Solve $2y''+y'=0$ using power series
$begingroup$
I got the answer $C_0e^{-x/2}$ by doing
$$sum_{n=1}^{infty }:left(x^{n-1}right) left(2nleft(n+1right)C_{n+1}:right)+left(nC_nright) = 0$$
thus giving
$$C_{n+1}= -dfrac{C_n}{2(n+1)}.$$
This leads me to the answer $y = C_0e^{-x/2}$.
However, the answer in my textbook is $C_1e^{-x/2} + C_0$. I've found this to be true, but still couldn't figure out a method of solving the equation so that there are two constants.
Edit: what my textbook did was $$c_{n+2}= -dfrac{1}{2(n+2)}c_{n+1}.$$ , and then substitute $$c_{0}=C_0-2c_1$$ and $$c_1=dfrac{-1}{2}C_1$$ into $y=c_0 +c_1 - dfrac{1}{2*2!}c_{1} x^2. + dfrac{1}{2^2*3!}c_{1} x^3........$
Where did the substitutions come from?
ordinary-differential-equations power-series
asked Dec 8 '18 at 15:59
Fourth Fourth
514
$endgroup$
|
show 6 more comments
$begingroup$
I got the answer $C_0e^{-x/2}$ by doing
$$sum_{n=1}^{infty }:left(x^{n-1}right) left(2nleft(n+1right)C_{n+1}:right)+left(nC_nright) = 0$$
thus giving
$$C_{n+1}= -dfrac{C_n}{2(n+1)}.$$
This leads me to the answer $y = C_0e^{-x/2}$.
However, the answer in my textbook is $C_1e^{-x/2} + C_0$. I've found this to be true, but still couldn't figure out a method of solving the equation so that there are two constants.
Edit: what my textbook did was $$c_{n+2}= -dfrac{1}{2(n+2)}c_{n+1}.$$ , and then substitute $$c_{0}=C_0-2c_1$$ and $$c_1=dfrac{-1}{2}C_1$$ into $y=c_0 +c_1 - dfrac{1}{2*2!}c_{1} x^2. + dfrac{1}{2^2*3!}c_{1} x^3........$
Where did the substitutions come from?
ordinary-differential-equations power-series
asked Dec 8 '18 at 15:59
Fourth Fourth
514
$endgroup$
3
$begingroup$
If you substitute $y(x) = C_0 + C_1 x + C_2 x^2 + ldots$ then the equation reads $2(2C_2 + 3C_3 x + ldots) + (C_1 + 2C_2 x + ldots) = 0$. The recurssion you get only holds from $n=1$ so $C_0$ can be anything so and $C_n$ is determined by $C_1$ giving you a solution on the form $Ae^{-x/2} + B$.
$endgroup$
– Winther
Dec 8 '18 at 16:05
$begingroup$
You can fix the underdetermination with regularization.
$endgroup$
– mathreadler
Dec 8 '18 at 16:11
1
$begingroup$
Also don't omit the question in the text-body even if you have it in title.
$endgroup$
– mathreadler
Dec 8 '18 at 16:19
$begingroup$
Use Picard's rule.
$endgroup$
– Cesareo
Dec 8 '18 at 16:27
2
$begingroup$
The problem is that you translated each identity $$2nleft(n+1right)C_{n+1}+nC_n = 0$$ as $$2left(n+1right)C_{n+1}+C_n = 0$$ This holds true for every $nne0$ but not for $n=0$.
$endgroup$
– Did
Dec 8 '18 at 16:27
|
show 6 more comments
$begingroup$
I got the answer $C_0e^{-x/2}$ by doing
$$sum_{n=1}^{infty }:left(x^{n-1}right) left(2nleft(n+1right)C_{n+1}:right)+left(nC_nright) = 0$$
thus giving
$$C_{n+1}= -dfrac{C_n}{2(n+1)}.$$
This leads me to the answer $y = C_0e^{-x/2}$.
However, the answer in my textbook is $C_1e^{-x/2} + C_0$. I've found this to be true, but still couldn't figure out a method of solving the equation so that there are two constants.
