Question

Solve the initial value problem, if possible.

yy'(t) = t8, y(0) = 9

Answer #1

Solve the initial value problem
3y'(t)y''(t)=16y(t) , y(0)=1,
y'(0)=2

Solve the initial value problem: y′′ − y = t + sint , y(0) = 2 ,
y′(0) = 3.

2. Solve the initial-value problem: y′′′ + 4y′ = t, y(0) = y′(0)
= 0, y′′(0) = 1.

Solve the Initial Value Problem:
dydx+2y=9,
y(0)=0
dydx+ycosx=5cosx,
y(0)=7d
Find the general solution, y(t)y(t), which solves the problem
below, by the method of integrating factors.
8tdydt+y=t3,t>08tdydt+y=t3,t>0
Put the problem in standard form.
Then find the integrating factor,
μ(t)=μ(t)= ,__________
and finally find y(t)=y(t)= __________ . (use C as the
unkown constant.)
Solve the following initial value problem:
tdydt+6y=7ttdydt+6y=7t
with y(1)=2.y(1)=2.
Put the problem in standard form.
Then find the integrating factor, ρ(t)=ρ(t)= _______ ,
and finally find y(t)=y(t)= _________ .

Solve the initial value problem:
y”+16y’+64y=7e^(-t)
y(0)=0, y’(0)=0

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

Solve the initial value problem 9(t+1) dy dt −6y=18t,
9(t+1)dydt−6y=18t, for t>−1 t>−1 with y(0)=14. y(0)=14. Find
the integrating factor, u(t)= u(t)= , and then find y(t)= y(t)=

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'+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

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