Find the general solution of y''(e^x+1) + y' = 0

Find the general solution of $y''(e^x+1) + y' = 0$.

Step 1: Substitution

Let $p = y'$, so $p' = y''$. The equation becomes: \[(e^x+1)\,p' + p = 0. \tag{1}\]

This is a first-order separable ODE in $p(x)$.

Step 2: Separate variables

\[ \frac{dp}{p} = -\frac{dx}{e^x+1}. \tag{2} \]

Step 3: Compute the integral on the right

Write $\dfrac{1}{e^x+1} = \dfrac{e^{-x}}{1+e^{-x}}$. Use $u = 1 + e^{-x}$, $du = -e^{-x}\,dx$: $\int \frac{dx}{e^x+1} = \int \frac{e^{-x}\,dx}{1+e^{-x}} = -\ln|1+e^{-x}| + \mathrm{const}.$

Rewrite using $1 + e^{-x} = \dfrac{e^x + 1}{e^x}$: $\int \frac{dx}{e^x+1} = x - \ln(e^x+1) + \mathrm{const}.$

Step 4: Integrate equation (2)

\[ \ln|p| = -x + \ln(e^x+1) + \ln|C_1|,\qquad p = C_1\,\frac{e^x+1}{e^x} = C_1\left(1 + e^{-x}\right). \]

Thus: \[y' = C_1\left(1 + e^{-x}\right). \tag{3}\]

Step 5: Second integration

\[ y = C_1\int\left(1 + e^{-x}\right)dx + C_2 = C_1\left(x - e^{-x}\right) + C_2. \]

Step 6: General solution

where $C_1$ and $C_2$ are arbitrary constants.

Step 7: Verification

Differentiate: $y' = C_1\left(1 + e^{-x}\right),\qquad y'' = -C_1\,e^{-x}.$

Check $y''(e^x+1) + y'$: $-C_1\,e^{-x}(e^x+1) + C_1(1+e^{-x}) = -C_1(1+e^{-x}) + C_1(1+e^{-x}) = 0.$ $\checkmark$