solving Differential Equation $y''+x^2 y'+(2x+1)y=0$
I tried to solve this problem with power series method,
but it became so complicated.
like this: $na_{n-2}+n(n+1)a_{n+1}+a_{n-1}=0 $for $n>=2$
And I cannot solve this a_{n}
How can I get this?
$y''+x^2 y'+(2x+1)y=0$
differential-equations
asked Nov 27 at 7:34
HiReaper_
161
add a comment |
I tried to solve this problem with power series method,
but it became so complicated.
like this: $na_{n-2}+n(n+1)a_{n+1}+a_{n-1}=0 $for $n>=2$
And I cannot solve this a_{n}
How can I get this?
$y''+x^2 y'+(2x+1)y=0$
differential-equations
asked Nov 27 at 7:34
HiReaper_
161
add a comment |
I tried to solve this problem with power series method,
but it became so complicated.
like this: $na_{n-2}+n(n+1)a_{n+1}+a_{n-1}=0 $for $n>=2$
And I cannot solve this a_{n}
How can I get this?
$y''+x^2 y'+(2x+1)y=0$
differential-equations
asked Nov 27 at 7:34
HiReaper_
161
I tried to solve this problem with power series method,
but it became so complicated.
like this: $na_{n-2}+n(n+1)a_{n+1}+a_{n-1}=0 $for $n>=2$
And I cannot solve this a_{n}
How can I get this?
$y''+x^2 y'+(2x+1)y=0$
differential-equations
differential-equations
asked Nov 27 at 7:34
HiReaper_
161
asked Nov 27 at 7:34
HiReaper_
161
asked Nov 27 at 7:34
HiReaper_
161
asked Nov 27 at 7:34
HiReaper_
161
asked Nov 27 at 7:34
HiReaper_
161
161
add a comment |
add a comment |
1 Answer
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Plugging in $y = sum_{n=0}^infty a_nx^n$ gives
$$ sum_{n=2}^infty n(n-1)a_nx^{n-2} + sum_{n=0}^infty (n+2)a_nx^{n+1} + sum_{n=0}^infty a_nx^n = 0 $$
Notice the middle sum's lowest-order term is $x^1$, so we need to first take out the constant term from the other sums, and shift the remaining indexes to get
$$ (2a_2 + a_0) + sum_{n=0}^infty left[(n+3)(n+2)a_{n+3} + (n+2)a_n + a_{n+1} right] x^{n+1} = 0 $$
This gives
begin{cases} a_2 = -dfrac{a_0}{2} \ a_{n+3} = -dfrac{(n+2)a_n + a_{n+1}}{(n+3)(n+2)} end{cases}
In this arrangement, $a_0$ and $a_1$ define $a_3$, $a_1$ and $a_2$ define $a_4$, and so on. Every following coefficient will depend on $a_0$ and $a_1$, which you're free to set.
answered Nov 27 at 9:26
Dylan
12.2k31026
add a comment |
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1 Answer
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Plugging in $y = sum_{n=0}^infty a_nx^n$ gives
$$ sum_{n=2}^infty n(n-1)a_nx^{n-2} + sum_{n=0}^infty (n+2)a_nx^{n+1} + sum_{n=0}^infty a_nx^n = 0 $$
Notice the middle sum's lowest-order term is $x^1$, so we need to first take out the constant term from the other sums, and shift the remaining indexes to get
$$ (2a_2 + a_0) + sum_{n=0}^infty left[(n+3)(n+2)a_{n+3} + (n+2)a_n + a_{n+1} right] x^{n+1} = 0 $$
This gives
begin{cases} a_2 = -dfrac{a_0}{2} \ a_{n+3} = -dfrac{(n+2)a_n + a_{n+1}}{(n+3)(n+2)} end{cases}
In this arrangement, $a_0$ and $a_1$ define $a_3$, $a_1$ and $a_2$ define $a_4$, and so on. Every following coefficient will depend on $a_0$ and $a_1$, which you're free to set.
answered Nov 27 at 9:26
Dylan
12.2k31026
add a comment |
Plugging in $y = sum_{n=0}^infty a_nx^n$ gives
$$ sum_{n=2}^infty n(n-1)a_nx^{n-2} + sum_{n=0}^infty (n+2)a_nx^{n+1} + sum_{n=0}^infty a_nx^n = 0 $$
Notice the middle sum's lowest-order term is $x^1$, so we need to first take out the constant term from the other sums, and shift the remaining indexes to get
$$ (2a_2 + a_0) + sum_{n=0}^infty left[(n+3)(n+2)a_{n+3} + (n+2)a_n + a_{n+1} right] x^{n+1} = 0 $$
This gives
begin{cases} a_2 = -dfrac{a_0}{2} \ a_{n+3} = -dfrac{(n+2)a_n + a_{n+1}}{(n+3)(n+2)} end{cases}
In this arrangement, $a_0$ and $a_1$ define $a_3$, $a_1$ and $a_2$ define $a_4$, and so on. Every following coefficient will depend on $a_0$ and $a_1$, which you're free to set.
answered Nov 27 at 9:26
Dylan
12.2k31026
add a comment |
Plugging in $y = sum_{n=0}^infty a_nx^n$ gives
$$ sum_{n=2}^infty n(n-1)a_nx^{n-2} + sum_{n=0}^infty (n+2)a_nx^{n+1} + sum_{n=0}^infty a_nx^n = 0 $$
Notice the middle sum's lowest-order term is $x^1$, so we need to first take out the constant term from the other sums, and shift the remaining indexes to get
$$ (2a_2 + a_0) + sum_{n=0}^infty left[(n+3)(n+2)a_{n+3} + (n+2)a_n + a_{n+1} right] x^{n+1} = 0 $$
This gives
begin{cases} a_2 = -dfrac{a_0}{2} \ a_{n+3} = -dfrac{(n+2)a_n + a_{n+1}}{(n+3)(n+2)} end{cases}
In this arrangement, $a_0$ and $a_1$ define $a_3$, $a_1$ and $a_2$ define $a_4$, and so on. Every following coefficient will depend on $a_0$ and $a_1$, which you're free to set.
answered Nov 27 at 9:26
Dylan
12.2k31026
Plugging in $y = sum_{n=0}^infty a_nx^n$ gives
$$ sum_{n=2}^infty n(n-1)a_nx^{n-2} + sum_{n=0}^infty (n+2)a_nx^{n+1} + sum_{n=0}^infty a_nx^n = 0 $$
Notice the middle sum's lowest-order term is $x^1$, so we need to first take out the constant term from the other sums, and shift the remaining indexes to get
$$ (2a_2 + a_0) + sum_{n=0}^infty left[(n+3)(n+2)a_{n+3} + (n+2)a_n + a_{n+1} right] x^{n+1} = 0 $$
This gives
begin{cases} a_2 = -dfrac{a_0}{2} \ a_{n+3} = -dfrac{(n+2)a_n + a_{n+1}}{(n+3)(n+2)} end{cases}
In this arrangement, $a_0$ and $a_1$ define $a_3$, $a_1$ and $a_2$ define $a_4$, and so on. Every following coefficient will depend on $a_0$ and $a_1$, which you're free to set.
answered Nov 27 at 9:26
Dylan
12.2k31026
edited Nov 27 at 9:43
answered Nov 27 at 9:26
Dylan
12.2k31026
answered Nov 27 at 9:26
Dylan
12.2k31026
answered Nov 27 at 9:26
Dylan
12.2k31026
12.2k31026
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