ODE Homogeneous Equation Algebra

Question

Assignment Statement

Given \begin{equation} \frac{dy}{dx} = \frac{(y/x)-4}{1-(y/x)} \end{equation} Let $v = y/x$ a dependent variable such that $y(x)=x\ v(x)$ Express $dy/dx$ in terms of $x$ , $v$ and $dv/dx$ , then replace $y$ and $dy/dx$ in Equation.1 by the expression found which involves $v$ and $dv/dx$. Show the resulting differential equation is $$ x\frac{dv}{dx} = \frac{v^{2}-4}{1-v}$$

Attempt at Solution

First and foremost, since $y(x) = x v(x)$ then , \begin{align*} \frac{dv}{dx} &= \frac{d}{dx}\left(\frac{y(x)}{x}\right) \\ &=\frac{dy/dx}{x}-\frac{y(x)}{x^{2}}\\ x^{2}\frac{dv}{dx} &= x(dy/dx) - y(x)\\ \implies &\frac{dy}{dx} = x\frac{dv}{dx} + \frac{y(x)}{x} \end{align*} Here i'm then assuming that $y(x)/x = v \iff y/x = v$. Then I have an expression of $dy/dx$ involving $x$ , $v$ and $dv/dx$.

I then replace $y$ and $dy/dx$ in Equation 1 by this result as I'm asked to : $$ x\frac{dv}{dx} + v = \frac{\left(\frac{dv}{dx}+\frac{v}{x}\right)-4}{1-\left(\frac{dv}{dx}+\frac{v}{x}\right)}$$. So from here I must show that: $$ \frac{\left(\frac{dv}{dx}+\frac{v}{x}\right)-4}{1-\left(\frac{dv}{dx}+\frac{v}{x}\right)} + v =\frac{v^{2}-4}{1-v}$$ After many pages of algebra I'm stuck , the farthest I've got is $$ \frac{x-4x^{2}-vx^{2}+v^{2}}{x^{2}-v}$$

Can someone help me ? Thank you very much.

Answer 1

You wrote: \begin{align*} \frac{dv}{dx} &= \frac{d}{dx}\left(\frac{y(x)}{x}\right) \\ &=\frac{dy/dx}{x}-\frac{y(x)}{x^{2}}\\ x^{2}\frac{dv}{dx} &= x(dy/dx) - y(x)\\ \implies &\frac{dy}{dx} = x\frac{dv}{dx} + \frac{y(x)}{x} \end{align*} These lines are correct but you should conclude that: $$\frac{dy}{dx} = x\frac{dv}{dx} + \frac{y(x)}{x} $$ $$ \implies \frac{dy}{dx} = x\frac{dv}{dx} + v$$

Then transform your original equation into:

$$ \implies x\frac{dv}{dx} + v = \frac{v-4}{1-v}$$ Which becomes a separable DE.

Answer 2

Why try to compute $v'$ in terms of $y'$ when the expression provided begins with $y'$? We should get

$$y'=(xv)'=xv'+v$$

After this, when substituting expressions in, we should have $v=y/x$ so that

$$xv'+v=\frac{v-4}{1-v}$$

which, after a bit of algebra, becomes

$$xv'=\frac{v^2-4}{1-v}$$

as desired.

Answer 3

If $y=xv$ then by the product rule, $\dfrac{dy}{dx} = x\dfrac{dv}{dx} + v.$

\begin{align} \frac{dy}{dx} = {} & \frac{v-4}{1-v} \\[8pt] x \frac{dv}{dx} +v = {} & \frac{v-4}{1-v} \\[8pt] x\frac{dv}{dx} = {} & \frac{v-4}{1-v} -v \\[8pt] = {} & \frac{v^2 - 4}{1-v}. \end{align} After that you can separate variables: $$ \frac{1-v}{v^2-4} \, dv = \frac{dx} x. $$ Then integrate both sides, using partial fractions on the left side.