Solving Sin(x) Equation with ln(y), exp(y), and a Constant k

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The equation sin(x) = ln(y) + (y^2)/2 + k presents a challenge in isolating y due to the presence of both ln(y) and y in an exponential form. To address this, taking the exponential of both sides leads to the equation ye^(y^2/2) = e^(sin(x) - k). The Lambert W function, which is the inverse of the function xe^x, may be applicable for solving this type of equation, although its direct adaptation to isolate y is uncertain. Generally, when a variable appears both inside and outside a transcendental function, algebraic isolation is not feasible. Thus, the discussion highlights the complexity of isolating y in this context.
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Ive got the equation: sin(x) = ln(y) + (y^2)/2 + k

The k is a constant from an earlier integration. How do I isolate y? What makes it hard for me is that if i want to get rid of ln() i need to use exp() but then the other y is in exp() and if i want to get rid of that, the first y is in ln() again.

Sorry, for the bad english =) I hope you can help.
 
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I'm afraid it's not possible to isolate y algebraically.

Taking the exponential of both sides: y=e^{sinx-\frac{y^2}{2}-k}=\frac{e^{sinx-k}}{e^{\frac{y^2}{2}}}

So now you have: ye^{\frac{y^2}{2}}=e^{sinx-k}

There is some method to solving for x in: xe^x=y
but the name of it has slipped my mind and whether it can be adapted to solve for y in your problem I'm unsure of as well. Hopefully someone else can help you with this.
 
Generally speaking, when you have a variable both "inside" and "outside" a transcendental function, there is no algebraic way to isolate that variable.

The "method of solving xe^x= y" is "Lambert's W function" which is defined as the inverse function to xe^x. That is, W(xe^x)= x so W(xe^x)= W(y) and x= W(y).
 
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