Suppose the Wronskian of ff and gg is W (t) = 5e12tW (t) = 5e12t, where f (t) = e14tf (t) = e14t. Find the general formula for g (t) g (t), writing C for the arbitrary constant. g (t) = g (t) = 512e-2t+Ce14t512e-2t+Ce14t 5 12 e-2t + C e14t .

g = (-5/16) e^ (-2t) + Ce^ (14t)

Step-by-step explanation:

Given f (t) and g (t).

The Wronskian of f (t) and g (t) is given as

W (t) = 5e^ (12t)

If f (t) = e^ (14t)

We need to find g (t).

First, Wronskian of f and g is the determinant

W (t) = W (f, g) = Det (f, g; f', g')

= fg' - f'g

But f = e^ (14t) and W (t) = 5e^ (12t)

So

5e^ (12t) = e^ (14t). g' - 14e^ (14t). g

Multiplying through by e^ (-14t)

5e^ (-2t) = g' - 14g

Now, we have the first order ordinary differential equation

g' - 14g = 5e^ (-2t)

Multiply through by the integrating factor

IF = e^ (integral of - 14dt)

= e^ (-14t)

So, we have

e^ (-14t) [g' - 14g] = 5e^ (-2t) e^ (-14t)

d[ge^ (-14t) ] = 5e^ (-16t)

Integrate both sides

ge^ (-14t) = (-5/16) e^ (-16t) + C

Multiplying through by e^ (14t)

g = (-5/16) e^ (-2t) + Ce^ (14t)

And this is what we want.