Consider the differential equation x2y'' + xy' + y = 0; cos(ln(x)), sin(ln(x)), (0, [infinity]). Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. The functions satisfy the differential equation and are linearly independent since W(cos(ln(x)), sin(ln(x))) = Correct: Your answer is correct. ≠ 0 for 0 < x < [infinity]. Form the general solution. y =

The Wronskian

W(cos(ln(x)), sin(ln(x))) = 1/x

And

The general solution is

y = C1cos(ln(x)) + C2sin(ln(x)

Where C1 and C2 are constants .

Step-by-step explanation:

To verify if the given functions form a fundamental set of solutions to the given differential equation, we find the Wronskian of the two functions.

The Wronskian of functions y1 and y2 is given as

W(y1, y2) = Det (y1, y2; y1', y2')

Where Det(A) signifies the determinant of matrix A

y1' and y2' are the derivatives of y1 and y2 with respect to x respectively.

y1 = cos(ln(x))

y2 = sin(ln(x))

y1' = -(1/x)sin(ln(x))

y2' = (1/x)cos(ln(x))

W(cos(ln(x)), sin(ln(x))) =

Det (cos(ln(x)), sin(ln(x)); -(1/x)sin(ln(x)), (1/x)cos(ln(x)))

= (1/x)cos²(ln(x)) + (1/x)sin²(ln(x))

= (1/x)[cos²(ln(x)) + sin²(ln(x))]

= 1/x(1)

= 1/x

≠ 0

Because the Wronskian is not zero, we say the solutions are linearly independent

The general solution is therefore,

y = C1cos(ln(x)) + C2sin(ln(x)

Where C1 and C2 are constants .

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

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