Question

Let be the problem with initial value

f(t,y) = yt^{3} , y(0)=1

Write the general formula for Picard iterations. Then start with
the function y_{0} (t) = y (0) = 1 and calculate the
iterations y_{1} (t) and y_{2} (t).

Answer #1

Let y(t) be the solution of the initial-value problem y'=
sin(y)e^(y2+1); y(0) = 1
Calculate limt->INF y(t). (Hint: do not attempt to solve the
ODE). THE ANSWER IS PI. PLEASE EXPLAIN CAUSE IM CONFUSED!

Consider the following Initial Value Problem (IVP)
y' = 2xy, y(0) = 1.
Does the IVP exists unique solution? Why? If it does, ﬁnd the
solution by Picard iteration with y0(x) = 1.

Consider the initial value problem: y' - (7/2)y = 7t + 2e^t
Initial condition: y(0) = y0
a) Find the value of y0 that separates
solutions that grow positively as t → ∞ from those that
grow negatively. (A computer algebra system is recommended. Round
your answer to three decimal places.)
b) How does the solution that corresponds to this critical value
of y0 behave as t → ∞?
Will the corresponding solution increase without bound, decrease
without bound, converge...

Find the function y1(t) which is the solution of 4y″+32y′+64y=0
with initial conditions y1(0)=1,y′1(0)=0.
y1(t)=?
Find the function y2(t) which is the solution of 4y″+32y′+64y=0
with initial conditions y2(0)=0, y′2(0)=1.
y2(t)= ?
Find the Wronskian of these two solutions you have found:
W(t)=W(y1,y2).
W(t)=?

Consider the initial value problem
y' +
5
4
y = 1 −
t
5
, y(0) =
y0.
Find the value of
y0
for which the solution touches, but does not cross, the
t-axis. (A computer algebra system is recommended. Round
your answer to three decimal places.)
y0 =

Find the solution of the given initial value problem: y " + y =
f(t); y(0) = 6, y' (0) = 3 where f(t) = 1, 0 ≤ t < π 2 0, π 2 ≤
t < ∞

Choose C so that y(t) = −1/(t + C) is a solution to the initial
value problem
y' = y2 y(2) = 3.
Verify that the given formula is a solution to the initial value
problem.
x′ = −y, y′ = x, x(0) = 1, y(0) = 0: x(t) = cost, y(t) = sin
t

For 2y' = -tan(t)(y^2-1) find general solution (solve for y(t))
and solve initial value problem y(0) = -1/3

Let y = y ( t
) be the solution to the initial value problem
t
d y d t + 2 y = sin t , y ( π ) = 0
Find the value
of

Use the Laplace transform to solve the given initial-value
problem. y'' + y = f(t), y(0) = 0, y'(0) = 1, where f(t) = 0, 0 ≤ t
< π 5, π ≤ t < 2π 0, t ≥ 2π

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