If we allow complex solutions, then we get more solutions from u=1 too: 3^x = t = ⅓ => x ln 3 = log t + 2nπi = -ln 3 + 2nπi 3^x = t = -⅔ => x ln 3 = log t + 2nπi = ln 2 - ln 3 + (2n+1)πi So x = -1 + 2nπi/ln 3, or x = -1 + (ln 2 + (2n+1)πi)/ln 3
Can you do a video explaining how to solve a second order differential equation that's not reducible? As in there are both x and y terms and you can't do a t=dy/dx, t'=dt/dx or t'=t*dt/dy substitution?
That floats my boat. I am an ex-Navy submarine officer. I used long division also. I think it is easier, but math has many methods of solving problems. I prefer to use the old school long division method.
writing z for 3 ^x one gets z^3 + z^2 = 1/27 + 1/9 ( z - 1/3) (z^2 + z/3 + 1/9 + z + 1/3) = 0 ( z - 1/3) ( z + 2/3) ^ 2 = 0 3 ^ x = z = 1/3, - 2/3 Hereby x = -1 is only feasible solution in real domain
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If we allow complex solutions, then we get more solutions from u=1 too:
3^x = t = ⅓ => x ln 3 = log t + 2nπi = -ln 3 + 2nπi
3^x = t = -⅔ => x ln 3 = log t + 2nπi = ln 2 - ln 3 + (2n+1)πi
So x = -1 + 2nπi/ln 3, or x = -1 + (ln 2 + (2n+1)πi)/ln 3
2npii works miracles!
Excellent and nice solution thanks.
You're welcome!
👍
Can you do a video explaining how to solve a second order differential equation that's not reducible? As in there are both x and y terms and you can't do a t=dy/dx, t'=dt/dx or t'=t*dt/dy substitution?
Since the left hand side of the original equation is an increasing function, couldn’t we just stop once we found the first real solution?
Yes, if you only care about real solutions.
Since you know u=1, why not use long division to get the other 2 solutions?
Yes, long division works beautifully. I prefer long division to the "factor when you already know one solution" method.
Whatever floats your boat!
That floats my boat. I am an ex-Navy submarine officer. I used long division also. I think it is easier, but math has many methods of solving problems. I prefer to use the old school long division method.
I got x=-1 for the real solution, and ((2n+1)/2)-(ln_9(2)/(2*pi))i for the complex solution which is a double root, and where n is an integer.
I came up with x = (ln2 - ln3 +i*pi*(1+2n) )/ln3. Your solution has a factor of 1/(2*pi*i) that my soultion does not have.
@@stephenshefsky5201 i also got your answer
writing z for 3 ^x one gets
z^3 + z^2 = 1/27 + 1/9
( z - 1/3) (z^2 + z/3 + 1/9 + z + 1/3) = 0
( z - 1/3) ( z + 2/3) ^ 2 = 0
3 ^ x = z = 1/3, - 2/3
Hereby x = -1 is only feasible solution in real domain
x = -1