Find g(t)?

Question

If the Wronskian W of f and g is 3e^4t, and if f(t)=e^2t, find g(t).

Answer: 3te^2t+ce^2t

Answer 1

W = fg' - f'g

So,

3e^4t = e^2t g'(t) - [e^2t]' g(t)

==> 3e^4t = e^2t g'(t) - 2e^2t g(t)

Divide both sides by e^2t,

g'(t) - 2g(t) = 3e^2t

This is a typical first order linear differential equation, with

p = -2

e^∫p = e^-2t

and the solution is,

g(t)

= e^2t [∫e^-2t*3e^2t dt + c]

= e^2t (3t + c)

= 3t e^2t + ce^2t

 

Answer 2

The Wronskian of two differentiable functions f  and g is W(f, g) = f g′ – g f ′. 

Given: W(f,g)=3*e^(4*t) and f(t)=e^(2*t), we get f'(t)=2*e^(2*t)

Plug into the Wronskian formula we get an ODE: g'(t) -2g=3*e^(2*t)

1st order, linear, and non-homogeneous ODE, so first we find the homogeneous solution

solve g'-2g=0

can look at this as dg/dt-2g=0 -> dg/dt=2g -> (1/g)dg=2dt, integrate and get ln(g) = 2t +C, solve for g

homogeneous solution for g=e^(2*t +C), which can also be written as g=A*e^(2*t), where A=e^C (constant)

Now looking at the non-homogeneous part 3*e^(2*t), so we can guess the form of B*e^(2*t) as the particular solution. BUT, after observation, we'll find that our guess is also a homogeneous solution and will give us 3e^2t=0, which is incorrect. So, we need to adjust the guess for the particular solution as B*t*e^(2*t).

Now we have g=B*t*e^(2*t)+A*e^(2*t)

Take the derivative we get g'=Be^2t+2Bte^2t+2Ae^2t, then plug into the ODE (g'(t) -2g=3*e^(2*t))

g'(t) -2g= (Be^2t+2Bte^2t+2Ae^2t) -2*(B*t*e^(2*t)+A*e^(2*t)) 

2Bte^2t and 2Ae^2t cancels

so we get g'(t) -2g= Be^2t=3e^2t, which means B must equal to 3

So we get the solution g=3*t*e^(2*t)+A*e^(2*t).

A is a constant and is unknown unless we have more information.

Hope this helps!