Question

Verify that the function ϕ(t)=c1e^−t+c2e^−2t is a solution of the linear equation

y′′+3y′+2y=0

for any choice of the constants c1c1 and c2c2. Determine c1c1 and c2c2 so that each of the following initial conditions is satisfied:

**(a)** y(0)=−1,y′(0)=4

**(b)** y(0)=2,y′(0)=0

Answer #1

Suppose x=c1e−t+c2e^5t. Verify that x=c1e^−t+c2e^5t is a
solution to x′′−4x′−5x=0 by substituting it into the differential
equation. (Enter the terms in the order given. Enter c1 as c1 and
c2 as c2.)

In Exercises 16-25, use any method to solve each nonhomogeneous
equation.
y'''+3y''+2y'=cos(t)
Answer is apparently y=(1/10)
sin(t)-(3/10)cos(t)+c1e^(-2t)+c2e^(-t)+c3

Verify that the indicated function is a solution of the given
dierentail
equation. In some cases assume an appropriate interval of validity
for the
solution.
y''+y'-12y=0; y=c1e^3x+c2e^-4x

The general solution of the equation
y′′+6y′+13y=0 is
y=c1e-3xcos(2x)+c2e−3xsin(2x)
Find values of c1 and c2 so that y(0)=1
and y′(0)=−9.
c1=?
c2=?
Plug these values into the general solution to obtain
the unique solution.
y=?

y''-2y'-3y=15te^2t y(0)=2 y'(0)=0 find the solution

Given the differential equation to the right y''-3y'+2y=0
a) State the auxiliary equation.
b) State the general solution.
c) Find the solution given the following initial conditions
y(0)=4 and y'(0)=5

Differential equation for y'=2y-t+g(y) that has a solution
y(t)=e^(2t)

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

Verify that the function
y=x^2+c/x^2
is a solution of the differential equation
xy′+2y=4x^2, (x>0).
b) Find the value of c for which the solution satisfies the initial
condition y(4)=3.
c=

For problems 3-6, note that if y(t) is a solution of the
homogeneous problem, then y(t−t0) is a solution as well, where t0
is a fixed constant. So, for example, the general solution of a
problem with complex roots can be expressed as y(x) = c1e µ(t−t0)
cos(ω(t − t0)) + c1e µ(t−t0) sin(ω(t − t0)) When initial conditions
are given at time t = t0 and not t = 0, expressing the general
solution in terms of t −...

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