Find the function f2 when the function f1, the wronskian w[f1,f2], and f2 are given.
f1(t)=t-2, w[f1,f2](t)=(3-2t)e^-2t, f2(0)=1

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below: