Question

Solve y'+ 1/4y = 3 + 2cos(2t) with initial condition y(0) = 0.

Solve y'+ 1/4y = 3 + 2cos(2t) with initial condition y(0) = 0.

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Use Laplace Transforms to solve the given initial value problem y''-4y'+4y=t^3e6(2t) y(0)=1 and y'(0)=-2
Use Laplace Transforms to solve the given initial value problem y''-4y'+4y=t^3e6(2t) y(0)=1 and y'(0)=-2
y'' + 4y' + 4y = 12e^-2t y(0) = 3 y'(0) = -1 find particular soln.
y'' + 4y' + 4y = 12e^-2t y(0) = 3 y'(0) = -1 find particular soln.
y'' - 4y' + 4y = (6)(e^(2t)) y(0)=y'(0)=0 Use Laplace Transforms to solve. Sketch the solution...
y'' - 4y' + 4y = (6)(e^(2t)) y(0)=y'(0)=0 Use Laplace Transforms to solve. Sketch the solution or use matlab to show the graph.
y'' + 4y' +4y = e^(-2t) y(0)=0 y'(0)=4 Use Laplace Transforms to solve. Sketch the solution...
y'' + 4y' +4y = e^(-2t) y(0)=0 y'(0)=4 Use Laplace Transforms to solve. Sketch the solution or use matlab to show the graph.
Solve the initial value problem y''+4y'+4y=0, y(-1)=5, y'(-1)=5
Solve the initial value problem y''+4y'+4y=0, y(-1)=5, y'(-1)=5
find t? x= 2cos(t)+ sin(2t) y=2sin(t)+cos(2t) when x= 0, y= -3
find t? x= 2cos(t)+ sin(2t) y=2sin(t)+cos(2t) when x= 0, y= -3
2. Solve the initial-value problem: y′′′ + 4y′ = t, y(0) = y′(0) = 0, y′′(0)...
2. Solve the initial-value problem: y′′′ + 4y′ = t, y(0) = y′(0) = 0, y′′(0) = 1.
Solve the initial value problem. Explain and show all steps. y'' - 4y' +4y = 0...
Solve the initial value problem. Explain and show all steps. y'' - 4y' +4y = 0 where y(0) = 1 and y'(0) = 2
use the laplace transform to solve the initial value problem: y"+4y=4t, y(0)=1, y'(0)=0
use the laplace transform to solve the initial value problem: y"+4y=4t, y(0)=1, y'(0)=0
solve the following initial value problem x' = 2xy y' = y -2t +1 x(0) =...
solve the following initial value problem x' = 2xy y' = y -2t +1 x(0) = x0 y(0) = y0