Solving (simple?) differential eqn: (a - x)^2 - y t'(a) (1 - e^{-yt(a)}) = 0

[gadelke]

This this the first time that I have encountered a differential equation - and after watching a few videos on the topic, I still do not know where to start. The equation that I have encountered in one of my derivations is

. . . . .\(\displaystyle \large{(a\, -\, x)^2\, -\, yt'(a)\, \left(1\, -\, e^{-yt(a)}\right)\, =\, 0}\)

I have never taken a class on differential equations, nor have I seen any examples that look like this. In an attempt to make it look more 'linear', I get

. . . . .\(\displaystyle \large{-yt(a)\, =\, \ln\left(1\, -\, \dfrac{(a\, -\, x)^2}{yt'(a)}\right)}\)

Any ideas on how to solve this?
Thank you

 

[Deleted member 4993]

This this the first time that I have encountered a differential equation - and after watching a few videos on the topic, I still do not know where to start. The equation that I have encountered in one of my derivations is

. . . . .\(\displaystyle \large{(a\, -\, x)^2\, -\, yt'(a)\, \left(1\, -\, e^{-yt(a)}\right)\, =\, 0}\)

I have never taken a class on differential equations, nor have I seen any examples that look like this. In an attempt to make it look more 'linear', I get

. . . . .\(\displaystyle \large{-yt(a)\, =\, \ln\left(1\, -\, \dfrac{(a\, -\, x)^2}{yt'(a)}\right)}\)

Any ideas on how to solve this?
Thank you

What is yt(a)?

Is the name of the function yt, where as it is a function of 'x'? Is yt(a) the value of yt at x =a?

What is yt'(a)?

 

[HallsofIvy]

If yt is the dependent variable and x is the independent variable, then yt(a) and yt'(a) are simply numbers, not functions, and there is no way to find yt as a function of x.

If yt is the dependent variable and a is the independent variable, x a given parameter, then we have \(\displaystyle (1- e^{yt(a)})yt'(a)= (a- x)^2\) so that \(\displaystyle (1- e^{yt(a)}dyt= (a- x)^2da\).


Integrating both sides \(\displaystyle yt- e^{yt}= \frac{(a- x)^3}{3}+ C\) where "C" is the constant of integration. If "yt" means the parameter y times the function t, then we would have \(\displaystyle t- \frac{e^{yt}}{y}= \frac{(a- x)^3}{3}+ C\)

 

[gadelke]

This this the first time that I have encountered a differential equation - and after watching a few videos on the topic, I still do not know where to start. The equation that I have encountered in one of my derivations is

​

I have never taken a class on differential equations, nor have I seen any examples that look like this. In an attempt to make it look more 'linear', I get

​

Any ideas on how to solve this?
Thank you

Sorry, I should clarify
The name of the function I am solving for is t(a), and this is multiplied by y. In other words

So in that case, I think a is the independent variable and t(a) is dependent?

Also I should clarify that x and y are constants. Thanks

 

[HallsofIvy]

Taking "x" and "y" as constants and "a" as the independent variable is pretty much the opposite of standard notation but, okay!

So we have \(\displaystyle yt'(a)\left(1- e^{-yt(a)}\right)= (a- x)^2\).

We can separate variables as \(\displaystyle y\left(1- e^{-yt(a)}\right)dt= (a- x)^2 da\) and integrate both sides:
\(\displaystyle y\left(t+ \frac{1}{y}e^{-yt}\right)= \frac{1}{3}(a- x)^3+ C\) for some constant, C. That is what Subhotosh Kahn got before.