Finding a First Integral of a System of First-Order DEs
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I am trying to show that the system $$fracdx_1dt=ax_1, fracdx_2dt=-x_2$$ has a first integral of the form $$K(x_1,x_2)=ln(f(x_1))+ln(g(x_2))a$$
My attempt:
I will use the following method.
$$fracdx_1dt=ax_1iff fracdx_1ax_1=dt (1)$$
$$fracdx_2dt=-x_2iff fracdx_2-x_2=dt (2)$$
Equating $(1)$ and $(2)$ yields
beginalign
fracdx_1ax_1&=fracdx_2-x_2 \
frac1aln|x_1|&=-ln|x_2|+C \
ln|x_1|+aln|x_2|&=0 (C=0)
endalign
I do not know where to go from here. Is my working correct so far? Any advice would be greatly appreciated.
differential-equations proof-verification systems-of-equations
asked Sep 2 at 7:33
Bell
835314
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up vote
4
down vote
favorite
I am trying to show that the system $$fracdx_1dt=ax_1, fracdx_2dt=-x_2$$ has a first integral of the form $$K(x_1,x_2)=ln(f(x_1))+ln(g(x_2))a$$
My attempt:
I will use the following method.
$$fracdx_1dt=ax_1iff fracdx_1ax_1=dt (1)$$
$$fracdx_2dt=-x_2iff fracdx_2-x_2=dt (2)$$
Equating $(1)$ and $(2)$ yields
beginalign
fracdx_1ax_1&=fracdx_2-x_2 \
frac1aln|x_1|&=-ln|x_2|+C \
ln|x_1|+aln|x_2|&=0 (C=0)
endalign
I do not know where to go from here. Is my working correct so far? Any advice would be greatly appreciated.
differential-equations proof-verification systems-of-equations
asked Sep 2 at 7:33
Bell
835314
3
$frac1aln|x_1|=-ln|x_2|+C$ is equivalent to $hatC=ln|x_1|+aln|x_2|$, where $hatC$ is the first integral.
– W. mu
Sep 2 at 7:40
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am trying to show that the system $$fracdx_1dt=ax_1, fracdx_2dt=-x_2$$ has a first integral of the form $$K(x_1,x_2)=ln(f(x_1))+ln(g(x_2))a$$
My attempt:
I will use the following method.
$$fracdx_1dt=ax_1iff fracdx_1ax_1=dt (1)$$
$$fracdx_2dt=-x_2iff fracdx_2-x_2=dt (2)$$
Equating $(1)$ and $(2)$ yields
beginalign
fracdx_1ax_1&=fracdx_2-x_2 \
frac1aln|x_1|&=-ln|x_2|+C \
ln|x_1|+aln|x_2|&=0 (C=0)
endalign
I do not know where to go from here. Is my working correct so far? Any advice would be greatly appreciated.
differential-equations proof-verification systems-of-equations
asked Sep 2 at 7:33
Bell
835314
I am trying to show that the system $$fracdx_1dt=ax_1, fracdx_2dt=-x_2$$ has a first integral of the form $$K(x_1,x_2)=ln(f(x_1))+ln(g(x_2))a$$
My attempt:
I will use the following method.
$$fracdx_1dt=ax_1iff fracdx_1ax_1=dt (1)$$
$$fracdx_2dt=-x_2iff fracdx_2-x_2=dt (2)$$
Equating $(1)$ and $(2)$ yields
beginalign
fracdx_1ax_1&=fracdx_2-x_2 \
frac1aln|x_1|&=-ln|x_2|+C \
ln|x_1|+aln|x_2|&=0 (C=0)
endalign
I do not know where to go from here. Is my working correct so far? Any advice would be greatly appreciated.
differential-equations proof-verification systems-of-equations
differential-equations proof-verification systems-of-equations
asked Sep 2 at 7:33
Bell
835314
asked Sep 2 at 7:33
Bell
835314
asked Sep 2 at 7:33
Bell
835314
asked Sep 2 at 7:33
Bell
835314
asked Sep 2 at 7:33
Bell
835314
835314
3
$frac1aln|x_1|=-ln|x_2|+C$ is equivalent to $hatC=ln|x_1|+aln|x_2|$, where $hatC$ is the first integral.
– W. mu
Sep 2 at 7:40
add a comment |Â
3
$frac1aln|x_1|=-ln|x_2|+C$ is equivalent to $hatC=ln|x_1|+aln|x_2|$, where $hatC$ is the first integral.
