riccati sub-methods: <- Riccati particular polynomial solution successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 22
dsolve(diff(y(x),x) = (y(x)-exp(x))^2+exp(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-1+\left (x -c_{1} \right ) {\mathrm e}^{x}}{x -c_{1}} \]
✓ Solution by Mathematica
Time used: 0.307 (sec). Leaf size: 24
DSolve[y'[x] == E^x + (-E^x + y[x])^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to e^x+\frac {1}{-x+c_1} \\ y(x)\to e^x \\ \end{align*}