2.5.1.2 Solved using first_order_ode_parametric method

Entering first order ode parametric solver

\[\begin {aligned} x {y^{\prime }}^{2}-y+2 y^{\prime }&=0 \end {aligned}\]

Let \(y^{\prime }\) be a parameter \(\lambda \). The ode becomes

\begin{align*} x \,\lambda ^{2}+2 \lambda -y = 0 \end{align*}

Isolating \(y\) gives

\begin{align*} y&=x \,\lambda ^{2}+2 \lambda \\ &=x \,\lambda ^{2}+2 \lambda \\ &=F \left (x , \lambda \right ) \end{align*}

Now we generate an ode in \(x \left (\lambda \right )\) using

\begin{align*} \frac {d}{d \lambda }x \left (\lambda \right ) &= \frac { \frac {\partial F}{\partial \lambda }} { \lambda -\frac {\partial F}{\partial x} } \\ &= \frac {2 \lambda x +2}{-\lambda ^{2}+\lambda }\\ &= \frac {-2-2 \lambda x \left (\lambda \right )}{\lambda \left (\lambda -1\right )} \end{align*}

Which is now solved for \(x\).

Entering first order ode linear solverIn canonical form a linear first order is

\begin{align*} \frac {d}{d \lambda }x \left (\lambda \right ) + q(\lambda )x \left (\lambda \right ) &= p(\lambda ) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(\lambda ) &=\frac {2}{\lambda -1}\\ p(\lambda ) &=-\frac {2}{\lambda \left (\lambda -1\right )} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,d\lambda }}\\ &= {\mathrm e}^{\int \frac {2}{\lambda -1}d \lambda }\\ &= \left (\lambda -1\right )^{2} \end{align*}

The ode becomes

\begin{gather*} \begin {aligned} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\lambda }}\left ( \mu x\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\lambda }}\left ( \mu x\right ) &= \left (\mu \right ) \left (-\frac {2}{\lambda \left (\lambda -1\right )}\right )\\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\lambda }} \left (x \left (\lambda -1\right )^{2}\right ) &= \left (\left (\lambda -1\right )^{2}\right ) \left (-\frac {2}{\lambda \left (\lambda -1\right )}\right )\\ \mathrm {d} \left (x \left (\lambda -1\right )^{2}\right ) &= \left (-\frac {2 \left (\lambda -1\right )}{\lambda }\right )\, \mathrm {d} \lambda \end {aligned} \end{gather*}

Integrating gives

\begin{align*} x \left (\lambda -1\right )^{2}&= \int {-\frac {2 \left (\lambda -1\right )}{\lambda } \,d\lambda } \\ &=-2 \lambda +2 \ln \left (\lambda \right ) + c_1 \end{align*}

Dividing throughout by the integrating factor \(\left (\lambda -1\right )^{2}\) gives the final solution

\[ x \left (\lambda \right ) = \frac {-2 \lambda +2 \ln \left (\lambda \right )+c_1}{\left (\lambda -1\right )^{2}} \]

Now that we found solution \(x\) we have two equations with parameter \(\lambda \). They are

\begin{gather*} \begin {aligned} y &= x \,\lambda ^{2}+2 \lambda \\ x &= \frac {-2 \lambda +2 \ln \left (\lambda \right )+c_1}{\left (\lambda -1\right )^{2}} \end {aligned} \end{gather*}

Eliminating \(\lambda \) gives the solution for \(y\).

\begin{align*} -y+2 \operatorname {RootOf}\left (x \,\textit {\_Z}^{2}+2 \textit {\_Z} -y\right ) x +2 \ln \left (\operatorname {RootOf}\left (x \,\textit {\_Z}^{2}+2 \textit {\_Z} -y\right )\right )+c_1 -x \end{align*}

Which can be written as

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

Simplifying \(-y-2-2 \sqrt {y x +1}+2 \ln \left (-\frac {1+\sqrt {y x +1}}{x}\right )+c_1 -x = 0\) gives

\[ -y-2-2 \sqrt {y x +1}+2 \ln \left (\frac {-1-\sqrt {y x +1}}{x}\right )+c_1 -x = 0 \]

Isoclines

Summary of solutions found

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