Question

y"-y= 6t ; y_p = -6t y(t) =?

y"-y= 6t ; y_p = -6t

y(t) =?

Homework Answers

Answer #1

Please feel free to ask any query in the comment box and don't forget to rate it if you like

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
Initial value problem dy/dt=(6t^5/1+t^6)y+7(1+t^6)^2 y(1)=8
Initial value problem dy/dt=(6t^5/1+t^6)y+7(1+t^6)^2 y(1)=8
If u(t) = sin(6t), cos(2t), t and v(t) = t, cos(2t), sin(6t) , use Formula 4...
If u(t) = sin(6t), cos(2t), t and v(t) = t, cos(2t), sin(6t) , use Formula 4 of this theorem to find d dt u(t) · v(t) .
find the solution to the following ivp dy/dy-2ty=6t^2e^(t^2),y(0)=5
find the solution to the following ivp dy/dy-2ty=6t^2e^(t^2),y(0)=5
Find derivative for the following  : 1- m(t) =-3t (6t^4 -1)^5 m'(t) = 2- y= (3x+2)^5 (4x+1)^-3...
Find derivative for the following  : 1- m(t) =-3t (6t^4 -1)^5 m'(t) = 2- y= (3x+2)^5 (4x+1)^-3 dy/dx 3- y= (10x^2 - 20x +20) e^-4x
Find the arc length of a(t)=(2cosh3t,-2sinh3t,6t) for 0≤t≤5
Find the arc length of a(t)=(2cosh3t,-2sinh3t,6t) for 0≤t≤5
The position of a particle is given by s = f(t) = t^3 − 6t^2 +...
The position of a particle is given by s = f(t) = t^3 − 6t^2 + 9t. The total distance travelled by the particle in the first 5 seconds is : A. 4 B. 20 C. 28 D. None of the above The maximum vertical distance between the line y=x+2 and the parabola y=x^2 for −1 ≤ x ≤ 2 is A. 9/4 B. 1/4 C. 3/4 D. None of the above
Find the unit tangent vector T(t) and the curvature κ(t) for the curve r(t) = <6t^3...
Find the unit tangent vector T(t) and the curvature κ(t) for the curve r(t) = <6t^3 , t, −3t^2 >.
5. Find a general solution of t^2x′′ − 5tx′ + 5x = 6t^3, t > 0...
5. Find a general solution of t^2x′′ − 5tx′ + 5x = 6t^3, t > 0 using the method of undetermined coefficients
4. Find a general solution of t^2x′′ − 5tx′ + 5x = 6t^3, t > 0...
4. Find a general solution of t^2x′′ − 5tx′ + 5x = 6t^3, t > 0 using the method of variation of parameters.
The equation of motioj of a partucle is given by s(t)= t^3 -6t^2 +9t where s...
The equation of motioj of a partucle is given by s(t)= t^3 -6t^2 +9t where s is in meters and t is in seconds. a) Find when the particle is moving in the negatuve direction b) Find when the particle is accelerating in the positive direction
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT