2.1.23 problem 24
Internal
problem
ID
[8161]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
24
Date
solved
:
Sunday, November 10, 2024 at 03:06:05 AM
CAS
classification
:
[_separable]
Solve
\begin{align*} y&=x y^{\prime }+x^{2} {y^{\prime }}^{2} \end{align*}
Solving for the derivative gives these ODE’s to solve
\begin{align*}
\tag{1} y^{\prime }&=\frac {-\frac {1}{2}+\frac {\sqrt {1+4 y}}{2}}{x} \\
\tag{2} y^{\prime }&=\frac {-\frac {1}{2}-\frac {\sqrt {1+4 y}}{2}}{x} \\
\end{align*}
Now each of the above is solved
separately.
Solving Eq. (1)
The ode \(y^{\prime } = \frac {-1+\sqrt {1+4 y}}{2 x}\) is separable as it can be written as
\begin{align*} y^{\prime }&= \frac {-1+\sqrt {1+4 y}}{2 x}\\ &= f(x) g(y) \end{align*}
Where
\begin{align*} f(x) &= \frac {1}{x}\\ g(y) &= -\frac {1}{2}+\frac {\sqrt {1+4 y}}{2} \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx}\\ \int { \frac {1}{-\frac {1}{2}+\frac {\sqrt {1+4 y}}{2}}\,dy} &= \int { \frac {1}{x} \,dx}\\ \sqrt {1+4 y}+\ln \left (-1+\sqrt {1+4 y}\right )&=\ln \left (x \right )+c_1 \end{align*}
We now need to find the singular solutions, these are found by finding for what values \(g(y)\) is
zero, since we had to divide by this above. Solving \(g(y)=0\) or \(-\frac {1}{2}+\frac {\sqrt {1+4 y}}{2}=0\) for \(y\) gives
\begin{align*} y&=0 \end{align*}
Now we go over each such singular solution and check if it verifies the ode itself and
any initial conditions given. If it does not then the singular solution will not be
used.
Therefore the solutions found are
\begin{align*} \sqrt {1+4 y}+\ln \left (-1+\sqrt {1+4 y}\right ) = \ln \left (x \right )+c_1\\ y = 0 \end{align*}
Solving for \(y\) gives
\begin{align*}
y &= \frac {\operatorname {LambertW}\left (x \,{\mathrm e}^{c_1 -1}\right )^{2}}{4}+\frac {\operatorname {LambertW}\left (x \,{\mathrm e}^{c_1 -1}\right )}{2} \\
\end{align*}
We now need to find the singular solutions, these are found by finding for
what values \((\frac {-\frac {1}{2}+\frac {\sqrt {1+4 y}}{2}}{x})\) is zero. These give
\begin{align*}
y&=0 \\
\end{align*}
Now we go over each such singular solution and check if it
verifies the ode itself and any initial conditions given. If it does not then the singular solution
will not be used.
The solution \(y = 0\) satisfies the ode and initial conditions.
