What Type of Differential Equation Is This?

Writing $z = y+x$ we have $\frac{dz}{dx} \; = \; e^{z-x}$ so that $e^{-z} \frac{dz}{dx} \; = \; e^{-x}$ and hence $e^{-z} = e^{-x} + c$ Since $y(0)=-2$ , we deduce that $c = e^2-1$ , and so $\begin{aligned} e^{-y} & = \; 1 + (e^2-1)e^x \\ -y & \approx \; x + \ln(e^2-1) \end{aligned}$ so that $m=-1$ and $n = -\ln(e^2-1)$ , making the answer $\ln(e^2-1) = \boxed{1.854586542}$

This could be solve in the traditional way:

$\frac{dy}{dx}\; = \; e^{y}-1$ $\frac{dy}{e^{y}-1}\; = \; dx$

Now integral the right side we can get x, but for the left side, we need to do the following u-substitution:

Let $u=e^{y}-1$ , $du=e^{y}dy$ , $dy=\frac{du}{e^{y}}\; = \frac{du}{u+1}$

Therefore the equation becomes $\frac{du}{u(u+1)}\; = \; dx$

Applying Praction Fraction, we get $\frac{1}{u}-\frac{1}{u+1}\ du = \; dx$

Integrate both side, we have $ln\frac{u}{u+1}=x+C$

Substitute $u=e^y-1$ , and simplify a bit, we get $ln(1-e^{-y})=x+C$

By putting $x=0$ and $y=-2$ , we can easily see $C=ln(1-e^2)$

Therefore, $ln(1-e^{-y})=x+ln(1-e^2)=ln[(1-e^2)e^x]$

Keep simplifying, we finally get $e^{-y}=1-(1-e^2)e^x$

$y=-ln[1-(1-e^2)e^x]$

Now we start to find out $m$ and $n$ to answer the question:

$m=lim_{x \to \infty} \frac{-ln[1-(1-e^2)e^x]}{x}$

Applying L'Hospital Rule, we get $m=lim_{x \to \infty} \frac{(1-e^2)e^x}{1-(1-e^2)e^x}=lim_{x \to \infty} [-\frac{1-(1-e^2)e^x-1}{1-(1-e^2)e^x}]=lim_{x \to \infty} [-1+\frac{1}{1-(1-e^2)e^x}]=-1$

Now solve for $n$ :

$n=lim_{x \to \infty} [-ln[1-(1-e^2)e^x]+x]=lim_{x \to \infty} ln\frac{e^x}{1-(1-e^2)e^x}=ln [lim_{x \to \infty} \frac{e^x}{1-(1-e^2)e^x}]$

Notice that we don't have to re-comput the limit since this limit is almost the same as the limit with m, just a factor of $1-e^2$ off.

Therefore, $n=ln \frac{-1}{1-e^2}=-ln(e^2-1)$

Finally, $mn=(-1)(-ln(e^2-1))=ln(e^2-1)=\boxed{1.854586542}$

P.S. What Type of Differential Equation IS This? This is a Separable First-order Differential Equation :)

$\begin{aligned} e^y&=\frac{dy}{dx}+1\\ y&=\ln\left(\frac{dy}{dx}+1\right) \end{aligned}$

Now, differentiate both sides and substitute $v=\frac{dy}{dx}$ : $\begin{aligned} \frac{dy}{dx}&=\frac{d^2y}{dx^2}\cdot\frac{1}{\frac{dy}{dx}+1}\\ v&=\frac{dv}{dx}\cdot\frac{1}{v+1}\\ v\left(v+1\right)&=\frac{dv}{dx} \end{aligned}$

Now, this is simply a separable differential equation. Solve it to find out that $v=\frac{Ce^x}{1+Ce^x}$

Integrate this to get that $y=D-\ln\left(1+Ce^x\right)$

Substitute this into the original equation to see that $D=0$ . In addition, since $y(0)=-2$ , $C=e^2-1$ : $y=-\ln\left(1+\left(e^2-1\right)e^x\right)$

To find the oblique asymptotes $y=mx+n$ for some function $f(x)$ , use the following formulas (where $a$ is one of $\infty$ or $-\infty$ ): $\begin{aligned} \displaystyle m&=\lim_{x\to a}\frac{f(x)}{x}\\ n&=\lim_{x\to a}\left(f(x)-mx\right) \end{aligned}$ (These formulas basically state that $m$ is the slope of the function $f(x)$ as $x\to\pm\infty$ , and the y-intercept $n$ is the difference between $f(x)$ and $mx$ as $x\to\pm\infty$ .)

to get that $m=-1,n=-\ln\left(e^2-1\right)$ . Thus, $mn=\ln\left(e^2-1\right)$ .