Question

Solve using first order methods
\[\frac{\text d^2 y}{\text dx^2}+n^2y=0\]

Answer 1

\[\text {let } \frac{\text dy}{\text dx}=p\]\[\frac{\text d^2y}{\text dx^2}=\frac{\text dp}{\text dx}=\frac{\text dp}{\text dy}\frac{\text dy}{\text dx}=\frac{\text dp}{\text dy}p\]

Answer 2

\[\frac{\text dp}{\text dy}p+n^2y=0\]\[p\frac{\text dp}{\text dy}=n^2y\]\[p\text dp=n^2y\text dy\]\[\int p\cdot\text dp=n^2\int y\cdot\text dy\]

Answer 3

\[\frac {p^2} 2=\frac{n^2y^2+c}2\]\[p=\sqrt{n^2y^2+c}\]\[\frac{\text dy}{\text dx}=\sqrt{n^2y^2+c}\]\[\frac{\text dy}{\sqrt{n^2y^2+c}}={\text dx}\]\[\int\frac{\text dy}{\sqrt{n^2y^2+c}}=\int{\text dx}\]

Answer 4

\[\frac1n\int\frac{\text dy}{\sqrt{y^2+c/n^2}}=\int{\text dx}\]