– W. mu
Sep 2 at 7:40
3
3
$frac1aln|x_1|=-ln|x_2|+C$ is equivalent to $hatC=ln|x_1|+aln|x_2|$, where $hatC$ is the first integral.
– W. mu
Sep 2 at 7:40
$frac1aln|x_1|=-ln|x_2|+C$ is equivalent to $hatC=ln|x_1|+aln|x_2|$, where $hatC$ is the first integral.
– W. mu
Sep 2 at 7:40
add a comment |Â
1 Answer
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Your solution is correct. At the line before the end, simply multiply by $a$ and observe :
$$frac1aln|x_1|=-ln|x_2|+C Leftrightarrow aC =ln|x_1| + aln|x_2| $$
$$implies$$
$$hatC = ln|x_1| + aln|x_2|$$
$$implies$$
$$C(x_1,x_2) = ln|x_1| + aln|x_2|$$
which is indeed a first integral of the form $K(x_1,x_2)=ln(f(x_1))+ln(g(x_2))a$ with $K(x_1,x_2) = C(x_1,x_2), ; f(x_1) = |x_1|, ; f(x_2) = |x_2|$.
answered Sep 2 at 7:43
Rebellos
10.2k21039
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Your solution is correct. At the line before the end, simply multiply by $a$ and observe :
$$frac1aln|x_1|=-ln|x_2|+C Leftrightarrow aC =ln|x_1| + aln|x_2| $$
$$implies$$
$$hatC = ln|x_1| + aln|x_2|$$
$$implies$$
$$C(x_1,x_2) = ln|x_1| + aln|x_2|$$
which is indeed a first integral of the form $K(x_1,x_2)=ln(f(x_1))+ln(g(x_2))a$ with $K(x_1,x_2) = C(x_1,x_2), ; f(x_1) = |x_1|, ; f(x_2) = |x_2|$.
answered Sep 2 at 7:43
Rebellos
10.2k21039
add a comment |Â
up vote
3
down vote
accepted
Your solution is correct. At the line before the end, simply multiply by $a$ and observe :
$$frac1aln|x_1|=-ln|x_2|+C Leftrightarrow aC =ln|x_1| + aln|x_2| $$
$$implies$$
$$hatC = ln|x_1| + aln|x_2|$$
$$implies$$
$$C(x_1,x_2) = ln|x_1| + aln|x_2|$$
which is indeed a first integral of the form $K(x_1,x_2)=ln(f(x_1))+ln(g(x_2))a$ with $K(x_1,x_2) = C(x_1,x_2), ; f(x_1) = |x_1|, ; f(x_2) = |x_2|$.
answered Sep 2 at 7:43
Rebellos
10.2k21039
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Your solution is correct. At the line before the end, simply multiply by $a$ and observe :
$$frac1aln|x_1|=-ln|x_2|+C Leftrightarrow aC =ln|x_1| + aln|x_2| $$
$$implies$$
$$hatC = ln|x_1| + aln|x_2|$$
$$implies$$
$$C(x_1,x_2) = ln|x_1| + aln|x_2|$$
which is indeed a first integral of the form $K(x_1,x_2)=ln(f(x_1))+ln(g(x_2))a$ with $K(x_1,x_2) = C(x_1,x_2), ; f(x_1) = |x_1|, ; f(x_2) = |x_2|$.
answered Sep 2 at 7:43
Rebellos
10.2k21039
Your solution is correct. At the line before the end, simply multiply by $a$ and observe :
$$frac1aln|x_1|=-ln|x_2|+C Leftrightarrow aC =ln|x_1| + aln|x_2| $$
$$implies$$
$$hatC = ln|x_1| + aln|x_2|$$
$$implies$$
$$C(x_1,x_2) = ln|x_1| + aln|x_2|$$
which is indeed a first integral of the form $K(x_1,x_2)=ln(f(x_1))+ln(g(x_2))a$ with $K(x_1,x_2) = C(x_1,x_2), ; f(x_1) = |x_1|, ; f(x_2) = |x_2|$.
answered Sep 2 at 7:43
Rebellos
10.2k21039
answered Sep 2 at 7:43
Rebellos
10.2k21039
answered Sep 2 at 7:43
Rebellos
10.2k21039
answered Sep 2 at 7:43
Rebellos
10.2k21039
10.2k21039
add a comment |Â
add a comment |Â
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3
$frac1aln|x_1|=-ln|x_2|+C$ is equivalent to $hatC=ln|x_1|+aln|x_2|$, where $hatC$ is the first integral.
– W. mu
Sep 2 at 7:40