Parametric solution of yy'-y=A

Question

I am learning the basics of differential equations and in a course, we were asked to solve parametrically the equation, $$yy'-y=A,\ y=y(x)$$ using the substitution $y'=t$.

I know the above equation can be solved by separating variables. I think that the substitution suggests that $t$ is a function of $x$ just as $y'$ and so, the equation becomes $$y(x)t(x)-y(x)=A\rightarrow y(x)=\frac{A}{t(x)-1}\rightarrow y(t)=\frac{A}{t-1}$$

From here one should propably use the chain rule to find $x(t)$, $$\frac{dy(x)}{dx}\frac{dx(t)}{dt}=t\frac{dx(t)}{dt}$$ but I do not see how. I could use some help to finish the problem and some insight into what is really going on with the variables, which are independent, which are dependent and if the above method for finding $y(t)$ is valid.

Answer 1

$$yy'-y=A$$ $$y=\frac{A}{y'-1}$$ With $t=y'=\frac{dy}{dx}$ $$y=\frac{A}{t-1}\quad\implies\quad \frac{dy}{dt}=-\frac{A}{(t-1)^2}$$ $$ \frac{dy}{dt}= \frac{dy}{dx} \frac{dx}{dt}=t \frac{dx}{dt}$$ $$t \frac{dx}{dt}=-\frac{A}{(t-1)^2}$$ This is a first order separable ODE with respect to the unknown $x(t)$. $$x=-A\int \frac{dt}{t(t-1)^2}+\text{constant}$$ The solution on parametric form is : $$\boxed{\begin{cases} x(t)=\frac{A}{t-1}+A\ln\left|\frac{t-1}{t}\right|+C \\ y(t)=\frac{A}{t-1} \end{cases}}$$ NOTE :

The explicite form of the solution $y(x)$ involves the Lambert W function : $$y(x)=-A\left(1+W\left(-\frac{1 }{A}e^{\frac{C-x}{A}} \right)\right)$$

Answer 2

You are looking for a parametric solution $\left(x\left(t\right),y\left(t\right)\right)$ with $\frac{{\rm d}y}{{\rm d}x}\left(t\right)=t$. This is a valid assumption because normally you can reparametrize a curve to fulfill this conditon. This means $y^{\prime}=\frac{\dot{y}}{\dot{x}}=t$ (where $\dot{x}$ means $\frac{{\rm d}x}{{\rm d}t}$). Moreover, from the equation itself

$$yy'-y=A\Longrightarrow yt-y=A\Longrightarrow y=\dfrac{A}{t-1}$$

Plug this into $\dot{x}=\frac{\dot{y}}{t}$ to have

$$\dot{x}=-\dfrac{A}{t\left(t-1\right)^{2}}\Longrightarrow x-C=-A\ln\left(\dfrac{t}{t-1}\right)-\dfrac{A}{t-1}$$

You can easily substitute $t-1=\frac{A}{y}$ inside to get rid of $t$

$$x-C=-A\ln\left(\dfrac{\frac{A}{y}+1}{\frac{A}{y}}\right)+y=-A\ln\left(1+\frac{y}{A}\right)+y$$

Inverting this expression requires Lambert $W$ function (as attested by Wolfram and confirmed to be a solution) and is therefore highly not recommended. You can, however, use implicit differentiation to show

$$1=-A\dfrac{1}{A}\dfrac{1}{1+\frac{y}{A}}y^{\prime}+y^{\prime}$$

$$1=\left[-\dfrac{A}{A+y}+1\right]y^{\prime}$$

$$y+A=yy^{\prime}$$

$$yy^{\prime}-y=A$$

which is the same differential equation we started with.