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Solving ODE using power series
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I have to solve
$$y''+y'-2y=0$$
with $y(0)=0$ and $y'(0)=1$.
Using
$$y(x)=sum_n=0^inftya_nx^n$$
I got $a_0=0,a_1=1$ and
$$a_n=frac(n+1)(a_n+1+(n+2)a_n+2)2$$
And I really don't know what can I do now. I mean: is there some recursive thing? I calculated the value of $a_2,a_3,a_4,a_5$ and $a_6$ and I couldn't see anything.
differential-equations power-series
asked Jul 16 at 2:27
mvfs314
422210
add a comment |Â
up vote
0
down vote
favorite
I have to solve
$$y''+y'-2y=0$$
with $y(0)=0$ and $y'(0)=1$.
Using
$$y(x)=sum_n=0^inftya_nx^n$$
I got $a_0=0,a_1=1$ and
$$a_n=frac(n+1)(a_n+1+(n+2)a_n+2)2$$
And I really don't know what can I do now. I mean: is there some recursive thing? I calculated the value of $a_2,a_3,a_4,a_5$ and $a_6$ and I couldn't see anything.
differential-equations power-series
asked Jul 16 at 2:27
mvfs314
422210
What are the values for $a_2, a_3, a_4, a_5$ and $a_6$? Not necessarilly you must find a pattern
– Mateus Rocha
Jul 16 at 2:47
Well, the are "weird", I mean, I really couldn't find anything between them. So could I leave the solution as I described?
– mvfs314
Jul 16 at 2:54
1
Perhaps having a solution would help. The characteristic equation is $r^2 + r - 2 = 0$ so that $r = -2$ or $r = 1$ so that a general solution is $c_1 e^-2x + c_2 e^x$. Since the solution is so simple, I presume you are supposed to simplify the coefficients you get
– Brevan Ellefsen
Jul 16 at 3:36
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have to solve
$$y''+y'-2y=0$$
with $y(0)=0$ and $y'(0)=1$.
Using
$$y(x)=sum_n=0^inftya_nx^n$$
I got $a_0=0,a_1=1$ and
$$a_n=frac(n+1)(a_n+1+(n+2)a_n+2)2$$
And I really don't know what can I do now. I mean: is there some recursive thing? I calculated the value of $a_2,a_3,a_4,a_5$ and $a_6$ and I couldn't see anything.
differential-equations power-series
asked Jul 16 at 2:27
mvfs314
422210
I have to solve
$$y''+y'-2y=0$$
with $y(0)=0$ and $y'(0)=1$.
Using
$$y(x)=sum_n=0^inftya_nx^n$$
I got $a_0=0,a_1=1$ and
$$a_n=frac(n+1)(a_n+1+(n+2)a_n+2)2$$
And I really don't know what can I do now. I mean: is there some recursive thing? I calculated the value of $a_2,a_3,a_4,a_5$ and $a_6$ and I couldn't see anything.
differential-equations power-series
asked Jul 16 at 2:27
mvfs314
422210
asked Jul 16 at 2:27
mvfs314
422210
asked Jul 16 at 2:27
mvfs314
422210
asked Jul 16 at 2:27
mvfs314
422210
422210
What are the values for $a_2, a_3, a_4, a_5$ and $a_6$? Not necessarilly you must find a pattern
– Mateus Rocha
Jul 16 at 2:47
Well, the are "weird", I mean, I really couldn't find anything between them. So could I leave the solution as I described?
– mvfs314
Jul 16 at 2:54
1
Perhaps having a solution would help. The characteristic equation is $r^2 + r - 2 = 0$ so that $r = -2$ or $r = 1$ so that a general solution is $c_1 e^-2x + c_2 e^x$. Since the solution is so simple, I presume you are supposed to simplify the coefficients you get
– Brevan Ellefsen
Jul 16 at 3:36
add a comment |Â
What are the values for $a_2, a_3, a_4, a_5$ and $a_6$? Not necessarilly you must find a pattern
– Mateus Rocha
Jul 16 at 2:47
Well, the are "weird", I mean, I really couldn't find anything between them. So could I leave the solution as I described?
