Question

Verify that y is an explicit solution to the differential equation, y''+y=tan(x) , y =-cos(x)ln(secx+tanx).... just help with differentiation

Answer 1

wolframalpha on rescue mission
http://www.wolframalpha.com/input/?i=d%5E2%2Fdx%5E2+%28-cos%28x%29ln%28secx%2Btanx%29%29+%2B+%28-cos%28x%29ln%28secx%2Btanx%29%29

Answer 2

that's cheating dawg

Answer 3

wouldn't it be product rule with chain

Answer 4

this is what my teach will put on our test lol for sure... feel my pain

Answer 5

yeah ... you can do that ... use product rule to differentiate ...
then again differentiate again ... it will take lot of time!!

Answer 6

yes but we don't have wolfram alpha at test time haha anyways i'll do that for a test problem next week

Answer 7

yeah .. best of luck!!

Answer 8

y =-cos(x)ln(secx+tanx)


y' = d/dx[ -cos(x)ln( sec(x) + tan(x) ) ]


y' = d/dx[ -cos(x)]*ln( sec(x) + tan(x) ) -cos(x)*d/dx[ln( sec(x) + tan(x) )]


y' = sin(x)*ln( sec(x) + tan(x) ) - cos(x)*(1/( sec(x) + tan(x) ))*d/dx[ sec(x) + tan(x) ]


y' = sin(x)*ln( sec(x) + tan(x) ) - cos(x)*(1/( sec(x) + tan(x) ))*(sec(x)tan(x) + sec^2(x))


y' = sin(x)*ln( sec(x) + tan(x) ) - cos(x)*(1/( sec(x) + tan(x) ))*(sec(x)(tan(x) + sec(x)))


y' = sin(x)*ln( sec(x) + tan(x) ) - cos(x)*sec(x)


y' = sin(x)*ln( sec(x) + tan(x) ) - 1


Now repeat this to find y''


y'' = d/dx[ sin(x)*ln( sec(x) + tan(x) ) - 1 ]


y'' = d/dx[ sin(x)*ln( sec(x) + tan(x) )] - d/dx[ 1 ]


y'' = d/dx[ sin(x)*ln( sec(x) + tan(x) )]


y'' = d/dx[ sin(x)]*ln( sec(x) + tan(x) ) + sin(x)*d/dx[ln( sec(x) + tan(x) ) ]


y'' = cos(x)*ln( sec(x) + tan(x) ) + sin(x)*(1/(sec(x) + tan(x)) )*d/dx[sec(x) + tan(x)]


y'' = cos(x)*ln( sec(x) + tan(x) ) + sin(x)*(1/(sec(x) + tan(x)) )*(sec(x)*tan(x) + sec^2(x))


y'' = cos(x)*ln( sec(x) + tan(x) ) + sin(x)*(1/(sec(x) + tan(x)) )*(sec(x)(tan(x) + sec(x)))


y'' = cos(x)*ln( sec(x) + tan(x) ) + sin(x)*sec(x)


y'' = cos(x)*ln( sec(x) + tan(x) ) + tan(x)


===========================================


y''+y=tan(x)


( cos(x)*ln( sec(x) + tan(x) ) + tan(x) )+ ( -cos(x)ln(secx+tanx) )=tan(x)


tan(x) = tan(x)

Answer 9

It's a lot of work, but once you get the hang of derivatives and you've mastered them (and perhaps memorized them), then it shouldn't be too bad.