Question

Solve the initial value problem. 5d^2y/dt^2 + 5dy/dt - y = 0; y(0)=0, y'(0)=1

Answer #1

solve the given initial value problem
dx/dt=7x+y x(0)=1
dt/dt=-6x+2y y(0)=0
the solution is x(t)= and y(t)=

1. Solve the following initial value problem using Laplace
transforms.
d^2y/dt^2+ y = g(t) with y(0)=0 and dy/dt(0) = 1 where g(t) = t/2
for 0<t<6 and g(t) = 3 for t>6

For the initial value problem
• Solve the initial value problem.
y' = 1/2−t+2y withy(0)=1

Solve the initial value problem: y''−2y'+y=e^t/(1+t^2), y(0) =
1, y'(0) = 0.

Solve the Initial Value Problem:
?x′ = 2y−x
y′ = 5x−y
Initial Conditions:
x(0)=2
y(0)=1

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''-2y'+5y=u(t-2) y(0)=0
y'(0)=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 initial value problem.
(?2+1)?′+2?=1+?where?(0)=2(x2+1)y′+2y=1+xwherey(0)=2

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