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First Order Differential Equations by Separating Variables

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Original post by Silentieyes
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What's your question, and what have you tried so far?
Yes I could help
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Silentieyes
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so far i only did this muchIMG_7684.JPG
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Silentieyes
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Original post by 15Characters...
What's your question, and what have you tried so far?

to prove that R = e^1/250 θ
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mqb2766
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Original post by Silentieyes
to prove that R = e^1/250 θ

There is only one constant. They (it) do not cancel
It's found using the initial conditions.

Also remember ln is the inverse of an exponential so the left hand side is ...
(edited 3 years ago)
Original post by Silentieyes
so far i only did this muchIMG_7684.JPG

I'm not sure if the other poster resolved your difficulties, but I'll add something just incase.


You are not correct that the constants cancel out, because the constants on the left and right hand side need not be equal. Instead you could have written something like

[br]ln⁡∣R∣+C1=1250θ+C2[br][br]\ln|R| + C_1 = \frac{1}{250}\theta + C_2[br]

where C1,C2C_1, C_2 are arbitrary constants, not necessarily equal. However we do not really have two free constants to set, as we can move C1C_1 to the right hand side to get

[br]ln⁡∣R∣=1250θ+C   (∗)[br][br]\ln|R| = \frac{1}{250}\theta + C \>\>\>(*)[br]

where C=C2−C1C = C_2 - C_1 is our single arbitrary constant, which is then set by applying the initial conditions.


In practice you should just put a constant on one side in the first place, like in Eq. (*).
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Solved it! thank you so much everyoneIMG_7703.JPG
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Original post by 15Characters...
I'm not sure if the other poster resolved your difficulties, but I'll add something just incase.


You are not correct that the constants cancel out, because the constants on the left and right hand side need not be equal. Instead you could have written something like

[br]ln⁡∣R∣+C1=1250θ+C2[br][br]\ln|R| + C_1 = \frac{1}{250}\theta + C_2[br]

where C1,C2C_1, C_2 are arbitrary constants, not necessarily equal. However we do not really have two free constants to set, as we can move C1C_1 to the right hand side to get

[br]ln⁡∣R∣=1250θ+C   (∗)[br][br]\ln|R| = \frac{1}{250}\theta + C \>\>\>(*)[br]

where C=C2−C1C = C_2 - C_1 is our single arbitrary constant, which is then set by applying the initial conditions.


In practice you should just put a constant on one side in the first place, like in Eq. (*).

ohh sorry i didnt know about this
[br]ln⁡∣R∣+C1=1250θ+C2[br][br]\ln|R| + C_1 = \frac{1}{250}\theta + C_2[br]

but i will take your advice for upcoming practices. thank you so much for helping me :smile:

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