New Notes Blog: `y = 5lnx` Determine whether the function is a solution of the differential equation `y^((4)) - 16y = 0`

Friday, 27 March 2015

$\displaystyle{y}={5}{\ln{{x}}}$ Determine whether the function is a solution of the differential equation $\displaystyle{{y}}^{{{\left({4}\right)}}}-{16}{y}={0}$

Given

$\displaystyle{y}={5}{\ln{{\left({x}\right)}}}$

$\displaystyle{y}'=\frac{{5}}{{x}}$

$\displaystyle{y}''=-\frac{{5}}{{{x}}^{{2}}}$

$\displaystyle{y}'''=\frac{{10}}{{{x}}^{{3}}}$

$\displaystyle{y}''''=-\frac{{30}}{{{x}}^{{4}}}$

so we have to check whether $\displaystyle{{y}}^{{{\left({4}\right)}}}-{16}{y}={0}$

$\displaystyle{y}''''-{16}{y}=-\frac{{30}}{{{x}}^{{4}}}-{16}{\left({5}{\ln{{\left({x}\right)}}}\right)}\ne{0}$

So it's not a solution to the differential equation.

Given

$\displaystyle{y}={5}{\ln{{\left({x}\right)}}}$

$\displaystyle{y}'=\frac{{5}}{{x}}$

$\displaystyle{y}''=-\frac{{5}}{{{x}}^{{2}}}$

$\displaystyle{y}'''=\frac{{10}}{{{x}}^{{3}}}$

$\displaystyle{y}''''=-\frac{{30}}{{{x}}^{{4}}}$

so we have to check whether $\displaystyle{{y}}^{{{\left({4}\right)}}}-{16}{y}={0}$

$\displaystyle{y}''''-{16}{y}=-\frac{{30}}{{{x}}^{{4}}}-{16}{\left({5}{\ln{{\left({x}\right)}}}\right)}\ne{0}$

So it's not a solution to the differential equation.

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)