Ordinary Differential Equation Problem
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Let $$g$$ and $$ h$$ be solutions of the second-order linear differential equation
$$a(x)y'' +b(x)y' + c(x)y = 0$$
on an interval where a(x) is never
zero.
Show that the Wronskian W of g and h satisfies
$$a(x)W'(x) + b(x)W(x) = 0$$
Now $$W(x) = gh' - hg$$
and
$$W'(x) = h''g - g''h$$
Then I get,
$$a(x)W'(x) + b(x)W(x) = g(a(x)h'' + b(x)h') - h(a(x)g'' + b(x)g')$$
To get this expression to equal 0 I need to show $$c(x) = 0$$ so that the h and g satisfy the ODE to bring 0 but how?
differential-equations
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Let $$g$$ and $$ h$$ be solutions of the second-order linear differential equation
$$a(x)y'' +b(x)y' + c(x)y = 0$$
on an interval where a(x) is never
zero.
Show that the Wronskian W of g and h satisfies
$$a(x)W'(x) + b(x)W(x) = 0$$
Now $$W(x) = gh' - hg$$
and
$$W'(x) = h''g - g''h$$
Then I get,
$$a(x)W'(x) + b(x)W(x) = g(a(x)h'' + b(x)h') - h(a(x)g'' + b(x)g')$$
To get this expression to equal 0 I need to show $$c(x) = 0$$ so that the h and g satisfy the ODE to bring 0 but how?
differential-equations
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $$g$$ and $$ h$$ be solutions of the second-order linear differential equation
$$a(x)y'' +b(x)y' + c(x)y = 0$$
on an interval where a(x) is never
zero.
Show that the Wronskian W of g and h satisfies
$$a(x)W'(x) + b(x)W(x) = 0$$
Now $$W(x) = gh' - hg$$
and
$$W'(x) = h''g - g''h$$
Then I get,
$$a(x)W'(x) + b(x)W(x) = g(a(x)h'' + b(x)h') - h(a(x)g'' + b(x)g')$$
To get this expression to equal 0 I need to show $$c(x) = 0$$ so that the h and g satisfy the ODE to bring 0 but how?
differential-equations
Let $$g$$ and $$ h$$ be solutions of the second-order linear differential equation
$$a(x)y'' +b(x)y' + c(x)y = 0$$
on an interval where a(x) is never
zero.
Show that the Wronskian W of g and h satisfies
$$a(x)W'(x) + b(x)W(x) = 0$$
Now $$W(x) = gh' - hg$$
and
$$W'(x) = h''g - g''h$$
Then I get,
$$a(x)W'(x) + b(x)W(x) = g(a(x)h'' + b(x)h') - h(a(x)g'' + b(x)g')$$
To get this expression to equal 0 I need to show $$c(x) = 0$$ so that the h and g satisfy the ODE to bring 0 but how?
differential-equations
edited Aug 19 at 6:56
asked Aug 19 at 6:46
Gordon Ramsey
545
545
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1 Answer
1
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You haven't used the fact that $h$ and $g$ satisfy the given ODE. This gives you the following equalities:
$$
beginalign*
a(x)h'' + b(x)h' &= -c(x)h \
a(x)g'' + b(x)g' &= -c(x)g.
endalign*
$$
Substituting into your last equation,
$$
beginalign*
a(x)W'(x) + b(x)W(x) &= g(-c(x)h) - h(-c(x)g) \
&= c(x)[-gh + hg] \
&= 0.
endalign*
$$
All good, thanks.
â Gordon Ramsey
Aug 19 at 7:00
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You haven't used the fact that $h$ and $g$ satisfy the given ODE. This gives you the following equalities:
$$
beginalign*
a(x)h'' + b(x)h' &= -c(x)h \
a(x)g'' + b(x)g' &= -c(x)g.
endalign*
$$
Substituting into your last equation,
$$
beginalign*
a(x)W'(x) + b(x)W(x) &= g(-c(x)h) - h(-c(x)g) \
&= c(x)[-gh + hg] \
&= 0.
endalign*
$$
All good, thanks.
â Gordon Ramsey
Aug 19 at 7:00
add a comment |Â
up vote
0
down vote
accepted
You haven't used the fact that $h$ and $g$ satisfy the given ODE. This gives you the following equalities:
$$
beginalign*
a(x)h'' + b(x)h' &= -c(x)h \
a(x)g'' + b(x)g' &= -c(x)g.
endalign*
$$
Substituting into your last equation,
$$
beginalign*
a(x)W'(x) + b(x)W(x) &= g(-c(x)h) - h(-c(x)g) \
&= c(x)[-gh + hg] \
&= 0.
endalign*
$$
All good, thanks.
â Gordon Ramsey
Aug 19 at 7:00
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You haven't used the fact that $h$ and $g$ satisfy the given ODE. This gives you the following equalities:
$$
beginalign*
a(x)h'' + b(x)h' &= -c(x)h \
a(x)g'' + b(x)g' &= -c(x)g.
endalign*
$$
Substituting into your last equation,
$$
beginalign*
a(x)W'(x) + b(x)W(x) &= g(-c(x)h) - h(-c(x)g) \
&= c(x)[-gh + hg] \
&= 0.
endalign*
$$
You haven't used the fact that $h$ and $g$ satisfy the given ODE. This gives you the following equalities:
$$
beginalign*
a(x)h'' + b(x)h' &= -c(x)h \
a(x)g'' + b(x)g' &= -c(x)g.
endalign*
$$
Substituting into your last equation,
$$
beginalign*
a(x)W'(x) + b(x)W(x) &= g(-c(x)h) - h(-c(x)g) \
&= c(x)[-gh + hg] \
&= 0.
endalign*
$$
answered Aug 19 at 6:54
youngsmasher
35537
35537
All good, thanks.
â Gordon Ramsey
Aug 19 at 7:00
add a comment |Â
All good, thanks.
â Gordon Ramsey
Aug 19 at 7:00
All good, thanks.
â Gordon Ramsey
Aug 19 at 7:00
All good, thanks.
â Gordon Ramsey
Aug 19 at 7:00
add a comment |Â
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