2.1.14.3 Solved using first_order_ode_homog_type_G

Entering first order ode homog type G solver

\[\begin {aligned} y^{\prime } c&=a x \end {aligned}\]

Multiplying the right side of the ode, which is \(\frac {x a}{c}\) by \(\frac {x}{y}\) gives

\begin{align*} y^{\prime } &= \left (\frac {x}{y}\right ) \frac {x a}{c}\\ &= \frac {x^{2} a}{y c}\\ &= F(x,y) \end{align*}

Since \(F \left (x , y\right )\) has \(y\), then let

\begin{align*} f_x&= x \left (\frac {\partial }{\partial x}F \left (x , y\right )\right )\\ &= \frac {2 x^{2} a}{y c}\\ f_y&= y \left (\frac {\partial }{\partial y}F \left (x , y\right )\right )\\ &= -\frac {x^{2} a}{y c}\\ \alpha &= \frac {f_x}{f_y} \\ &=-2 \end{align*}

Since \(\alpha \) is independent of \(x,y\) then this is Homogeneous type G. Let

\begin{align*} y&=\frac {z}{x^ \alpha }\\ &=\frac {z}{\frac {1}{x^{2}}} \end{align*}

Substituting the above back into \(F(x,y)\) gives

\begin{align*} F \left (z \right ) &=\frac {a}{z c} \end{align*}

The above shows that \(F \left (z \right )\) does not depend on \(x\) nor on \(y\). If this was not the case, then this method will not work.

Therefore, the implicit solution is given by

\begin{align*} \ln \left (x \right )- c_1 - \int ^{y x^\alpha } \frac {1}{z \left (\alpha + F(z)\right ) } \,dz & = 0 \end{align*}

Which gives

\begin{align*} \ln \left (x \right )-c_1 +\int _{}^{\frac {y}{x^{2}}}\frac {1}{z \left (2-\frac {a}{z c}\right )}d z = 0 \end{align*}

The value of the above is

\begin{align*} \ln \left (x \right )-c_1 +\frac {\ln \left (\frac {-x^{2} a +2 c y}{x^{2}}\right )}{2} = 0 \end{align*}

Solving for \(y\) from \(\ln \left (x \right )-c_1 +\frac {\ln \left (\frac {-x^{2} a +2 c y}{x^{2}}\right )}{2} = 0\) gives

\begin{align*} y &= \frac {x^{2} a +{\mathrm e}^{2 c_1}}{2 c} \\ \end{align*}

Summary of solutions found

\[ y = \frac {x^{2} a +{\mathrm e}^{2 c_1}}{2 c} \]