Question

Solve the initial value problem

t^(13) (dy/dt) +2t^(12) y =t^25 with t>0 and y(1)=0

(y'-e^-t+4)/y=-4, y(0)=-1

Answer #1

solve the given initial value problem. y(cos2t)e^ty -
2(sin2t)e^ty + 2t + (t(cos2t)e^ty - 3) dy/dt = 0, y(0)=0

Solve the initial value problem 8(t+1)dy/dt - 6y = 12t
for t > -1 with y(0) = 7
7 =

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

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)=

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 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)=

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...

1. Solve the given initial value problem. dy/dt = (t^3 +
t)/(y^2); y(0) = 2 .
2. We know from Newton’s Law of Cooling that the rate at which a
cold soda warms up is proportional to the difference between the
ambient temperature of the room and the temperature of the drink.
The differential equation corresponding to this situation is given
by y' = k(M − y) where k is a positive constant. The solution to
this equation is given...

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

Solve the initial-value problem.
(x2 + 1)
dy
dx
+ 3x(y − 1) = 0,
y(0) = 4

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