Question

Find only the particular solution of the given differential equation by using variation of parameters and Wronskians.

y ' ' - y = csc x cot x

Answer #1

Find the general solution to the following differential equation
using the method of variation of parameters.
y"-2y'+2y=ex csc(x)

3. Find the general solution if the given differential equation
by using the variation of parameters method. y''' + y'= 2 tan x, −
π /2 < x < π/2

Use the method of variation of parameters to find a particular
solution of the differential equation y′′−8y′+15y=32et.

Use the method of variation parameters to find the
general solution of the differential equation
y'' +16y = csc 4x

By using method of variation of parameters the particular
solution of the following differential equation
y″+y=sec2(x)
is

Find a particular solution for the differential equation by
variation of parameters.
y''- y' -2y = e^3x , y(0) = -3/4 , y'(0)=15/4

Using variation of parameters, find a particular solution of the
given differential equations:
a.) 2y" + 3y' - 2y = 25e-2t (answer should be: y(t) =
2e-2t (2e5/2 t - 5t - 2)
b.) y" - 2y' + 2y = 6 (answer should be: y = 3 + (-3cos(t) +
3sin(t))et )

Use variation of parameters to find a general solution to the
differential equation given that the functions y 1 and y 2 are
linearly independent solutions to the corresponding homogeneous
equation for t>0.
ty"-(t+1)y'+y=30t^2 ; y1=e^t , y2=t+1
The general solution is y(t)= ?

Use variation of parameters to find a general solution to the
differential equation given that the functions y1 and y2 are
linearly independent solutions to the corresponding homogeneous
equation for t>0.
y1=et y2=t+1
ty''-(t+1)y'+y=2t2

Find a solution to y^''-4y^'-5y=2e^2t using variation of
parameters. Find the solution to the differential equation in
problem 6, this time using the method of undetermined
coefficients.

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