Question

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)= ?

Answer #1

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

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

B. a non-homogeneous differential equation, a complementary
solution, and a particular solution are given. Find a solution
satisfying the given initial conditions.
y''-2y'-3y=6 y(0)=3 y'(0) = 11 yc=
C1e-x+C2e3x
yp = -2
C. a third-order homogeneous linear equation and three linearly
independent solutions are given. Find a particular solution
satisfying the given initial conditions
y'''+2y''-y'-2y=0, y(0) =1, y'(0) = 2, y''(0) = 0
y1=ex, y2=e-x,,
y3= e-2x

The indicated functions are known linearly independent solutions
of the associated homogeneous differential equation on (0, ∞). Find
the general solution of the given nonhomogeneous equation.
x2y'' + xy' + y = sec(ln(x))
y1 = cos(ln(x)), y2 = sin(ln(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

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

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

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.

The indicated function y1(x) is a solution of the
given differential equation. Use reduction of order or formula (5)
in Section 4.2,
y2 = y1(x)
e−∫P(x) dx
y
2
1
(x)
dx (5) as instructed, to find a
second solution y2(x).
y'' + 36y = 0; y1 =
cos(6x)
y2 =
2) The indicated function y1(x) is a solution of the
given differential equation. Use reduction of order or formula (5)
in Section 4.2,
y2 = y1(x)
e−∫P(x) dx
y
2
1...

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

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