Answer: (a) \(\displaystyle \frac{\mathrm d}{\mathrm dx}(\ln(\tanh x))=2\,\operatorname{cosech}(2x)\).

(b) \(\displaystyle y\tanh x=\ln(\cosh x)+3-\ln\!\left(\frac54\right)\).

Hence an exact form for the solution is

\(\displaystyle y=\frac{\ln(\cosh x)+3-\ln\!\left(\frac54\right)}{\tanh x}.\)

(a) Using the chain rule,

\(\displaystyle \frac{\mathrm d}{\mathrm dx}(\ln(\tanh x))=\frac{1}{\tanh x}\cdot \frac{\mathrm d}{\mathrm dx}(\tanh x)=\frac{\operatorname{sech}^2 x}{\tanh x}.\)

Now

\(\displaystyle \frac{\operatorname{sech}^2 x}{\tanh x}=\frac{1/\cosh^2 x}{\sinh x/\cosh x}=\frac{1}{\sinh x\cosh x}.\)

Since \(\sinh 2x=2\sinh x\cosh x\),

\(\displaystyle \frac{1}{\sinh x\cosh x}=\frac{2}{\sinh 2x}=2\,\operatorname{cosech}(2x).\)

Therefore

\(\displaystyle \frac{\mathrm d}{\mathrm dx}(\ln(\tanh x))=2\,\operatorname{cosech}(2x).\)

(b) Start with

\(\displaystyle \sinh 2x\,\frac{\mathrm dy}{\mathrm dx}+2y=\sinh 2x.\)

Divide through by \(\sinh 2x\):

\(\displaystyle \frac{\mathrm dy}{\mathrm dx}+\frac{2y}{\sinh 2x}=1.\)

From part (a), \(\displaystyle \frac{2}{\sinh 2x}=\frac{\mathrm d}{\mathrm dx}(\ln(\tanh x))\), so the integrating factor is

\(\displaystyle e^{\int \frac{2}{\sinh 2x}\,\mathrm dx}=e^{\ln(\tanh x)}=\tanh x.\)

Multiplying through by \(\tanh x\) gives

\(\displaystyle \frac{\mathrm d}{\mathrm dx}(y\tanh x)=\tanh x.\)

Integrating,

\(\displaystyle y\tanh x=\int \tanh x\,\mathrm dx=\ln(\cosh x)+C.\)

Use \(y=5\) when \(x=\ln 2\).

\(\displaystyle \cosh(\ln 2)=\frac{2+1/2}{2}=\frac54, \qquad \sinh(\ln 2)=\frac{2-1/2}{2}=\frac34,\)

so

\(\displaystyle \tanh(\ln 2)=\frac{3/4}{5/4}=\frac35.\)

Substitute into \(y\tanh x=\ln(\cosh x)+C\):

\(\displaystyle 5\left(\frac35\right)=\ln\!\left(\frac54\right)+C.\)

Hence

\(\displaystyle 3=\ln\!\left(\frac54\right)+C,\)

so

\(\displaystyle C=3-\ln\!\left(\frac54\right).\)

Therefore

\(\displaystyle y\tanh x=\ln(\cosh x)+3-\ln\!\left(\frac54\right),\)

and hence

\(\displaystyle y=\frac{\ln(\cosh x)+3-\ln\!\left(\frac54\right)}{\tanh x}.\)