– mvfs314
Jul 16 at 2:54
1
Perhaps having a solution would help. The characteristic equation is $r^2 + r - 2 = 0$ so that $r = -2$ or $r = 1$ so that a general solution is $c_1 e^-2x + c_2 e^x$. Since the solution is so simple, I presume you are supposed to simplify the coefficients you get
– Brevan Ellefsen
Jul 16 at 3:36
What are the values for $a_2, a_3, a_4, a_5$ and $a_6$? Not necessarilly you must find a pattern
– Mateus Rocha
Jul 16 at 2:47
What are the values for $a_2, a_3, a_4, a_5$ and $a_6$? Not necessarilly you must find a pattern
– Mateus Rocha
Jul 16 at 2:47
Well, the are "weird", I mean, I really couldn't find anything between them. So could I leave the solution as I described?
– mvfs314
Jul 16 at 2:54
Well, the are "weird", I mean, I really couldn't find anything between them. So could I leave the solution as I described?
– mvfs314
Jul 16 at 2:54
1
1
Perhaps having a solution would help. The characteristic equation is $r^2 + r - 2 = 0$ so that $r = -2$ or $r = 1$ so that a general solution is $c_1 e^-2x + c_2 e^x$. Since the solution is so simple, I presume you are supposed to simplify the coefficients you get
– Brevan Ellefsen
Jul 16 at 3:36
Perhaps having a solution would help. The characteristic equation is $r^2 + r - 2 = 0$ so that $r = -2$ or $r = 1$ so that a general solution is $c_1 e^-2x + c_2 e^x$. Since the solution is so simple, I presume you are supposed to simplify the coefficients you get
– Brevan Ellefsen
Jul 16 at 3:36
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Beside the solution suggested by Brevan Ellefsen, you could rewrite
$$a_n=frac(n+1)(a_n+1+(n+2)a_n+2)2$$ as
$$2a_n=(n+1)a_n+1+(n+1)(n+2)a_n+2$$ Muliply each side by $n!$ and simplify to get
$$2,n!, a_n=(n+1)!,a_n+1+(n+2)!, a_n+2$$Now, define $b_n=n! , a_n$ to get
$$2 b_n=b_n+1+b_n+2$$ for which the characteristic equation is $$2=r+r^2implies (r-1)(r+2)=0$$ I am sure that you can take it from here and get a simple expression for $a_n$.
answered Jul 16 at 3:49
Claude Leibovici
112k1055126
But why did you get a characteristic equation to the coeficients? Can you do that?
– mvfs314
Jul 16 at 11:05
@mvfs314. How do you solve $2 b_n=b_n+1+b_n+2$ ?
– Claude Leibovici
Jul 16 at 11:10
I don't know hahaha is this an ODE?
– mvfs314
Jul 16 at 11:13
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Beside the solution suggested by Brevan Ellefsen, you could rewrite
$$a_n=frac(n+1)(a_n+1+(n+2)a_n+2)2$$ as
$$2a_n=(n+1)a_n+1+(n+1)(n+2)a_n+2$$ Muliply each side by $n!$ and simplify to get
$$2,n!, a_n=(n+1)!,a_n+1+(n+2)!, a_n+2$$Now, define $b_n=n! , a_n$ to get
$$2 b_n=b_n+1+b_n+2$$ for which the characteristic equation is $$2=r+r^2implies (r-1)(r+2)=0$$ I am sure that you can take it from here and get a simple expression for $a_n$.
answered Jul 16 at 3:49
Claude Leibovici
112k1055126
But why did you get a characteristic equation to the coeficients? Can you do that?
– mvfs314
Jul 16 at 11:05
@mvfs314. How do you solve $2 b_n=b_n+1+b_n+2$ ?
– Claude Leibovici
Jul 16 at 11:10
I don't know hahaha is this an ODE?
– mvfs314
Jul 16 at 11:13
add a comment |Â
up vote
3
down vote
accepted
Beside the solution suggested by Brevan Ellefsen, you could rewrite
$$a_n=frac(n+1)(a_n+1+(n+2)a_n+2)2$$ as
$$2a_n=(n+1)a_n+1+(n+1)(n+2)a_n+2$$ Muliply each side by $n!$ and simplify to get
$$2,n!, a_n=(n+1)!,a_n+1+(n+2)!, a_n+2$$Now, define $b_n=n! , a_n$ to get
$$2 b_n=b_n+1+b_n+2$$ for which the characteristic equation is $$2=r+r^2implies (r-1)(r+2)=0$$ I am sure that you can take it from here and get a simple expression for $a_n$.
answered Jul 16 at 3:49
Claude Leibovici
112k1055126
But why did you get a characteristic equation to the coeficients? Can you do that?
