1.23 problem 24

Internal problem ID [7715]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 24
Date solved : Monday, October 21, 2024 at 03:59:03 PM
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 \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y&=0\\ 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 \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\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*}

1.23.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y=x y^{\prime }+x^{2} {y^{\prime }}^{2} \\ \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} \\ {} & {} & \left [y^{\prime }=\frac {-\frac {1}{2}-\frac {\sqrt {1+4 y}}{2}}{x}, y^{\prime }=\frac {-\frac {1}{2}+\frac {\sqrt {1+4 y}}{2}}{x}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {-\frac {1}{2}-\frac {\sqrt {1+4 y}}{2}}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{-\frac {1}{2}-\frac {\sqrt {1+4 y}}{2}}=\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{-\frac {1}{2}-\frac {\sqrt {1+4 y}}{2}}d x =\int \frac {1}{x}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y\right )}{2}-\sqrt {1+4 y}-\frac {\ln \left (-1+\sqrt {1+4 y}\right )}{2}+\frac {\ln \left (1+\sqrt {1+4 y}\right )}{2}=\ln \left (x \right )+\textit {\_C1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {-\frac {1}{2}+\frac {\sqrt {1+4 y}}{2}}{x} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{-\frac {1}{2}+\frac {\sqrt {1+4 y}}{2}}=\frac {1}{x} \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{-\frac {1}{2}+\frac {\sqrt {1+4 y}}{2}}d x =\int \frac {1}{x}d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y\right )}{2}+\sqrt {1+4 y}+\frac {\ln \left (-1+\sqrt {1+4 y}\right )}{2}-\frac {\ln \left (1+\sqrt {1+4 y}\right )}{2}=\ln \left (x \right )+\textit {\_C1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\frac {\ln \left (y\right )}{2}-\sqrt {1+4 y}-\frac {\ln \left (-1+\sqrt {1+4 y}\right )}{2}+\frac {\ln \left (1+\sqrt {1+4 y}\right )}{2}=\ln \left (x \right )+\mathit {C1} , \frac {\ln \left (y\right )}{2}+\sqrt {1+4 y}+\frac {\ln \left (-1+\sqrt {1+4 y}\right )}{2}-\frac {\ln \left (1+\sqrt {1+4 y}\right )}{2}=\ln \left (x \right )+\mathit {C1} \right \} \end {array} \]

1.23.2 Maple trace

Methods for first order ODEs:
 

1.23.3 Maple dsolve solution

Solving time : 0.121 (sec)
Leaf size : 97

dsolve(y(x) = x*diff(y(x),x)+x^2*diff(y(x),x)^2, 
       y(x),singsol=all)
 

\begin{align*} \ln \left (x \right )-\sqrt {1+4 y}-\frac {\ln \left (-1+\sqrt {1+4 y}\right )}{2}+\frac {\ln \left (1+\sqrt {1+4 y}\right )}{2}-\frac {\ln \left (y\right )}{2}-c_1 &= 0 \\ \ln \left (x \right )+\sqrt {1+4 y}+\frac {\ln \left (-1+\sqrt {1+4 y}\right )}{2}-\frac {\ln \left (1+\sqrt {1+4 y}\right )}{2}-\frac {\ln \left (y\right )}{2}-c_1 &= 0 \\ \end{align*}

1.23.4 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*}