Prove that y = - ln(1+x) satisfies the differential
Prove that $y = -\ln(1+x)$ satisfies the differential equation $xy' + 1 = e^{y}$.
Explanation of Method
Substituting a candidate solution into a differential equation.
Step 1: Compute $y'$
\[ y = -\ln(1+x) \;\Longrightarrow\; y' = -\frac{1}{1+x}. \]
Step 2: Compute $e^{y}$
The exponential undoes the logarithm, and the minus sign inverts: \[e^{y} = e^{-\ln(1+x)} = \frac{1}{e^{\ln(1+x)}} = \frac{1}{1+x}.\]
Step 3: Evaluate the left-hand side
\[ xy' + 1 = x\left(-\frac{1}{1+x}\right) + 1 = \frac{-x}{1+x} + \frac{1+x}{1+x} = \frac{-x+1+x}{1+x} = \frac{1}{1+x}. \]
Step 4: Compare
Both sides equal $\dfrac{1}{1+x}$, so the equation holds identically on the domain $x \gt -1$.
Step 5: Verification
\[x = 0:\quad y = 0,\ y' = -1 \;\Longrightarrow\; xy'+1 = 0+1 = 1, \quad e^{y} = e^{0} = 1 \;\;\checkmark\]
\[x = 1:\quad y = -\ln 2,\ y' = -\tfrac12 \;\Longrightarrow\; xy'+1 = \tfrac12, \quad e^{-\ln 2} = \tfrac12 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,xy' + 1 = e^{y} \ \text{ is satisfied by }\ y = -\ln(1+x), \qquad x > -1\,}\]
Remark
Remark. Verifying a solution is a one-way check: it confirms this particular $y$ works, but says nothing about whether others do. In fact substituting $u = e^{-y}$ turns the equation into the linear equation $xu' = u - 1$, whose general solution is $u = 1 + Cx$, i.e. $y = -\ln(1+Cx)$ — the given function is the case $C = 1$.