Question
Show that the general solution of the differential equation $$y^{\prime \prime}-\omega^{2} y=0 \quad(\omega>0)$$ can be written $$y=C_{1} \cosh \omega x+C_{2} \sinh \omega x$$
Step 1
The derivative of $\cosh \omega x$ is $\omega \sinh \omega x$ and the derivative of $\sinh \omega x$ is $\omega \cosh \omega x$. So, we have: $$y^{\prime}=C_{1} \omega \sinh \omega x+C_{2} \omega \cosh \omega x$$ Show more…
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