Answer:
44 and 22
Step-by-step explanation:
The question is:
Check whether the function:
y = [cos(2x)]/x
is a solution of
xy' + y = -2sin(2x)
with the initial condition y(π/4) = 0
Answer:
To check if the function y = [cos(2x)]/x is a solution of the differential equation xy' + y = -2sin(2x), we need to substitute the value of y and the value of the derivative of y on the left hand side of the differential equation and see if we obtain the right hand side of the equation.
Let us do that.
y = [cos(2x)]/x
y' = (-1/x²) [cos(2x)] - (2/x) [sin(2x)]
Now,
xy' + y = x{(-1/x²) [cos(2x)] - (2/x) [sin(2x)]} + ([cos(2x)]/x
= (-1/x)cos(2x) - 2sin(2x) + (1/x)cos(2x)
= -2sin(2x)
Which is the right hand side of the differential equation.
Hence, y is a solution to the differential equation.
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Answer:
828.75
Step-by-step explanation:
If you input the 975 and take 15 percent away you would get 828.75
