# First Order Differential Equation

• I
Basically, I am confused by one question in a practice paper in which the equation is given as follows:

dy/dx = e^-2y

and I know the general solution is equal to : y = -0.5e^-2y + C

which would make sense if it was direct integration however it seems to me it is in fact separable integration as there is a lone function of y. Below is a picture of my attempt at a solution through separable integration and would appreciate if anyone has any input? Am I wrong in thinking it's seperable?

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Mark44
Mentor
Basically, I am confused by one question in a practice paper in which the equation is given as follows:

dy/dx = e^-2y

and I know the general solution is equal to : y = -0.5e^-2y + C
That's not the solution, as it does not give y as a function of x.
santeria13 said:
which would make sense if it was direct integration however it seems to me it is in fact separable integration as there is a lone function of y. Below is a picture of my attempt at a solution through separable integration and would appreciate if anyone has any input? Am I wrong in thinking it's seperable?
The image you posted is terrible! It goes from barely legible to completely unreadable because of stuff you have crossed out. There are several errors in the lines at the end.

S.G. Janssens
HallsofIvy
Homework Helper
Basically, I am confused by one question in a practice paper in which the equation is given as follows:

dy/dx = e^-2y then e
the
and I know the general solution is equal to : y = -0.5e^-2y + C
I hope you don't 'know' that because it is not true!
which would make sense if it was direct ingration however it seems to me it is in fact separable integration as there is a lone function of y. Below is a picture of my attempt at a solution through separable integration and would appreciate if anyone has any input? Am I wrong in thinking it's seperable?
You are not "wrong in thinking it's separable" but you are "separating" it wrong.
dy/dx= e^(-2y) can be written in "differential" form as e^(2y)dy= dx and then integrating to get e^(2y)/2= x+ C which could then be solved for y as a logarithm of x.
But you seem to be "separating" it as dy= e^(-2y)dx. You can't integrate that because on the right you have a function of y that you cannot integrate with respect to x.

santeria13
S.G. Janssens