Hi this problem was explained to me before but I did not quite get it so I will give the problem and answer and if someone can clarify that would be great , the problem was : if dy / dt = ky is a constant and k is a non-zero constant , then y could be ?
The answer was 2e^(kt) but I don't get how or why.
Here was the answer : dy / dt = ky
dy / y = k dt
∫ dy/ y = ∫ kdt
lny = kt + C
e^lny = e^ (kt+c)
y = e^(kt + c)
y = e^kt + e^c
y = e^kt + C1
I don't get how this answer would get me to the answer of : 2e^(kt) , and also how to find the antiderivative of : kdt
The answer was 2e^(kt) but I don't get how or why.
Here was the answer : dy / dt = ky
dy / y = k dt
∫ dy/ y = ∫ kdt
lny = kt + C
e^lny = e^ (kt+c)
y = e^(kt + c)
y = e^kt + e^c
y = e^kt + C1
I don't get how this answer would get me to the answer of : 2e^(kt) , and also how to find the antiderivative of : kdt
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Because this answer is incorrect.
e^(kt+C) does NOT equal e^kt + e^C. I explained this in the first answer to your original post.
e^(kt+C) = e^kt * e^C. Remember when you multiply terms, you add exponents.
e^C is a constant. It could be anything. In this case, let it equal 2.
This gives you 2e^kt
For the antiderivative of kdt, k is a constant. This is simply k times the antiderivative of dt
The antiderivative of dt is t, just like the antiderivative of dx is x.
e^(kt+C) does NOT equal e^kt + e^C. I explained this in the first answer to your original post.
e^(kt+C) = e^kt * e^C. Remember when you multiply terms, you add exponents.
e^C is a constant. It could be anything. In this case, let it equal 2.
This gives you 2e^kt
For the antiderivative of kdt, k is a constant. This is simply k times the antiderivative of dt
The antiderivative of dt is t, just like the antiderivative of dx is x.
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I believe they gave you initial condition of 2. (y=2).
Usually when i solve these problems, after your 5th step i write it as:
y=Ce^(kt) instead of y=e^(kt)+e^c
because if you plug in values of t along with the initial condition which will be what 'C' equals, you will get the same answer. It would pretty much be e^(constant) which will equal a new constant which in your case, will be what "C1" equals.
so, using the form i gave you on the second line, C just equals the constant of 2. This is where you get 2e^(kt) from.
Also, to find the antiderivative of kdt, imagine 'k' equals a number, for example 'k' equals 5.
then you would have 5dt. the antiderivative of this would simply be 5t+C. This is how you get 'kt+C', where 'k' is not defined as a number yet. Hope this helps!
Usually when i solve these problems, after your 5th step i write it as:
y=Ce^(kt) instead of y=e^(kt)+e^c
because if you plug in values of t along with the initial condition which will be what 'C' equals, you will get the same answer. It would pretty much be e^(constant) which will equal a new constant which in your case, will be what "C1" equals.
so, using the form i gave you on the second line, C just equals the constant of 2. This is where you get 2e^(kt) from.
Also, to find the antiderivative of kdt, imagine 'k' equals a number, for example 'k' equals 5.
then you would have 5dt. the antiderivative of this would simply be 5t+C. This is how you get 'kt+C', where 'k' is not defined as a number yet. Hope this helps!
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dy / dt = ky
∫ dy/y = k∫ dt
ln(y) = kt
y = e^(kt) e^C
y =Ce^(kt)
when y(0) = 2
2 = Ce^[k(0)]
C = 2
y =2e^(kt) answer//
∫ dy/y = k∫ dt
ln(y) = kt
y = e^(kt) e^C
y =Ce^(kt)
when y(0) = 2
2 = Ce^[k(0)]
C = 2
y =2e^(kt) answer//
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dy / dt = ky
dy / y = k dt
∫ dy/ y = ∫ kdt
lny = kt + C
y = e^(kt+C) = Ae^(kt) where A is a constant
Let A = 2 for example
y = 2e^(kt)
dy / y = k dt
∫ dy/ y = ∫ kdt
lny = kt + C
y = e^(kt+C) = Ae^(kt) where A is a constant
Let A = 2 for example
y = 2e^(kt)
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your line before the last one should be y---> e^kt x e^c also you should have been given y--->2 when t is 0 in which case 2--> e^0 x e^c so e^c -->2
so your final answer is 2e^kt
so your final answer is 2e^kt