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differential equation ! HELP

Given that k is a constant, find the solution of the differential equation
dy/dx+ky=2k

for which y=3 when t=0

I did:
dy/dt+ky=2k
integrated ky dy= integrated 2k dt

???? is that right?
Reply 1
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BabyMaths
Original post by 0utdoorz
Given that k is a constant, find the solution of the differential equation
dy/dx+ky=2k

for which y=3 when t=0

I did:
dy/dt+ky=2k
integrated ky dy= integrated 2k dt

???? is that right?


Start with

dydx=2k−ky\frac{dy}{dx}=2k-ky
Reply 2
Avatar for 0utdoorz
0utdoorz
OP
1
I got up to
dy/dt=2k-y
dy/dt= k(2-y)

integrated 1/(2-y) dy= integrated k dt
ln (2-y) = kt +c

e both sides:

2-y= e^kt x e^c

let A= e^c

2-y = e^kt
y= 2- e^kt

but the answer should be y= 2+e^(-kt)
Original post by 0utdoorz
I got up to
dy/dt=2k-y
dy/dt= k(2-y)

integrated 1/(2-y) dy= integrated k dt
ln (2-y) = kt +c

e both sides:

2-y= e^kt x e^c

let A= e^c

2-y = e^kt
y= 2- e^kt

but the answer should be y= 2+e^(-kt)


Be careful!

What do you get when you differentiate

ln⁡(2−y)\ln(2-y)

?
Reply 4
Avatar for 0utdoorz
0utdoorz
OP
1
Original post by Indeterminate
Be careful!

What do you get when you differentiate

ln⁡(2−y)\ln(2-y)

?



when I integrate 1/2-y dy

is it meant to be - ln 2-y ?
Original post by 0utdoorz
when I integrate 1/2-y dy

is it meant to be - ln 2-y ?


Precisely :smile:
Reply 6
Avatar for 0utdoorz
0utdoorz
OP
1
Original post by Indeterminate
Precisely :smile:


thanks! why is the power negative though '-kt'?
Original post by 0utdoorz
thanks! why is the power negative though '-kt'?


Multiply through by that negative I just told you about, and let

A=e−cA=e^{-c}
Reply 8
Avatar for 0utdoorz
0utdoorz
OP
1
oh i see! thank you! :biggrin:

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