Internal problem ID [3342]
Internal file name [OUTPUT/2834_Sunday_June_05_2022_08_41_22_AM_22239509/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 3
Problem number: 78.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind"
Maple gives the following as the ode type
[_Abel]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-\left (a \,{\mathrm e}^{x}+y\right ) y^{2}=0} \]
This is Abel first kind ODE, it has the form \[ y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \] Comparing the above to given ODE which is \begin {align*} y^{\prime }&=y^{3}+{\mathrm e}^{x} y^{2} a\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= 0\\ f_1(x) &= 0\\ f_2(x) &= a \,{\mathrm e}^{x}\\ f_3(x) &= 1 \end {align*}
Since \(f_2(x)=a \,{\mathrm e}^{x}\) is not zero, then the first step is to apply the following transformation to remove \(f_2\). Let \(y = u(x) - \frac {f_2}{3 f_3}\) or \begin {align*} y &= u(x) - \left ( \frac {a \,{\mathrm e}^{x}}{3} \right ) \\ &= u \left (x \right )-\frac {a \,{\mathrm e}^{x}}{3} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} u^{\prime }\left (x \right ) = \frac {a \,{\mathrm e}^{x}}{3}-\frac {a^{2} {\mathrm e}^{2 x} u \left (x \right )}{3}+\frac {2 a^{3} {\mathrm e}^{3 x}}{27}+u \left (x \right )^{3}\tag {2} \end {align*}
This is Abel first kind ODE, it has the form \[ u^{\prime }\left (x \right )= f_0(x)+f_1(x) u \left (x \right ) +f_2(x)u \left (x \right )^{2}+f_3(x)u \left (x \right )^{3} \] Comparing the above to given ODE which is \begin {align*} u^{\prime }\left (x \right )&=\frac {a \,{\mathrm e}^{x}}{3}-\frac {a^{2} {\mathrm e}^{2 x} u \left (x \right )}{3}+\frac {2 a^{3} {\mathrm e}^{3 x}}{27}+u \left (x \right )^{3}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \frac {a \,{\mathrm e}^{x}}{3}+\frac {2 a^{3} {\mathrm e}^{3 x}}{27}\\ f_1(x) &= -\frac {{\mathrm e}^{2 x} a^{2}}{3}\\ f_2(x) &= 0\\ f_3(x) &= 1 \end {align*}
Since \(f_2(x)=0\) then we check the Abel invariant to see if it depends on \(x\) or not. The Abel invariant is given by \begin {align*} -\frac {f_{1}^{3}}{f_{0}^{2} f_{3}} \end {align*}
Which when evaluating gives \begin {align*} -\frac {\left (-\frac {a \,{\mathrm e}^{x}}{3}-\frac {2 a^{3} {\mathrm e}^{3 x}}{9}-\left (\frac {a \,{\mathrm e}^{x}}{3}+\frac {2 a^{3} {\mathrm e}^{3 x}}{27}\right ) {\mathrm e}^{2 x} a^{2}\right )^{3}}{27 \left (\frac {a \,{\mathrm e}^{x}}{3}+\frac {2 a^{3} {\mathrm e}^{3 x}}{27}\right )^{5}} \end {align*}
Since the Abel invariant depends on \(x\) then unable to solve this ode at this time.
Unable to complete the solution now.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\left (a \,{\mathrm e}^{x}+y\right ) y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (a \,{\mathrm e}^{x}+y\right ) y^{2} \end {array} \]
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 trying Abel <- Abel successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 62
dsolve(diff(y(x),x) = (a*exp(x)+y(x))*y(x)^2,y(x), singsol=all)
\[ \frac {a \,\operatorname {erf}\left (\frac {\left (a \,{\mathrm e}^{x} y \left (x \right )+1\right ) \sqrt {2}}{2 y \left (x \right )}\right ) \sqrt {2}\, \sqrt {\pi }+2 c_{1} a +2 \,{\mathrm e}^{-x -\frac {\left (a \,{\mathrm e}^{x} y \left (x \right )+1\right )^{2}}{2 y \left (x \right )^{2}}}}{2 a} = 0 \]
✓ Solution by Mathematica
Time used: 0.702 (sec). Leaf size: 78
DSolve[y'[x]==(a Exp[x]+y[x])y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-i a e^x=\frac {2 e^{\frac {1}{2} \left (-i a e^x-\frac {i}{y(x)}\right )^2}}{\sqrt {2 \pi } \text {erfi}\left (\frac {-i a e^x-\frac {i}{y(x)}}{\sqrt {2}}\right )+2 c_1},y(x)\right ] \]