I'm trying to find the explicit form of an equation after performing an integration. I'm a bit rusty.

ln|y| = C - cos(x)

C is an arbitrary constant.

Is it:

y = e^(c-cos(x))

or because it's an arbitrary constant, can it be written:

y = C - e^(cos(x))

Thanks.

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Trancescient Trancescient · Answer 1 · 2021-10-23T17:27:35+02:00

No, it should be written as

[tex]y = Ce^{-\cos x}[/tex]

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LammettHash LammettHash · Answer 2 · 2021-12-03T19:30:42+01:00

Provided that y ≠ 0, both solutions

[tex]y = e^{C-\cos(x)}[/tex]

and

[tex]y = Ce^{-\cos(x)}[/tex]

are correct. The other solution you provided,

[tex]y = C - e^{\cos(x)}[/tex]

is not.

The only difference between the "valid" solutions lies in the particular value of the arbitrary constant C.

Suppose y(0) = Y ≠ 0 is given as an initial condition. Then in the first solution,

[tex]Y = e^{C - \cos(0)} \implies \ln|Y| = C - 1 \implies C = \ln|Y| + 1[/tex]

In the second solution,

[tex]Y = C e^{-\cos(0)} \implies Y = Ce^{-1} \implies C = eY[/tex]

But both solutions are the same, since

[tex]y = e^{\ln|Y| + 1 - \cos(x)} = e^{\ln|Y|} e^1 e^{-\cos(x)} = eY e^{-\cos(x)}[/tex]