Question

I would appreciate some assistance \[\frac{\text d^2 y}{\text dx^2}+n^2y=0\]

Answer 1

i can solve easily using second order techniques:
\[\frac{\text d^2 y}{\text dx^2}+n^2y=0\]\[m^2+n^2=0\]\[m=\pm in\]\[y=A\cos nx+B\sin nx\]

Answer 2

but im finding it hard with first order techniques:

\[\frac{\text d^2 y}{\text dx^2}+n^2y=0\]\[\text {let } \frac{\text dy}{\text dx}=p\]\[\frac{\text d^2y}{\text dx^2}=\frac{\text dp}{\text dx}=\frac{\text dp}{\text dy}\frac{\text dy}{\text dx}=\frac{\text dp}{\text dy}p\]\[\frac{\text dp}{\text dy}+n^2y=0\]\[\frac{\text dp}{\text dy}=-n^2y\]\[\int \text dp=-n^2\int y\text dy\]\[p=-n^2\frac{y^2}{2}+c\]\[\frac{\text dy}{\text dx}=-n^2\frac{y^2}{2}+c\]\[-2\frac{\text dy}{n^2y^2+c}=\text dx\]\[-2\int\frac{\text dy}{n^2y^2+c}=\int\text dx\]\[-2\left(\frac{\arctan{\frac{ny}{\sqrt c}}}{\sqrt cn}\right)=x+d\]\[\arctan{\frac{ny}{\sqrt c}}=\frac{\sqrt c n(x+d)}2\]\[\frac{ny}{\sqrt c}=\tan\left(\frac{\sqrt c n(x+d)}2\right)\]\[\frac{ny}{\sqrt c}=\frac{\sin\left(\frac{\sqrt c n(x+d)}2\right)}{\cos \left(\frac{\sqrt c n(x+d)}2\right)}\]

Answer 3

I heard 42 was the answer to everything O.O

Answer 4

im not looking of an answer im looking for a method

Answer 5

lol have you tried PEMDAS ;D

Answer 6

better yet; try elmo counts

Answer 7

im not evaluating @rebeccaskell94

Answer 8

i cant count any elmos here,

Answer 9

i think i can see a mistake in the working