Math Scrap 4

Solve the ODE:

-y \,\diff{y} + \parens{x + \sqrt{xy}} \,\diff{y} = 0

Rearranging the ODE:

\frac{\diff{x}}{\diff{y}} = \frac{x + \sqrt{xy}}{y} = \frac{x}{y} + \frac{\sqrt{x}}{\sqrt{y}}

From this form, we can try the following substitution:

z := \frac{x}{y} \qquad \Longrightarrow \qquad x(y) = z(y) y \qquad \Longrightarrow \qquad \frac{\diff{x}}{\diff{y}} = z \xcancelto{1}{\frac{\diff{y}}{\diff{y}}} + y \frac{\diff{z}}{\diff{y}} = z + y \frac{\diff{z}}{\diff{y}}

Substituting into our ODE, rearranging, then integrating both sides:

\begin{gather*} z + y \frac{\diff{z}}{\diff{y}} = \frac{zy + \sqrt{z y^2}}{y} = \frac{z \cancel{y} + \cancel{y} \sqrt{z}}{\cancel{y}} = z + \sqrt{z} \\ y \frac{\diff{z}}{\diff{y}} = \sqrt{z} \\ z^{-\frac{1}{2}} \,\diff{z} = \frac{1}{y} \,\diff{y} \\ 2 \sqrt{z} = \ln{\abs{y}} + C \end{gather*}

Reverting the substitution, we get our solution to the ODE:

2 \frac{\sqrt{x}}{\sqrt{y}} = \ln{\abs{y}} + C, \qquad C \in \mathbb{R}

OPTIONAL: We shall now verify our solution.

Rearranging our solution:

\begin{gather*} \sqrt{x} = \frac{1}{2} \sqrt{y} \parens{\ln{\abs{y}} + C} \qquad \Longrightarrow \qquad x = \frac{1}{4} y \parens{\ln{\abs{y}} + C}^2 \\ \begin{aligned} 4 \frac{\diff{x}}{\diff{y}} &= y \frac{\diff{}}{\diff{y}}\parens{\parens{\ln{\abs{y}} + C}^2} + \parens{\ln{\abs{y}} + C}^2 \xcancelto{1}{\frac{\diff{y}}{\diff{y}}} \\ &= 2 y \parens{\ln{\abs{y}} + C} \frac{1}{y} + \parens{\ln{\abs{y}} + C}^2 \end{aligned} \end{gather*}

Thus, LHS of our ODE is:

\lhs = \frac{1}{2} \parens{\ln{\abs{y}} + C} + \frac{1}{4} \parens{\ln{\abs{y}} + C}^2

For the RHS, we substitute our solution:

\begin{align*} \rhs &= \frac{\frac{1}{4} \cancel{y} \parens{\ln{\abs{y}} + C}^2}{\cancel{y}} + \sqrt{\frac{\frac{1}{4} \cancel{y} \parens{\ln{\abs{y}} + C}^2}{\cancel{y}}} = \frac{1}{4} \parens{\ln{\abs{y}} + C}^2 + \sqrt{\frac{1}{4} \parens{\ln{\abs{y}} + C}^2} \\ &= \frac{1}{4} \parens{\ln{\abs{y}} + C}^2 + \frac{1}{2} \parens{\ln{\abs{y}} + C} = \lhs \end{align*}

Thus, our solution has been verified.