Question

Show that the following initial value problem has a unique solution that exists for all of t

x'' + sin(x')cos(xt) = cos(x)cos(2t)

x(0) = 1, x'(0) = 0

Answer #1

x'' + sin(x')cos(x*t) = cos(x)*cos(2*t)

x(0) = 1, x'(0) = 0

we know if p(t), q(t), and g(t) be continuous on [a,b], then the differential equation

y'' + p(t)y' + q(t)y
= g(t)
y(t_{0}) = y_{0}
y'(t_{0}) = y'_{0}

has a unique solution defined for all t in [a,b].

here g(t)=cos(x)*cos(2*t) is continuous function.

and sin(x')*cos(x*t) is also a continuous function so the following problem has unique solution.

Does the initial value problem y' = ( ((cos 4x) / (1 + 2 log
x)1/2) + esin2x ), y(0) = 1.7 have a solution in an open
interval? Specify the interval if it exists. If the solution
exists, is the solution unique?

Solve the given initial-value problem. d2x dt2 + 4x = −5 sin(2t)
+ 9 cos(2t), x(0) = −1, x'(0) = 1

determine if the given initial value problem has a unique
solution or not. why?
y' = 2xy^(3/4) , y(1) = 0

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.

Solve this Initial Value Problem using the Laplace
transform.
x''(t) - 9 x(t) = cos(2t),
x(0) = 1,
x'(0) = 3

transform the given initial value problem into an algebraic
equation for Y = L{y} in the s-domain. Then find the Laplace
transform of the solution of the initial value problem.
y'' + 4y = 3e^(−2t) * sin 2t,
y(0) = 2, y′(0) = −1

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

Consider the initial value problem
y′ = (1+t)/(1+y)
y(1) = 2
t is within [1,2]
Show that y(t) = -1+(t^2 + 2t + 6)^1/2 is the
solution to the initial value problem.

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

Initial Value Problem. Use Laplace transform.
m''(t) - 9*m(t) = cos(2t)
where m(0) = 1 and m'(0) = 3

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