Differential equation: dy/dx = e^(8x-3y), y(0)=16

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SUMMARY

The differential equation under discussion is dy/dx = e^(8x - 3y) with the initial condition y(0) = 16. The participant initially attempted to apply the natural logarithm to both sides, which is not a valid approach for this equation. Instead, the correct method involves using the property e^(a+b) = e^a * e^b to separate the variables effectively. This technique allows for a clearer path to solving the differential equation.

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  • Basic skills in separation of variables technique
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Lo.Lee.Ta.
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1. The next differential equation I'm working on is this:

dy/dx= e8x - 3y

Alright, I thought to cancel out the e, we could take the ln of both sides...?

ln(dy/dx) = 8x-3y

Is this right so far? It doesn't seem like it's right because... well, how can you have an ln(dy/dx)??

Would you please help me with this part?
Thank you very much! :)
 
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Lo.Lee.Ta. said:
1. The next differential equation I'm working on is this:

dy/dx= e8x - 3y

Alright, I thought to cancel out the e, we could take the ln of both sides...?

ln(dy/dx) = 8x-3y

Is this right so far? It doesn't seem like it's right because... well, how can you have an ln(dy/dx)??

Would you please help me with this part?
Thank you very much! :)

You can have ln(dy/dx) but there's not much you can do with it. Use e^(a+b)=e^a*e^b to separate the equation.
 
This relationship will help a lot

ea+b = ea*eb
 

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