How can i solve the differential equation y'' + 4y' + 5y = 0 where y' and y'' are the first and second derivatives of y(x) respectively?
Sep 15, 2020The correct answer is y= e^-2t (c1.cos(t) + c2.sin(t))
Sep 15, 2020How can i solve the differential equation y'' + 4y' + 5y = 0 where y' and y'' are the first and second derivatives of y(x) respectively?
Sep 15, 2020The correct answer is y= e^-2t (c1.cos(t) + c2.sin(t))
Sep 15, 2020Can you please explain how you got this? Can i message you,... if you find that comfortable?
Sep 16, 2020Dude here asking real questions Not some bitchy shit like my girlfriend hates me, boyfriend and I are getting apart blah blah blah...
Sep 15, 2020Guys don't make jokes he got a real question here If y" +4y + 5y= 0 hence y' and y" are the first and the second u will add 5 with 4 and y by y and then u will shove it up ur ass😂😂😂😂 The office fans By: Anonymous 🎖 5 rep a few seconds ago
Sep 15, 2020Its shove it up ur butt... dammit u ruined a good joke
Sep 15, 2020Use implicit differentiation. "Larson calculus" this book will help you alot
Sep 16, 2020thank you 🙏
Sep 16, 2020Try finding soft copy of applied mathematics 3,,it's all there
Sep 15, 2020I meant y" is one and 4y' is 4
Sep 15, 2020Y"is o and4y' is 4 then do the rest
Sep 15, 2020y" is equivalent to 45^y hence 4y' + 5y' is almost 1% of £lan 3 subtract 6y for sin x and add 0 so it will beome 4'y + 5' and fuck u too
Sep 15, 2020Why are u posting all this math questions ... go read a book
Sep 15, 2020Nerddd alert😜😜
Sep 15, 2020