Check whether the function yequalsStartFraction cosine 2 x Over x EndFraction is a solution of x y prime plus yequalsnegative 2 sine 2 x with the initial condition y (StartFraction pi Over 4 EndFraction )equals0.

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

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.

sorry can't

step-by-step explanation:

i googled though and something came up but it was in science langueage not common english sooo

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3