Please can someone help prove LHS=RHS so 0=0

Question

So originally; $y=Ae^{−kt}sin(tω)$

$y=Ae^{−kt}sin⁡(tω)$

$y′=Ae^{−kt}(ωcos⁡(tω)−ksin⁡(tω))$

$y′′=Ae^{−kt}((k^{2}−ω^2)sin(tω)−2kωcos(tω))$

So we are trying to satisfy and prove that:

$my′′+λy′+ω^2y=0$

where this is a damped harmonic motion equation, $m, λ $ and $ y$ are constants.

Therefore;

$Ae^{-kt}[((k^2m−λk−(m−1)ω^2)sin(tω))+(ω(λ−2km)cos(tω))]=0$

... (unknown steps)

$0=0$

$LHS=RHS$

$QED$

Any advise on the (unknown steps) would mean the world, Thankyou

Answer 1

For the equation to be zero for all values of $t$, the coefficients of $\cos(tw)$ and $\sin(tw)$ must be zero. That means that the constants $\lambda$, $m$ etc. must satisfy certain constraints, such as, $\lambda = 2 k m$. If, for example, $$m =\frac{w^2}{k^2+w^2},$$ and $\lambda = 2 k m$ then $m y''+\lambda y' + w^2 y =0$.