Solving Eq. (2)
The ode \(y^{\prime } = -\frac {1+\sqrt {1+4 y}}{2 x}\) is separable as it can be written as
\begin{align*} y^{\prime }&= -\frac {1+\sqrt {1+4 y}}{2 x}\\ &= f(x) g(y) \end{align*}
Where
\begin{align*} f(x) &= \frac {1}{x}\\ g(y) &= -\frac {\sqrt {1+4 y}}{2}-\frac {1}{2} \end{align*}
Integrating gives
\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx}\\ \int { \frac {1}{-\frac {\sqrt {1+4 y}}{2}-\frac {1}{2}}\,dy} &= \int { \frac {1}{x} \,dx}\\ -\sqrt {1+4 y}+\ln \left (1+\sqrt {1+4 y}\right )&=\ln \left (x \right )+c_2 \end{align*}
Solving for \(y\) gives
\begin{align*}
y &= \frac {\operatorname {LambertW}\left (-x \,{\mathrm e}^{c_2 -1}\right )^{2}}{4}+\frac {\operatorname {LambertW}\left (-x \,{\mathrm e}^{c_2 -1}\right )}{2} \\
\end{align*}
Maple step by step solution
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (x \right )=x \left (\frac {d}{d x}y \left (x \right )\right )+x^{2} \left (\frac {d}{d x}y \left (x \right )\right )^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=\frac {-\frac {1}{2}-\frac {\sqrt {1+4 y \left (x \right )}}{2}}{x}, \frac {d}{d x}y \left (x \right )=\frac {-\frac {1}{2}+\frac {\sqrt {1+4 y \left (x \right )}}{2}}{x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {-\frac {1}{2}-\frac {\sqrt {1+4 y \left (x \right )}}{2}}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{-\frac {1}{2}-\frac {\sqrt {1+4 y \left (x \right )}}{2}}=\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{-\frac {1}{2}-\frac {\sqrt {1+4 y \left (x \right )}}{2}}d x =\int \frac {1}{x}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y \left (x \right )\right )}{2}-\sqrt {1+4 y \left (x \right )}-\frac {\ln \left (-1+\sqrt {1+4 y \left (x \right )}\right )}{2}+\frac {\ln \left (1+\sqrt {1+4 y \left (x \right )}\right )}{2}=\ln \left (x \right )+\textit {\_C1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {-\frac {1}{2}+\frac {\sqrt {1+4 y \left (x \right )}}{2}}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{-\frac {1}{2}+\frac {\sqrt {1+4 y \left (x \right )}}{2}}=\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{-\frac {1}{2}+\frac {\sqrt {1+4 y \left (x \right )}}{2}}d x =\int \frac {1}{x}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y \left (x \right )\right )}{2}+\sqrt {1+4 y \left (x \right )}+\frac {\ln \left (-1+\sqrt {1+4 y \left (x \right )}\right )}{2}-\frac {\ln \left (1+\sqrt {1+4 y \left (x \right )}\right )}{2}=\ln \left (x \right )+\textit {\_C1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\frac {\ln \left (y \left (x \right )\right )}{2}-\sqrt {1+4 y \left (x \right )}-\frac {\ln \left (-1+\sqrt {1+4 y \left (x \right )}\right )}{2}+\frac {\ln \left (1+\sqrt {1+4 y \left (x \right )}\right )}{2}=\ln \left (x \right )+\mathit {C1} , \frac {\ln \left (y \left (x \right )\right )}{2}+\sqrt {1+4 y \left (x \right )}+\frac {\ln \left (-1+\sqrt {1+4 y \left (x \right )}\right )}{2}-\frac {\ln \left (1+\sqrt {1+4 y \left (x \right )}\right )}{2}=\ln \left (x \right )+\mathit {C1} \right \} \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 dAlembert
trying simple symmetries for implicit equations
<- symmetries for implicit equations successful`
Maple dsolve solution
Solving time : 0.048 (sec)
Leaf size : 97
dsolve(y(x) = diff(y(x),x)*x+x^2*diff(y(x),x)^2,
y(x),singsol=all)
\begin{align*}
\ln \left (x \right )-\sqrt {4 y+1}-\frac {\ln \left (-1+\sqrt {4 y+1}\right )}{2}+\frac {\ln \left (1+\sqrt {4 y+1}\right )}{2}-\frac {\ln \left (y\right )}{2}-c_{1} &= 0 \\
\ln \left (x \right )+\sqrt {4 y+1}+\frac {\ln \left (-1+\sqrt {4 y+1}\right )}{2}-\frac {\ln \left (1+\sqrt {4 y+1}\right )}{2}-\frac {\ln \left (y\right )}{2}-c_{1} &= 0 \\
\end{align*}
Mathematica DSolve solution
Solving time : 18.695 (sec)
Leaf size : 72
DSolve[{y[x]==x*D[y[x],x]+x^2*(D[y[x],x])^2,{}},
y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to \frac {1}{4} W\left (-e^{-1-2 c_1} x\right ) \left (2+W\left (-e^{-1-2 c_1} x\right )\right ) \\
y(x)\to \frac {1}{4} W\left (e^{-1+2 c_1} x\right ) \left (2+W\left (e^{-1+2 c_1} x\right )\right ) \\
y(x)\to 0 \\
\end{align*}