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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​​.

Respuesta :

Answer:

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.

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