Solving The Differential Equation y'=e^{x-y}

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#CalculusProblems #DifferentialEquations

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[seanfraser]

y = ln(e^x + C), using separation of variables.

[holyshit]

Two ways
1. Substitution u= x - y
2. Separation of variables

[rex_yourbud]

I did it myself and my method was same as your second method ! You rarely give easy questions.. got lucky this time 😂

[angelmendez-rivera]

I am fairly certain I already saw this video on the channel. Or maybe the exponent was y(x) + x instead of x – y(x) in that video.

y'(x) = exp[x – y(x)] <==> exp[y(x)]·y'(x) = exp(x) <==> exp[y(x)] = exp(x) + A <==> y(x) = ln[exp(x) + A]. This does come with the caveat that if we want the domain to be R, then we need A < 0 to be false. If A = 0, them y(x) = x, and if A > 0, then there is no further simplication. A < 0 restricts the domain such that exp(x) + A > 0, which means exp(x) > –A > 0, requiring x > ln(–A). If you allow A < 0, then you must restrict the domain to (ln(–A), ♾).

I assume the other method the video used is letting z(x) = x – y(x), hence y'(x) = 1 – z'(x) = exp[z(x)], which implies 1 – exp[z(x)] = z'(x). This requires caution. If z(x) = 0 for all x, then z'(x) = 0 for all x, and this is consistent with the equation. This implies y(x) = x. Assuming |z(x)| > 0, then, for all x, we have that z'(x)/(1 – exp[z(x)]) = 1, and this is equivalent to exp[–z(x)]·z'(x)/(exp[–z(x)] – 1) = 1, which is equivalent to –exp[–z(x)]·z'(x)/(exp[–z(x)] – 1) = –1, which is equivalent to ln(1 – exp[–z(x)]) = A – x or ln(exp[–z(x)] – 1) = A – x, which implies 1 – exp[–z(x)] = exp(A)·exp(–x) or exp[–z(x)] – 1 = exp(A)·exp(–x), which implies exp[–z(x)] = 1 – exp(A)·exp(–x) or exp[–z(x)] = 1 + exp(A)·exp(–x). For the former, it is required that exp(A)·exp(–x) < 1, which means that x > A. These mean that z(x) = x – y(x) = –ln[1 – exp(A)·exp(–x)] = x – ln[exp(x) – exp(A)] or x – y(x) = x – ln[exp(x) + exp(A)], implying y(x) = ln[exp(x) – exp(A)] for all x > A, or y(x) = ln[exp(x) + exp(A)] for all real x. This is equivalent to the solution for the earlier method, just exchanging A with –exp(A), 0, or exp(A), respectively.

[anishmathew]

dy/dx = e^x/e^y
e^y dy = e^x dx, variable seprable eqn.
Integrating, e^y = e^x +C,
So y= log[e^x +C], which is the solution

[kokainum]

Yeah, but k can be any real number and exp(-c) will always be positive. So question is when did you lose negative constants? You lost it at 3:15, because antiderivative of 1/u is ln(|u|) and you wrote ln(u) so you crossed out negative u's.

[Grassmpl]

The second method gave more solutions than the first. The first only matches the second when k>0, which is exactly when y has an unrestricted domain.

[emanuellandeholm]

First method: From dz/dx = 1 - e^z, we get the identity dz * 1 / (1 - e^z) = dx. Now 1 / (1 - e^z) is a geometric series sum(k>0) e^(zk) => dz * sum(k > 0) e^kz = 1. Integrate both sides: Int(z) dz sum(k > 0) e^kz = Int(x) dx = x + C. Now swap integration and summation to get sum(k > 0) Int(z) e^kz = x + C => sum(k > 0) 1/k e^kz = x + C. Now we have an implicit power series solution to our differential equation! :D

[kinshuksinghania]

At a glance, I'd say that y = x would satisfy the given differential equation

[MrWorschtsuppe]

Can you explain to me why in ln(e^x+k) ln and the e function do not resolve each other?

[turkishwagnerian]

Thirth way: ( Bu da benden gelsin hocam)
e^y. y’ =e^x =>(e^y)’=e^x
İntegral:
e^y=e^x+k = ln(e^x+k) ✔️

[threstytorres]

y= Ln(e^x + c), where c is a constant

Transform y'(x) into dy/dx
Dy/Dx = e^(x-y)

Multiply both sides by Dx
Dy = e^(x-y) Dx

Multiply both sides by e^y
e^y Dy = e^y( e^(x-y) ) Dx = e^x Dx

Integrate both sides
Int(e^y) Dy = Int(e^x) Dx

Calculate both sides
e^y = e^x + c

Take Ln both sides
y= Ln(e^x + c)

[krishnanadityan]

The second method is much easier than the first method.

[shazullahyusufzai]

Could you please solve this equation, x^x=e

[BlackwoodCompany]

That was just way too easy that time based on it being the natural exponetial.

[emanuellandeholm]

SIgh. I'm in YT moderation again for the sin of editing a comment...

[lokeshbansal]

I challenge ypu to solve the equation
x^2-1=3ln(x)

[johns.]

Only a psychopath would prefer the 1st method.

[moeberry]

I looked at it for 2 seconds and knew it was a seperable differential equation and I didn’t bother to solve it. It was 1st grade math.

[dibyojyotibhattacherjee]

no offence but this problem is sub standard compared to problems u share man...