Find the general solution for the given differential
equation
9y′′′−27y′′+4y′−12y=x+4xe3x9y′′′−27y′′+4y′−12y=x+4xe3x
NOTE: Write your answer clearly in...
Find the general solution for the given differential
equation
9y′′′−27y′′+4y′−12y=x+4xe3x9y′′′−27y′′+4y′−12y=x+4xe3x
NOTE: Write your answer clearly in below type:
yg=yc+ypyg=yc+yp
For the following differential equation
y'' + 9y = sec3x,
(a) Find the general solution yh, for the
corresponding homogeneous ODE.
(b) Use the variation of parameters to find the
particular solution yp.
Find the general solution to the differential equation below.
y′′ − 6y′ + 9y = 24t−5e3
Calculate the inverse Laplace transform of ((3s-2)
e^(-5s))/(s^2+4s+53)
Calculate the Laplace transform of y = cosh(at) using the
integral definition of the Laplace transform. Be sure to note any
restrictionson the domain of s. Recall that cosh(t)
=(e^t+e^(-t))/2
Partial Differential Equations
(a) Find the general solution to the given partial differential
equation and (b) use it to find the solution satisfying the given
initial data.
Exercise 1. 2∂u ∂x − ∂u ∂y = (x + y)u
u(x, x) = e −x 2
Exercise 2. ∂u ∂x = −(2x + y) ∂u ∂y
u(0, y) = 1 + y 2
Exercise 3. y ∂u ∂x + x ∂u ∂y = 0
u(x, 0) = x 4
Exercise 4. ∂u...
4)Find the general solution of the following differential
equation.
y''+4y=tan(x) -pi
5)A mass of 100 grams stretches a spring 98 cm in equilibrium. A
dashpot attached to the spring supplies a damping force of 600
dynes for each cm/sec of speed. The mass is initially displaced 10
cm above the equilibrium point before the mass is attached, and
given a downward velocity of 1 m/sec. Find its displacement for
t>0.
Use the one solution given below to find the general solution of
the differential equation below by reduction of order method:
(1 - 2x) y'' + 2y' + (2x - 3) y = 0
One solution: y1 = ex
A. Find a particular solution to the nonhomogeneous differential
equation y′′ + 4y′ + 5y = −15x
+ e-x
y =
B. Find a particular solution to
y′′ + 4y = 16sin(2t).
yp =
C. Find y as a function of x if
y′′′ − 10y′′ + 16y′ =
21ex,
y(0) = 15, y′(0) = 28,
y′′(0) = 17.
y(x) =