– mvfs314
Jul 16 at 11:05
@mvfs314. How do you solve $2 b_n=b_n+1+b_n+2$ ?
– Claude Leibovici
Jul 16 at 11:10
I don't know hahaha is this an ODE?
– mvfs314
Jul 16 at 11:13
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Beside the solution suggested by Brevan Ellefsen, you could rewrite
$$a_n=frac(n+1)(a_n+1+(n+2)a_n+2)2$$ as
$$2a_n=(n+1)a_n+1+(n+1)(n+2)a_n+2$$ Muliply each side by $n!$ and simplify to get
$$2,n!, a_n=(n+1)!,a_n+1+(n+2)!, a_n+2$$Now, define $b_n=n! , a_n$ to get
$$2 b_n=b_n+1+b_n+2$$ for which the characteristic equation is $$2=r+r^2implies (r-1)(r+2)=0$$ I am sure that you can take it from here and get a simple expression for $a_n$.
answered Jul 16 at 3:49
Claude Leibovici
112k1055126
Beside the solution suggested by Brevan Ellefsen, you could rewrite
$$a_n=frac(n+1)(a_n+1+(n+2)a_n+2)2$$ as
$$2a_n=(n+1)a_n+1+(n+1)(n+2)a_n+2$$ Muliply each side by $n!$ and simplify to get
$$2,n!, a_n=(n+1)!,a_n+1+(n+2)!, a_n+2$$Now, define $b_n=n! , a_n$ to get
$$2 b_n=b_n+1+b_n+2$$ for which the characteristic equation is $$2=r+r^2implies (r-1)(r+2)=0$$ I am sure that you can take it from here and get a simple expression for $a_n$.
answered Jul 16 at 3:49
Claude Leibovici
112k1055126
answered Jul 16 at 3:49
Claude Leibovici
112k1055126
answered Jul 16 at 3:49
Claude Leibovici
112k1055126
answered Jul 16 at 3:49
Claude Leibovici
112k1055126
112k1055126
But why did you get a characteristic equation to the coeficients? Can you do that?
– mvfs314
Jul 16 at 11:05
@mvfs314. How do you solve $2 b_n=b_n+1+b_n+2$ ?
– Claude Leibovici
Jul 16 at 11:10
I don't know hahaha is this an ODE?
– mvfs314
Jul 16 at 11:13
add a comment |Â
But why did you get a characteristic equation to the coeficients? Can you do that?
– mvfs314
Jul 16 at 11:05
@mvfs314. How do you solve $2 b_n=b_n+1+b_n+2$ ?
– Claude Leibovici
Jul 16 at 11:10
I don't know hahaha is this an ODE?
– mvfs314
Jul 16 at 11:13
But why did you get a characteristic equation to the coeficients? Can you do that?
– mvfs314
Jul 16 at 11:05
But why did you get a characteristic equation to the coeficients? Can you do that?
– mvfs314
Jul 16 at 11:05
@mvfs314. How do you solve $2 b_n=b_n+1+b_n+2$ ?
– Claude Leibovici
Jul 16 at 11:10
@mvfs314. How do you solve $2 b_n=b_n+1+b_n+2$ ?
– Claude Leibovici
Jul 16 at 11:10
I don't know hahaha is this an ODE?
– mvfs314
Jul 16 at 11:13
I don't know hahaha is this an ODE?
– mvfs314
Jul 16 at 11:13
add a comment |Â
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What are the values for $a_2, a_3, a_4, a_5$ and $a_6$? Not necessarilly you must find a pattern
– Mateus Rocha
Jul 16 at 2:47
Well, the are "weird", I mean, I really couldn't find anything between them. So could I leave the solution as I described?
– mvfs314
Jul 16 at 2:54
1
Perhaps having a solution would help. The characteristic equation is $r^2 + r - 2 = 0$ so that $r = -2$ or $r = 1$ so that a general solution is $c_1 e^-2x + c_2 e^x$. Since the solution is so simple, I presume you are supposed to simplify the coefficients you get
– Brevan Ellefsen
Jul 16 at 3:36