Find all functions f satisfying f(x)/f'(x) = tan((k-1)x), where k is a constant not equal to 1.

Step-by-step solution

1. 1

   Flip the ratio so the logarithmic derivative appears. Taking reciprocals of both sides, $\frac{f'(x)}{f(x)}=\cot\!\left((k-1)x\right).$ The left side is exactly $\dfrac{d}{dx}\ln|f(x)|$, which is why this rearrangement is worth doing before anything else - it converts an equation about $f$ and $f'$ into a plain antiderivative problem.

2. 2

   Integrate both sides. $\ln|f(x)|=\int\cot\!\left((k-1)x\right)dx.$ Write $m=k-1$ (nonzero by hypothesis) and substitute $u=mx$, $du=m\,dx$: $\int\cot(mx)\,dx=\frac{1}{m}\int\cot u\,du=\frac{1}{m}\ln|\sin u|+C=\frac{1}{m}\ln\left|\sin(mx)\right|+C.$ The standard result $\int\cot u\,du=\ln|\sin u|$ comes from $u$-substituting the sine itself.

3. 3

   Exponentiate to recover $f$. From $\ln|f(x)|=\frac{1}{m}\ln|\sin(mx)|+C$, $|f(x)|=e^{C}\left|\sin(mx)\right|^{1/m},$ and absorbing $\pm e^{C}$ into a single arbitrary nonzero constant gives $f(x)=C\left|\sin\!\left((k-1)x\right)\right|^{\frac{1}{k-1}}.$

4. 4

   Verify by differentiating. With $m=k-1$ and $f=\left(\sin mx\right)^{1/m}$ on an interval where $\sin mx>0$, the chain rule gives $f'=\frac{1}{m}\left(\sin mx\right)^{\frac1m-1}\cdot m\cos mx=\left(\sin mx\right)^{\frac1m-1}\cos mx,$ so $\frac{f}{f'}=\frac{\left(\sin mx\right)^{1/m}}{\left(\sin mx\right)^{\frac1m-1}\cos mx}=\frac{\sin mx}{\cos mx}=\tan mx.$ The exponent $1/m$ is precisely what makes the powers of $\sin$ cancel down to a single first power.

5. 5

   Check numerically as well. Take $k=3$, so $m=2$ and $f(x)=\sqrt{\sin 2x}$. At $x=0.5$ a centred difference gives $f'(0.5)=0.606140$ and $f(0.5)=0.943985$, so $f/f'=1.557408$, while $\tan(2\times 0.5)=\tan 1=1.557408$. They agree to six decimals.

6. 6

   Note the domain and the excluded case. The solution only makes sense where $\sin\!\left((k-1)x\right)\neq 0$, so $x$ must avoid the multiples of $\pi/(k-1)$; on each such interval $C$ can be chosen independently. If $k=1$ the right-hand side is $\tan 0=0$, forcing $f\equiv 0$, which is why $k\neq 1$ is assumed.