If y = me^x + ne^-x, then which of the relation is true for the given function

Concept:

Some useful formulas are:

\(\rm{ d(e^{ax})\over dx} = ae^{ax}\)

Calculation:

Given function is,

y = mex + ne^-x

Differentiating the function we get,

\(\rm{ dy\over dx} = me^x-ne^{-x}\)

Differentiating further with respect to x, we get

\(\rm{ d^2y\over dx^2} = me^x+ne^{-x}\)

Differentiating further with respect to x, we get

\(\rm{ d^3y\over dx^3} = me^x-ne^{-x}\)

Now, \(\rm{ d^3y\over dx^3}+{ d^2y\over dx^2}-{ dy\over dx}-y \)

= \(\rm me^x-ne^{-x}+me^x+ne^{-x}-me^x+ne^{-x}-me^x-ne^{-x}\)

= 0

Hence, \(\rm{ d^3y\over dx^3}+{ d^2y\over dx^2}-{ dy\over dx}-y =0\)