Menu Close

Solve-D-4-4-y-0-given-y-0-0-y-0-2-y-0-0-and-y-0-4-




Question Number 74367 by Learner-123 last updated on 23/Nov/19
Solve :  (D^4 +4)y=0   given: y(0)=0 , y′(0)=2 , y′′(0)=0 and  y′′′(0)=4.
$${Solve}\:: \\ $$$$\left({D}^{\mathrm{4}} +\mathrm{4}\right){y}=\mathrm{0}\: \\ $$$${given}:\:{y}\left(\mathrm{0}\right)=\mathrm{0}\:,\:{y}'\left(\mathrm{0}\right)=\mathrm{2}\:,\:{y}''\left(\mathrm{0}\right)=\mathrm{0}\:{and} \\ $$$${y}'''\left(\mathrm{0}\right)=\mathrm{4}. \\ $$
Commented by mr W last updated on 23/Nov/19
please check question and answer.  if question is correct, answer should be  y=((e^((√2)x) −e^(−(√2)x) )/( (√2)))
$${please}\:{check}\:{question}\:{and}\:{answer}. \\ $$$${if}\:{question}\:{is}\:{correct},\:{answer}\:{should}\:{be} \\ $$$${y}=\frac{{e}^{\sqrt{\mathrm{2}}{x}} −{e}^{−\sqrt{\mathrm{2}}{x}} }{\:\sqrt{\mathrm{2}}} \\ $$
Commented by Learner-123 last updated on 23/Nov/19
There is some  discrepancy.Pl ignore  the problem.  my apologies.
$${There}\:{is}\:{some}\:\:{discrepancy}.{Pl}\:{ignore} \\ $$$${the}\:{problem}. \\ $$$${my}\:{apologies}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *