Answer:
y = 2cos5x-9/5sin5x
Step-by-step explanation:
Given the solution to the differential equation y'' + 25y = 0 to be
y = c1 cos(5x) + c2 sin(5x). In order to find the solution to the differential equation given the boundary conditions y(0) = 1, y'(π) = 9, we need to first get the constant c1 and c2 and substitute the values back into the original solution.
According to the boundary condition y(0) = 2, it means when x = 0, y = 2
On substituting;
2 = c1cos(5(0)) + c2sin(5(0))
2 = c1cos0+c2sin0
2 = c1 + 0
c1 = 2
Substituting the other boundary condition y'(π) = 9, to do that we need to first get the first differential of y(x) i.e y'(x). Given
y(x) = c1cos5x + c2sin5x
y'(x) = -5c1sin5x + 5c2cos5x
If y'(π) = 9, this means when x = π, y'(x) = 9
On substituting;
9 = -5c1sin5π + 5c2cos5π
9 = -5c1(0) + 5c2(-1)
9 = 0-5c2
-5c2 = 9
c2 = -9/5
Substituting c1 = 2 and c2 = -9/5 into the solution to the general differential equation
y = c1 cos(5x) + c2 sin(5x) will give
y = 2cos5x-9/5sin5x
The final expression gives the required solution to the differential equation.
<span>Let the number of nickles be x
Let the number of pennies be y
Value Value
Type Number of of
of of EACH ALL
coin coins coin coins
-------------------------------------------
nickles x $0.05 $0.05x
pennies y $0.01 $0.01y
-------------------------------------------
TOTALS 56 ----- $1.52
The first equation comes from the second column.</span>
6/7 because only one marble is orange
i) The given function is
The domain is all real values except the ones that will make the denominator zero.
The domain is all real values except, x=2.5.
ii) To find the vertical asymptote, we equate the denominator to zero and solve for x.
iii) If we equate the numerator to zero, we get;
This implies that;
iv) To find the y-intercept, we put x=0 into the given function to get;
.
.
.
v)
The degrees of both numerator and the denominator are the same.
The ratio of the coefficient of the degree of the numerator to that of the denominator will give us the asymptote.
The horizontal asymptote is 
.
vi) The function has no common factors that are at least linear.
The function has no holes in it.
vii) This rational function has no oblique asymptotes because it is a proper rational function.
