Question

Solve differential: 8y' = x+y, change of variable u = x + y
Medal for showing me the solutions!

I got y = e^(1/8x)e^C-8-x
These confuse me, please help. If I just see how I can do it

Answer 1

To me, I would solve it as
8y'-y =x
for homogenous part, the characteristic equation is 8r-1=0 --> r = 1/8
so the general solution is y = Ce^(1/8x)
for partial, y = Ax +B
y' = A
so 8y'-y= 8A -Ax -B = x
--> A = -1
and B = -8
so partial solution is -x-8
combine them I get \(y = Ce^{x/8}-x-8\)

Answer 2

we're covering "separable equations" is that what you mean by partial? so C is in place for e^C right? if that's the case we have the same answer, why is it wrong without the constant C before e?

Answer 3

nope not separable. It is the form of y'+P(x) y = Q(x) where P(x) = 1 and Q(x) = x

Answer 4

I just don't understand why C must be a dummy variable for e^C? but thank you for helping :)

Answer 5

that's the formula. I don't know why C but not e^C. however, e^C = C, they are the same because they are some constant.

Answer 6

ok, I understand that. THank you again! :)