Question

The Wronskian of 2 functions f and g is (x^2)exp(x) . Find g(x) if f(x) = x

Note : X square exponential x

Answer 1

hmm i thought wronskian was something else but there is no f or g in
\[x^2e^x\]

Answer 2

By definition the Wronskian W of f and g is the determinant of this matrix
f' g'
f g

I.e., W(f,g) = f'g - fg'.

Use the information in your problem to write down a differential equation for g

Answer 3

i will be quiet and learn, but i thought it was
\[gf'=fg'\]

Answer 4

ok now i feel like a moron, i misinterpreted the question completely

Answer 5

JamesJ - But after that you will get a non homogeneous first order diff equation . Am i right? I'm stuck there

Satelite - Thanks anyway

Answer 6

what was the equation u got?

Answer 7

W(f,g) = f'g - fg'.
Thus
x^2e^x = g - xg'

or in more standard form,

g' - (1/x)g = -xe^x

Now you can solve that with an integrating factor.

Answer 8

xg'+g=(x^2)(exp(x))

x square exponential x

Answer 9

notice the left side is the product rule
(xg)' = x^2e^x
integrate both sides

Answer 10

Can someone help me out with the integrating factor formula . Is it the same formula we used for non homogenoeous?

Answer 11

x . g(x) = integral of <x^2 . e^x

Am i right?

Answer 12

@lalady, not quite, because of the minus sign. Or at least not in the correct equation.

@ghan: watch this lecture:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-3-solving-first-order-linear-odes/

Using Integrating Factors is a very important basic tool for ODEs and worth learning. The lecture does it well.

Answer 13

oh in his equation there was no minus, im sorry

Answer 14

I made a mistake . Sorry guys