# DE involving Hyperbolic Trig function (Quick check)

I'm working problem 15 on page 133 of Boyce/DiPrima's Elementary Differential Equations and Boundary Value Problems (10th ed.) Note: this is homework, but it is not graded/turned in.

I have arrived at a solution which I think is correct, but it doesn't match the answer in the back of the book. (Don't you hate it when that happens?)

My question to the Math.SE community is, would you please see if a) I made an error in my process, or b) did I just not simplify it enough?

The problem statement:

Solve: $$(e^x + 1)\frac{dy}{dx} = y-ye^x$$

My (possibly erroneous) solution:

Separate and integrate: $$\frac{1}{y}\frac{dy}{dx} = \frac{(1-e^x)}{(e^x+1)}$$ $$\ln|y| = \int\frac{dx}{e^x+1} - \int\frac{e^x}{e^x + 1}dx$$ The left-most integral is simple: $$\ln|y| = \int\frac{dx}{e^x+1} - \ln{|e^x + 1|}$$ The other one, I multiplied by $\frac{e^{-x}}{e^{-x}}$: $$\ln|y| = \int\frac{e^{-x}dx}{1+e^{-x}} - \ln{|e^x + 1|}$$ $$\ln|y| = -\int\frac{-e^{-x}dx}{1+e^{-x}} - \ln{|e^x + 1|}$$ Now it's simple again! $$\ln|y| = -\ln{|1+e^{-x}|} - \ln{|e^x + 1|} + C$$ Raise the equation as a power of $e$: (I'm ignoring the abs signs from here on out, just for simplicity's sake... I don't think it causes a problem) $$y = \left(\frac{1}{1+e^{-x}}\right)\left(\frac{1}{e^x + 1}\right)(C)$$ Simplify: $$y = \left(\frac{C}{e^x+e^{-x}+2}\right)$$ $$y = \left(\frac{C}{2\left(\frac{e^x+e^{-x}}{2}\right)+2}\right)$$ $$y = \left(\frac{C}{2\cosh(x)+2}\right)$$ I'll factor the 2 out, and merge into C, giving my solution:

$$y = \left(\frac{C}{\cosh(x)+1}\right)$$

Their solution:

Close, but not exactly like mine...

$$y = \frac{c}{\cosh^2(\frac{x}{2})}$$

Because of the squaring, I doubt it's just a simplification error, and I've gone wrong somewhere in the problem. Any help?

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Half-angle formula: dlmf.nist.gov/4.35 – Ron Gordon Jan 29 '13 at 2:18

The answers are equivalent, by the half-angle formula for hyperbolic cosine: $$\cosh^2 \frac{x}{2} = \frac{1 + \cosh x}{2}.$$ This formula is easy to verify from the definition of $\cosh$. So your $C$ is $2c$.
By the way, it's ok for you to drop the absolute value sign since $$\frac{1}{(e^{-x} + 1)(1 + e^x)} > 0$$ for all $x$.