Consider the second order differential equation $\frac{d^2f}{dt^2}+a\frac{df}{dt}+bf=0$, where $a,b\in\mathbb{R}$. For which values of $a,b$ do we have $f(t)=\sinh(At+B)$ being a solution of this DE?
Attempt: Differentiating via the chain rule we have $\frac{df}{dt}=A\cosh(At+B)$ and $\frac{d^f}{dt^2}=A^2\sinh(At+B)$. Substituting in these results to the original DE gives:
$(A^2+b)\sinh(At+B)+aA\cosh(At+B)=0$. However I don't know how to go any further to determine $a,b$.
On
In the same spirit as @Tavish, consider that $$f(t)=(A^2 +b)\sinh (At+B)+aA \cosh(At+B)=0$$ needing to be true for all $t$, use the Taylor series around $t=0$ $$f(t)=\left(a A \cosh (B)+\left(A^2+b\right) \sinh (B)\right)+A \left(a A \sinh(B)+\left(A^2+b\right) \cosh (B)\right)t+O\left(t^2\right)$$
So, from the first coefficient $$b=-\text{csch}(B) \left(a A \cosh (B)+A^2 \sinh (B)\right)$$ Plug in the second coefficient and you are left with $$a A^2 \text{csch}(B)=0$$ Assuming $A\neq0$ then $a=0$ and then $b=-A^2$.
For all $t$, you need $$(A^2 +b)\sinh (At+B) = -aA \cosh(At+B) $$ or $$(A^2 +b)^2 \sinh^2 (At+B) =a^2 A^2 (1+\sinh^2(At+B) ) \\ [ (A^2+b)^2 -a^2 A^2 ] \sinh^2(At+B) - a^2 A^2=0$$ Let $At+B=0$. Then assuming $A\ne 0$, $$-a^2 A^2 =0 \implies a=0 $$ So $$ (A^2 +b)^2 \sinh^2(At+B) =0 \implies A^2 +b =0 \implies b=-A^2 $$ Assuming that $A$ can be a purely imaginary number, $b$ can be any real number.