Question

Solve the following differential equations

1. cos(t)y' - sin(t)y = t^2

2. y' - 2ty = t

Solve the ODE

3. ty' - y = t^3 e^(3t), for t > 0

Compare the number of solutions of the following three initial value problems for the previous ODE

4. (i) y(1) = 1 (ii) y(0) = 1 (iii) y(0) = 0

Solve the IVP, and find the interval of validity of the solution

5. y' + (cot x)y = 5e^(cos x), y(pi/2) = -4

If you can, please show all steps! I don't understand alot of this.

Answer #1

Differential Equations
Solve for the IVP
( y2 - 2 sin (2t) )dt + ( 1 + 2ty)dy = 0
y (0)= 1

Solve the following Differential equations
a) x sin y dx + (x^2 + 1) cos y dy = 0

1. Solve the following differential equations.
(a) dy/dt +(1/t)y = cos(t) +(sin(t)/t) , y(2pie) = 1
(b)dy/dx = (2x + xy) / (y^2 + 1)
(c) dy/dx=(2xy^2 +1) / (2x^3y)
(d) dy/dx = y-x-1+(xiy+2) ^(-1)
2. A hollow sphere has a diameter of 8 ft. and is filled half way
with water. A circular hole (with a radius of 0.5 in.) is opened at
the bottom of the sphere. How long will it take for the sphere to
become empty?...

Differential equations
Given that x1(t) = cos t is a solution of (sin t)x′′ − 2(cos
t)x′ − (sin t)x = 0, find a second linearly independent solution of
this equation.

Solve the differential equation (3y^2+2ty)+(2ty+t^2)dy/dt=0

solve the ODE solving for the general solutions
y''+y'-12y=sin(t)e^(3t)

solve the differential equations by series of potentials:
a)y''(t)=ty(t)
b)y(t)''+ty(t)'+2y(t)=0

(Part 1) Find all of the solutions of the given differential
equations:
a.) y' = -2y (answer should be y = -(1 / ln2) * ln(t
* ln(2) + c))
(Part 2) Find the solution of the IVP:
b.) y' = -2y3, y(0) = 0
c.) y' = 1 + cos(y), y(0) = pi / 2 (answer should be y(t) =
2arctan(1 + t))
d.) y' = sqrt(1 - y2), y(0) = 0 (Hint: y' > 0)
Please show work!

This is a differential equations problem:
use variation of parameters to find the general solution to the
differential equation given that y_1 and y_2 are linearly
independent solutions to the corresponding homogeneous equation for
t>0. ty"-(t+1)y'+y=18t^3 ,y_1=e^t ,y_2=(t+1)
it said the answer to this was C_1e^t + C_2(t+1) -
18t^2(3/2+1/2t)
I don't understand how to get this answer at all

In this problem, x = c1 cos t + c2 sin t is a two-parameter
family of solutions of the second-order DE x'' + x = 0. Find a
solution of the second-order IVP consisting of this differential
equation and the given initial conditions.
x(π/6) = 1 2 , x'(π/6) = 0
x=

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