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differential equation ! HELP watch

1. 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?
2. (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?

3. 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)
4. (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

?
5. (Original post by Indeterminate)
Be careful!

What do you get when you differentiate

?

when I integrate 1/2-y dy

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

is it meant to be - ln 2-y ?
Precisely
7. (Original post by Indeterminate)
Precisely
thanks! why is the power negative though '-kt'?
8. (Original post by 0utdoorz)
thanks! why is the power negative though '-kt'?
Multiply through by that negative I just told you about, and let

9. oh i see! thank you!

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