Question

y^4-2y^2+y=0 find the set of solutions solve y^2+y^'=0 y(0)=0, y(0)=-10

y^4-2y^2+y=0

find the set of solutions

solve y^2+y^'=0

y(0)=0, y(0)=-10

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
solve differential equation y^(4) +2y''' +2y''=0
solve differential equation y^(4) +2y''' +2y''=0
Solve the initial value problem y' = 3x^2 − 2y, y(0) = 4
Solve the initial value problem y' = 3x^2 − 2y, y(0) = 4
Solve the initial value problem y = 3x^2 − 2y, y(0) = 4
Solve the initial value problem y = 3x^2 − 2y, y(0) = 4
Solve using transforms. y''' +2y''-y'-2y=24e^-3t + 48t^2; y(0) = y'(0) = y''(0) = 0
Solve using transforms. y''' +2y''-y'-2y=24e^-3t + 48t^2; y(0) = y'(0) = y''(0) = 0
y"-3y''''+2y= te^2t, y(0)=1, y''(0)=4 solve
y"-3y''''+2y= te^2t, y(0)=1, y''(0)=4 solve
For 2y' = -tan(t)(y^2-1) find general solution (solve for y(t)) and solve initial value problem y(0)...
For 2y' = -tan(t)(y^2-1) find general solution (solve for y(t)) and solve initial value problem y(0) = -1/3
Solve the given boundary-value problem. y'' − 2y' + 2y = 2x − 2,   y(0) =...
Solve the given boundary-value problem. y'' − 2y' + 2y = 2x − 2,   y(0) = 0, y(π) = π
solve the eqution 2y"+3y'-1y=8x+2 and given the initial condition y(0)=0, (dy/dt at t=0)=2 find y(2)
solve the eqution 2y"+3y'-1y=8x+2 and given the initial condition y(0)=0, (dy/dt at t=0)=2 find y(2)
Solve the initial value problem. x'=x+2y y'=7x+y with inition contitions x(0)=2 and y(0)=4.
Solve the initial value problem. x'=x+2y y'=7x+y with inition contitions x(0)=2 and y(0)=4.
1) Solve by power series around x=0: y"-2xy'-2y=0 (Find the first three nonzero terms of each...
1) Solve by power series around x=0: y"-2xy'-2y=0 (Find the first three nonzero terms of each of the LI solutions)
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT