Consider the initial value problem: y′′+36y=e−ty′′+36y=e−t, y(0)=y0y(0)=y0, y′(0)=y′0y′(0)=y0′. Suppose we know that y(t)→0y(t)→0 as t→[infinity]t→[infinity]. Determine the solution and the initial conditions.

This differential equation can be solved with the method of the undetermined coefficients

The general solution is yg = yh + yp

where yh is the solution to the homogeneous equation and yp is a particular solution

first of all, we calculate the homogeneous solution

y'' + 36y = 0

with the characteristic polynomial

m = ± 6i

here we have imaginary solutions, hence:

yh = C1*cos(6t) + C2*sin(6t)

With C1 and C2 constants that we have to compute with the initial conditions. For the calculation of yp we assume that one solution is yp = C3*e^(-t)

by calculating the derivatives and replacing:

yp' = -C3*e^(-t)

yp'' = C3*e^(-t)

yp'' + 36yp = e^(-t)

C3*e^(-t) + 36*C3*e^(-t) = e^(-t)

37C3 = 1   -->   C3 = 1/37

Hence, the solution is

y = yg = yh + yp = C1*cos(6t) + C2*sin(6t) + (1/37)e^(-t)

By using the initial condition we have

y(0) = C1*cos(0) + C2*sin(0) + (1/37)e^(0) = C1 + 1/37 = 0  -->  C1 = -1/37

y'(0) = -C1*sin(0) + C2*cos(0) - (1/37)e^(0) = C2 - 1/37 = 0  -->  C2= 1/37

Thus, finally

y = (1/37)( cos(6t) + sin(6t) + e^(-t) )

infinitely many solutions

step-by-step explanation:

we have been given the equation;

2(2x + 3) = -4 + 2(2x + 5)

we need to determine the value of x.

2(2x + 3), we expand this expression to obtain 4x + 6

similarly 2(2x + 5) will become 4x + 10

now we have;

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 10 - 4

4x + 6 = 4x + 6

the expression on the left hand side is identical to the one on the right hand side. therefore, we are looking at just a single straight line whose equation is;

y = 4x + 6

consequently, we shall have infinitely many solutions

l.ln.lln.k.nkl

step-by-step explanation: