Differential Equations - How to go from my answer to the book's answer?

Question

so I'm doing variation of parameters:

$$y'' - y = \frac{1}{e^x + e^{-x}}$$ Let the particular solution be the form of: $$y_p = u_1y_1 + u_2y_2$$

$$y_1 = e^x$$ $$y_2 = e^{-x}$$

Now we assume $u_1'y_1 + u_2'y_2 = 0$, let's call this equation (1). $$u_1'e^x + u_2'e^{-x}=0$$ Now we differentiate $y_1$ and $y_2$ to get the next equation, let's call it (2):

$$u_1'e^x - u_2'e^{-x} = \frac{1}{e^x + e^{-x}}$$

Right? So far so good. Now we add both equation (1) and (2) and we can solve for $u_1'$.

$$2u_1'e^x = \frac{1}{e^x + e^{-x}}$$

$$u_1' = \frac{1}{2} \bigg(\frac{1}{e^{2x} + 1}\bigg)$$

After integrating, I end up with $$u_1 = \frac{1}{4}\bigg(-\ln(e^{2x}+1)+2x\bigg)$$

So now I plug in $u_1'$ into either equation (1) or (2) to get $u_2'$ and to solve for $u_2$.

I will plug it into equation (1) since it's easier.

$$\frac{1}{2}\bigg(\frac{e^x}{e^{2x}+1}\bigg) + u_2'e^{-x} = 0$$

$$u_2'e^{-x} = -\frac{1}{2}\bigg(\frac{1}{e^x + e^{-x}}\bigg)$$

$$u_2' = -\frac{1}{2}\bigg(\frac{1}{1 + e^{-2x}}\bigg)$$

Now my professor said to integrate this, we should manipulate the integrand using rules of exponents, but I'm not sure how to do that...

So I just did this:

After integrating, I end up with:

$$\frac{1}{4}\bigg(-\ln(e^{-2x}+1) + \ln(e^{-2x})\bigg)$$

Now I have both $u_1$ and $u_2$, but the problem is I don't know how to get the answer to match the books.

My final answer is:

$$\frac{-\ln(e^{2x}+1)e^x + 2xe^x -\ln(e^{-2x}+1)e^{-x}-2xe^{-x}}{4}$$

The books answer is:

Is my answer equivalent to the books answer? If so, how can I make it match?

Answer 1

You have $$ \ln(e^{-2x}+1) = \ln(e^{-2x}) + \ln(1+e^{2x}) = -2x + \ln(1+e^{2x}) $$

Therefore $$ \frac{-e^x\ln(e^{2x}+1) - e^{-x}(-2x + \ln(e^{2x}+1))+2xe^x-2xe^{-x}}{4} \\ = \frac{2xe^x - (e^x+e^{-x})\ln(e^{2x}+1)}{4} $$

Here's an alternate way to integrate (likely what your prof meant) $$ -\frac{1}{2}\int \frac{1}{e^{-2x}+1}dx = -\frac{1}{2}\int \frac{e^{2x}}{e^{2x}+1}dx = -\frac{1}{4}\ln(e^{2x}+1) $$

Answer 2

Checking Op's answer

$$ \begin{align} Y_p=&\frac{-ln(e^{2x}+1)e^x + 2xe^x -ln(e^{-2x}+1)e^{-x}-2xe^{-x}}{4} \\ =&\frac{-ln(e^{2x}+1)e^x + 2xe^x -ln(e^{2x}+1)e^{-x}+2xe^{-x}-2xe^{-x}}{4} \\ =&\frac{-ln(e^{2x}+1)e^x + 2xe^x -ln(e^{2x}+1)e^{-x}}4 \\ Y_p=&\frac{-(e^x+e^{-x})\ln|(e^{2x}+1)| + 2xe^x }4 \\ \end{align} $$ It's the same answer as that of the book...

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A simple hint

Not an anwer. Just checking the book's answer in case...

$$y'' - y = \frac{1}{e^x + e^{-x}}$$ $$y''-y'+y' - y = \frac{1}{e^x + e^{-x}}$$ $$g'+g = \frac{1}{e^x + e^{-x}}$$ $$(ge^x)' = \frac{e^x}{e^x + e^{-x}}$$ $$ge^x = \int \frac{e^x}{e^x + e^{-x}}dx$$ $$ge^x = \int \frac{u}{u^2 + 1}du \implies ge^x = \frac 12\int \frac{2u}{u^2 + 1}du=\frac 12 \ln|u^2+1|+K_1$$ $$(ye^{-x})' =e^{-2x}(\frac 12 \ln|e^{2x}+1|+K_1)$$ $$y =K_1e^{-x}+K_2e^{x}+e^{x}\underbrace{\int e^{-2x}(\frac 12 \ln|e^{2x}+1|)dx}_{\text { Integrale }I_2} $$ I evaluated the last integral $I_2$ and got this $$e^xI_2=\frac {xe^x}2-\frac 14 (e^v+e^{-x})\ln|e^{2x}+1|$$ Therefore $$\boxed{y =K_1e^{-x}+e^{x}(K_2+\frac {x}2)-\frac 12 \cosh(x)\ln|e^{2x}+1|}$$