Question

Verify that the indicated expression is an implicit solution of the given first order differential equation.Find at least one explicit solution y=Φ(x) in each case. Use a graphing utility to obtain the graph of an explicit solution.
Give an interval I of definition of each solution Φ.

2xydx+(x^2-y)dy=0 , -2x^2 y + y^2 =1

Answer 1

Answer on Question #70253, Math / Differential Equations

Verify that the indicated expression is an implicit solution of the given first order differential equation. Find at least one explicit solution $y = \Phi(x)$ in each case. Use a graphing utility to obtain the graph of an explicit solution.

Give an interval $I$ of definition of each solution $\Phi$.

$$2xydx + (x^2 - y)dy = 0, \quad -2x^2y + y^2 = 1$$

Solution

Implicit differentiation with respect to $x$

$$\begin{array}{l} \frac{d}{dx} \left[ -2x^2y + y^2 = 1 \right] \Rightarrow -4xy - 2x^2 \frac{dy}{dx} + 2y \frac{dy}{dx} = 0 \\ 2xy + x^2 \frac{dy}{dx} - y \frac{dy}{dx} = 0 \Rightarrow 2xy + (x^2 - y) \frac{dy}{dx} = 0 \\ 2xydx + (x^2 - y)dy = 0 \\ \end{array}$$

Hence, the indicated expression $-2x^2y + y^2 = 1$ is an implicit solution of the given first order differential equation $2xydx + (x^2 - y)dy = 0$.

Find at least one explicit solution $y = \Phi(x)$

$$\begin{array}{c} -2x^2y + y^2 = 1 \\ y^2 - 2x^2y + x^4 = 1 + x^4 \\ (y - x^2)^2 = 1 + x^4 \\ y - x^2 = \pm \sqrt{1 + x^4} \\ \end{array}$$ $$y_1 = x^2 - \sqrt{1 + x^4}, \quad y_2 = x^2 + \sqrt{1 + x^4}$$

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