Recent content by pedro123

so you mean Yp''-Yp'+15Yp=cost4t to solve -A sin(t) - B cos(t) - 16C sin(4t) - 16D cos(4t) + 15A sin(t)+ 15B cos(t)+ 15C sin(4t)+ 15D cos(4t) = cos (4t ) + 2 sin(t) -14 A sin(t) +14 B cos(t) -C sin(4t) -D cos(4t) = cos (4t) + 2 sin t -C = 1 = -1 -D = 1 = -1 -14A = 1 = -1/14 14B =...

so you mean for the nonhomegenous part yp=A sin(t)+ B cos(t)+ C sin(4t)+ D cos(4t). yp'=A cos(t) - B sin(t) + 4C cos (4t) + - 4D sin (4t) yp''= -A sin(t) - B cos(t) - 16C sin(4t) - 16D cos(4t)

So yh= cos(t√(15)) + sin(t√(15)) and now I need to use the method of undetermined cooficients to come with a particular solution. is Yp= Acos 2t +B sin 2t Yp'=-2Asin 2t + 2B cos 2t Yp''= -4Acos 2t - 4B sin 2t right or did I make mistake and if it is right it will make d^2y/dt^2 +15Yp=...

Homework Statement The problem states d^2y/dt^2 +15y= cost4t + 2sin t initial conditions y(0)=y'(0)=0 Homework Equations The Attempt at a Solution All I have is this r^2+15=0 making r(+-)=√15 and making yh= C1cos√15+C2√15 the next part includes solve for...

I know that 2 is a constant but can my particular solution be Yh=Acos4t +Bsin4t

and thanks for helping me so far

thanks can help solve this problemI have a problem which in involves a second order differential equations with imaginary roots and I can seem to know how to finish the problem. d^2y/dt^2 +15y =cos 4t+2 sin t this is what I got so far r^2+15=0 for the homogeneous part r=+-(√15)...

making the differential equation y''''-2y'''-2y-1=0 is this alright and sorry about my mistake in factoring

yes I know but can you explain how i know i need to have a particular solution for cos 4t+2 sin t but i don't know how to set up

I have a problem which in involves a second order differential equations with imaginary roots and I can seem to know how to finish the problem. d^2y/dt^2 +15y =cos 4t+2 sin t this is what I got so far r^2+15=0 for the homogeneous part r=+-(√15) Yh=C1cos√15+C2sin√15 now is...

so if its factored can be written as (r-1)r^3 and alternate fom beign r^4-r^3 making the general solution y''''-y'''=0 is this alright thanks.

is the the differential equation -y''''-y''=0 I am not sure

I came with taking the fourt derivate of y''''=Ce^-x(C3(x2-8x+12)+C2(x-4)+C+e^2x so Idont know what to do next. help please asap.

I need help finding a linear homogenous constant-coefficient differential equation with the given general solution. y(x)=C1e^x+(C2+C3x+C4x^2)e-x 2. I tried to come with differential equation but this is it I can 't seem how to begin