Observe that equation (3) has constant coefficients. If y1(x) and y2(x) form a fun- damental set of solutions of equation (3), show that y1(ln t) and y2(ln t) form a fundamental set of solutions of equation (2). Show that they are both solutions and also form a fundamental set.

The equations (2) and (3) you referred to are unavailable, but it is clear that you are trying to show that two set of solutions y1 and y2, to a (second-order) differential equation are solutions, and form a fundamental set. This will be explained.

SOLUTION OF A DIFFERENTIAL EQUATION.

Two functions y1 and y2 are set to be solutions to a differential equation if they both satisfy the said differential equation.

Suppose we have a differential equation

y'' + py' + qy = r

If y1 satisfies this differential equation, then

y1'' + py1' + qy1 = r

FUNDAMENTAL SET OF DIFFERENTIAL EQUATION.

Two functions y1 and y2 are said to form a fundamental set of solutions to a second-order differential equation if they are linearly independent. The functions are linearly independent if their Wronskian is different from zero.

If W(y1, y2) ≠ 0

Then solutions y1 and y2 form a fundamental set of the given differential equation.

1.4

2.3

step-by-step explanation:

the formula for the constant of variation is

y= kx

divide each side by x

y/x =kx/x

y/x =k

we know a and y

1.   y/x = k

4/1 =4

check the other values to make sure they are the same

8/2 = 4   12/3 =4   and 16/4 =4  

the constant of variation is 4

2.   y/x = k

3/1 =3

check the other values to make sure they are the same

6/2 = 3   9/3 =3   and 12/4 =3  

the constant of variation is 3

75 tickets have been sold so far.

step-by-step explanation:

in this question it is clearly mentioned that 45 children's tickets have been sold and this number is 60% of the total tickets sold.

now let's assume the total number of tickets sold be x.

then 60% of x = 45

⇒

⇒ x(.6) = 45

by dividing with .6 on both the sides of the equation

⇒ x = 75

therefore 75 tickets have been sold so far.