Question

Solve the Initial Value Problem:

a) dydx+2y=9, y(0)=0 y(x)=_______________

b) dydx+ycosx=5cosx, y(0)=7d y(x)=______________

c) Find the general solution, y(t), which solves the problem below, by the method of integrating factors.

8t dy/dt +y=t^3, t>0

Put the problem in standard form.

Then find the integrating factor, μ(t)= ,__________

and finally find y(t)= __________ . (use *C* as the unkown
constant.)

d) Solve the following initial value problem:

t dy/dt+6y=7t

with y(1)=2

Put the problem in standard form.

Then find the integrating factor, ρ(t)= _______ ,

and finally find y(t)= _________

Answer #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 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)=

Solve the initial value problem
2(sin(t)dydt+cos(t)y)=cos(t)sin^3(t)
for 0<t<π0<t<π and y(π/2)=13.y(π/2)=13.
Put the problem in standard form.
Then find the integrating factor, ρ(t)=
and finally find y(t)=

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

Solve the initial value problem 8(t+1)dy/dt−6y=12t, for t>−1
with y(0)=11.

solve by the integrating facote method the following initial
value problem
dy/dx=y+x, y(0)=0

Initial value problem : Differential equations:
dx/dt = x + 2y
dy/dt = 2x + y
Initial conditions:
x(0) = 0
y(0) = 2
a) Find the solution to this initial value problem
(yes, I know, the text says that the solutions are
x(t)= e^3t - e^-t and y(x) = e^3t + e^-t
and but I want you to derive these solutions yourself using one
of the methods we studied in chapter 4) Work this part out on paper
to...

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

Solve the initial value problem
x′=x+y
y′= 6x+ 2y+et,
x(0) = 0, y(0) = 0.

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