Question

Find the general solution to the following differential equation using the method of variation of parameters.

y"-2y'+2y=e^{x} csc(x)

Answer #1

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

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

Find only the particular solution of the given differential
equation by using variation of parameters and Wronskians.
y ' ' - y = csc x cot x

a) Find the general solution of the differential equation
y''-2y'+y=0
b) Use the method of variation of parameters to find the general
solution of the differential equation y''-2y'+y=2e^t/t^3

differential equations!
find the Differential Equation General Solve by using
variation of parameters method...
y''' - 3y'' +3y' - y =12e^x

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

Now find the general solution of the equation using the method
of variation of parameters without using the formula from the book:
y''-y'=5sin(2x)

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

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 the method of variation of parameters to determine the
general solution of the given differential equation.
y′′′−y′=3t
Use C1, C2, C3, ... for the constants of integration.

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