Internal problem ID [4341]
Internal file name [OUTPUT/3834_Sunday_June_05_2022_11_18_03_AM_52578100/index.tex
]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 37
Problem number: 1149.
ODE order: 1.
ODE degree: 0.
The type(s) of ODE detected by this program : "separable", "homogeneousTypeD2"
Maple gives the following as the ode type
[_separable]
\[ \boxed {y \ln \left (y^{\prime }\right )+y^{\prime }-y \ln \left (y\right )-x y=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= {\mathrm e}^{-\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+x} y \end {align*}
Where \(f(x)={\mathrm e}^{-\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+x}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= {\mathrm e}^{-\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+x} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {{\mathrm e}^{-\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+x} \,d x}\\ \ln \left (y \right )&=\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\operatorname {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}+c_{1}\\ y&={\mathrm e}^{\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\operatorname {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}+c_{1}}\\ &=c_{1} {\mathrm e}^{\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\operatorname {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\operatorname {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\operatorname {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}} \] Verified OK.
Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} u \left (x \right ) x \ln \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right )+u^{\prime }\left (x \right ) x +u \left (x \right )-u \left (x \right ) x \ln \left (u \left (x \right ) x \right )-x^{2} u \left (x \right ) = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u \left (x \operatorname {LambertW}\left ({\mathrm e}^{x}\right )-1\right )}{x} \end {align*}
Where \(f(x)=\frac {x \operatorname {LambertW}\left ({\mathrm e}^{x}\right )-1}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= \frac {x \operatorname {LambertW}\left ({\mathrm e}^{x}\right )-1}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {\frac {x \operatorname {LambertW}\left ({\mathrm e}^{x}\right )-1}{x} \,d x}\\ \ln \left (u \right )&=\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\operatorname {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}-\ln \left (x \right )+c_{2}\\ u&={\mathrm e}^{\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\operatorname {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}-\ln \left (x \right )+c_{2}}\\ &=c_{2} {\mathrm e}^{\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\operatorname {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}-\ln \left (x \right )} \end {align*}
Therefore the solution \(y\) is \begin {align*} y&=u x\\ &=x c_{2} {\mathrm e}^{\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\operatorname {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}-\ln \left (x \right )} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= x c_{2} {\mathrm e}^{\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\operatorname {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}-\ln \left (x \right )} \\ \end{align*}
Verification of solutions
\[ y = x c_{2} {\mathrm e}^{\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\operatorname {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}-\ln \left (x \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \ln \left (y^{\prime }\right )+y^{\prime }-y \ln \left (y\right )-x y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{-\mathit {LambertW}\left ({\mathrm e}^{x}\right )+x} y \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y}={\mathrm e}^{-\mathit {LambertW}\left ({\mathrm e}^{x}\right )+x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y}d x =\int {\mathrm e}^{-\mathit {LambertW}\left ({\mathrm e}^{x}\right )+x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (y\right )=\mathit {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\mathit {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{\mathit {LambertW}\left ({\mathrm e}^{x}\right )+\frac {\mathit {LambertW}\left ({\mathrm e}^{x}\right )^{2}}{2}+c_{1}} \end {array} \]
Maple trace
`Methods for first order ODEs: -> Solving 1st order ODE of high degree, 1st attempt trying 1st order WeierstrassP solution for high degree ODE trying 1st order WeierstrassPPrime solution for high degree ODE trying 1st order JacobiSN solution for high degree ODE trying 1st order ODE linearizable_by_differentiation trying differential order: 1; missing variables trying simple symmetries for implicit equations --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 17
dsolve(y(x)*ln(diff(y(x),x))+diff(y(x),x)-y(x)*ln(y(x))-x*y(x) = 0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} {\mathrm e}^{\frac {\operatorname {LambertW}\left ({\mathrm e}^{x}\right ) \left (\operatorname {LambertW}\left ({\mathrm e}^{x}\right )+2\right )}{2}} \]
✓ Solution by Mathematica
Time used: 0.093 (sec). Leaf size: 24
DSolve[y[x] Log[y'[x]] + y'[x] -y[x] Log[y[x]] -x y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 e^{\frac {1}{2} W\left (e^x\right ) \left (W\left (e^x\right )+2\right )} \]