Integral curve passing through two point

Question

An integral curve of the differential equation $(y-x)(dy/dx) = 1$ passes through $(0,0)$, and $(a,1)$, then what is $a$ equal to?

I have four options:

1. $2-1/e$
2. $1-1/e$
3. 1/e
4. 1+e

I know how to draw direction field and integral curves, but I'm not sure how to draw this one. I realise that the slope would increase when I'm coming close to $y=x$.

Also, is it true that the function won't be differentiable close to $y = x$?

Answer 1

$$(y(x)-x)y'(x)=1$$ $$y-x(y)=x'(y)$$ This first order linear ODE is easy to solve : $$x(y)=c\,e^{-y}+y-1$$ The condition $x(0)=0$ implies $c=1$ $$x(y)=e^{-y}+y-1$$ The condition $x(1)=a$ implies $x(1)=a=e^{-1}+1-1$ $$a=\frac{1}{e}$$

Answer 2

Let $z:=y-x$ so that $z'=y'-1$. This turns the equation to

$$z(z'+1)=1$$ or $$\frac z{1-z}z'=1,$$

$$-z-\log(1-z)=x+C,$$

$$-y-\log(1-y+x)=C$$ and from the initial condition, $C=0$.

Now we have to solve

$$-1-\log(1-1+a)=0$$ and $a=e^{-1}$.

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Alternatively:

$$\frac{dx}{dy}+x=y$$ or

$$\left(\frac{dx}{dy}+x\right)e^y=\frac d{dy}(xe^y)=ye^y$$

and integrating,

$$\left.xe^y\right|_{(0,0)}^{(a,1)}=a\,e-0=\int_0^1 ye^ydy=1.$$

Answer 3

$$(y-x)(dy/dx) = 1$$ Multipy by $\mu(x)=e^y$ $$(y-x)e^ydy-e^ydx =0 $$ The differential is exact $$-x+y-1-Ke^{-y}=0$$ $$y(0)=0 \implies K=-1$$ $$\boxed{x(y)=y-1+e^{-y}}$$ we have also that $y(a)=1$ $$\implies a=e^{-1}$$