I'm working on this question,
$$x' + x\tanh t = \sinh t, \quad x(0)= 3/2.$$
How I tried was
$$x' + x \tanh t = 0$$
$$x = C\tanh t \to x = C(t)\tanh t$$
Then derivative... but I couldn't get until the end, am I doing it in wrong way? or did just I make a mistake?
Assuming $x=x(t)$. The problem is of the form $x' + P(t)x = Q(t)$. Such equations can be made exact with the integrating factor $$\mu (t) = \exp\left (-\int P(t)dt\right ) $$ which then implies $$ x = \frac{1}{\mu (t)} \left (\int \mu (t) Q(t) dt+C\right ) $$
So, your main task would be to compute $$\int \mu (t)\sinh t dt $$
Your approach works, too. Solve homogeneous eq. and couple with a particular solution to the initial problem. You have
$$ x' + x \tanh t = 0 \Rightarrow \ln x = -\int \tanh tdt + C $$