Question

Find the solution to the differential equation dy/dx+y/4=0

y=? Find the solution to the differential equation dy/dx+y/4=0

y=? @Mathematics

Answer 1

i think you need on the them there e^(x/4)

Answer 2

or just plain seperation of variables

Answer 3

but the e is sooo cool

Answer 4

This is a simple first-order homogenous equation. The characteristic equation of the solution is
\[y = C_1 e^{-at}\] where a is the coefficient of the y term.

Answer 5

dy/dx+y/4=0



dy/dx=-y/4


dy/y=-dx/4


Now integrate both sides and solve for y to get


ln(|y|) = -x/4


|y| = e^(-x/4)


y = +-e^(-x/4)

Answer 6

yessss.... i knew it lol

Answer 7

sorry wait

Answer 8

dont forget that constant or itll mess you up

Answer 9

Y(0)=12

Answer 10

c = 12 then when x=0

Answer 11

\[y = C_1 e^{-{t \over 4}}\] Set y=0 and t=0 and solve for C_1

Answer 12

Sorry y=12 and t = 0

Answer 13

no i want to do it jim thompsons way with Y(0)=12

Answer 14

umm ... its all the same

Answer 15

once you get to the y = Ce^-(x/4) you plug in and solve for C

Answer 16

Dude e^0=1. 12 = c1.

Answer 17

ok hold on, im trying to solve for C so give me a sec

Answer 18

and i dont think y=+- e^.. is a solution to a differential equation

Answer 19

something feels off there

Answer 20

C=11 right?

Answer 21

It could be but the sign depends on the sign of the coefficient. In this case, c is positive. It really should be written with a coefficient and not a sign.

Answer 22

I got the +- from the fact that |y| = e^(-x/4)

Answer 23

No. \[12 = C_1e^{-{0 \over 4}} = C_1e^0 = C_1\]

Answer 24

from what i recall reading; in order to be a solution it has to be a function; and y=+- .... might be taboo in the rigidity of it, just a thought

Answer 25

well that means y = e^(-x/4), y = -e^(-x/4)

Answer 26

isnt it plus C? not multiplied by the e^(0/4)

Answer 27

good luck ...

Answer 28

dy/dx+y/4=0


dy/dx=-y/4


dy/y=-dx/4


Now integrate both sides and solve for y to get


ln(|y|) = -x/4 + C


|y| = e^(-x/4+C)


y = e^(-x/4+C) or y = -e^(-x/4+C)


y = e^(-x/4)*e^C or y = -e^(-x/4)*e^C


y = e^C*e^(-x/4) or y = -e^C*e^(-x/4)


y = k*e^(-x/4) or y = -k*e^(-x/4)

Answer 29

Since y(0)=12, we know that the function y(x) is positive


So y(x) > 0 which means that we pick the first function y = k*e^(-x/4) and plug in x = 0 and y = 12. From here solve for k


y = k*e^(-x/4)


12 = k*e^(-0/4)


12 = k*e^(0)


12 = k*1


12 = k


k = 12


So the function is y = 12*e^(-x/4)

Answer 30

ya ok i got that

Answer 31

thanks