Putting y = vx in a homogeneous equation dy/dx = F(x, y) gives xdv/dx = v^2 - 1. What is F(x, y)?

Question 29

HardOriginal question

Putting $y = vx$ in a homogeneous equation $\frac{dy}{dx} = F(x, y)$ gives $x\frac{dv}{dx} = v^{2} - 1$ . What is $F(x, y)$ ?

\frac{y^{2} + xy - x^{2}}{x^{2}}

Correct answerYour answer

\frac{y^{2} - x^{2}}{x^{2}}

Correct answerYour answer

\frac{y^{2} - xy - x^{2}}{x^{2}}

Correct answerYour answer

\frac{x^{2} + xy - y^{2}}{x^{2}}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$\frac{dy}{dx} = v + x\frac{dv}{dx} = v + v^{2} - 1$ .
With $v = \frac{y}{x}$ : $\frac{y}{x} + \frac{y^{2}}{x^{2}} - 1 = \frac{y^{2} + xy - x^{2}}{x^{2}}$ .

Worth noting

$\frac{y^{2} - x^{2}}{x^{2}}$ is $v^{2} - 1$ alone; $\frac{dy}{dx}$ is $v$ plus $x\frac{dv}{dx}$ .

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