Edit: what my textbook did was $$c_{n+2}= -dfrac{1}{2(n+2)}c_{n+1}.$$ , and then substitute $$c_{0}=C_0-2c_1$$ and $$c_1=dfrac{-1}{2}C_1$$ into $y=c_0 +c_1 - dfrac{1}{2*2!}c_{1} x^2. + dfrac{1}{2^2*3!}c_{1} x^3........$
Where did the substitutions come from?
ordinary-differential-equations power-series
asked Dec 8 '18 at 15:59
Fourth Fourth
514
$endgroup$
I got the answer $C_0e^{-x/2}$ by doing
$$sum_{n=1}^{infty }:left(x^{n-1}right) left(2nleft(n+1right)C_{n+1}:right)+left(nC_nright) = 0$$
thus giving
$$C_{n+1}= -dfrac{C_n}{2(n+1)}.$$
This leads me to the answer $y = C_0e^{-x/2}$.
However, the answer in my textbook is $C_1e^{-x/2} + C_0$. I've found this to be true, but still couldn't figure out a method of solving the equation so that there are two constants.
Edit: what my textbook did was $$c_{n+2}= -dfrac{1}{2(n+2)}c_{n+1}.$$ , and then substitute $$c_{0}=C_0-2c_1$$ and $$c_1=dfrac{-1}{2}C_1$$ into $y=c_0 +c_1 - dfrac{1}{2*2!}c_{1} x^2. + dfrac{1}{2^2*3!}c_{1} x^3........$
Where did the substitutions come from?
ordinary-differential-equations power-series
ordinary-differential-equations power-series
asked Dec 8 '18 at 15:59
Fourth Fourth
514
asked Dec 8 '18 at 15:59
Fourth Fourth
514
edited Dec 16 '18 at 5:49
Fourth
asked Dec 8 '18 at 15:59
Fourth Fourth
514
asked Dec 8 '18 at 15:59
Fourth Fourth
514
asked Dec 8 '18 at 15:59
Fourth Fourth
514
514
3
$begingroup$
If you substitute $y(x) = C_0 + C_1 x + C_2 x^2 + ldots$ then the equation reads $2(2C_2 + 3C_3 x + ldots) + (C_1 + 2C_2 x + ldots) = 0$. The recurssion you get only holds from $n=1$ so $C_0$ can be anything so and $C_n$ is determined by $C_1$ giving you a solution on the form $Ae^{-x/2} + B$.
$endgroup$
– Winther
Dec 8 '18 at 16:05
$begingroup$
You can fix the underdetermination with regularization.
$endgroup$
– mathreadler
Dec 8 '18 at 16:11
1
$begingroup$
Also don't omit the question in the text-body even if you have it in title.
$endgroup$
– mathreadler
Dec 8 '18 at 16:19
$begingroup$
Use Picard's rule.
$endgroup$
– Cesareo
Dec 8 '18 at 16:27
2
$begingroup$
The problem is that you translated each identity $$2nleft(n+1right)C_{n+1}+nC_n = 0$$ as $$2left(n+1right)C_{n+1}+C_n = 0$$ This holds true for every $nne0$ but not for $n=0$.
$endgroup$
– Did
Dec 8 '18 at 16:27
|
show 6 more comments
3
$begingroup$
If you substitute $y(x) = C_0 + C_1 x + C_2 x^2 + ldots$ then the equation reads $2(2C_2 + 3C_3 x + ldots) + (C_1 + 2C_2 x + ldots) = 0$. The recurssion you get only holds from $n=1$ so $C_0$ can be anything so and $C_n$ is determined by $C_1$ giving you a solution on the form $Ae^{-x/2} + B$.
$endgroup$
– Winther
Dec 8 '18 at 16:05
$begingroup$
You can fix the underdetermination with regularization.
$endgroup$
– mathreadler
Dec 8 '18 at 16:11
1
$begingroup$
Also don't omit the question in the text-body even if you have it in title.
$endgroup$
– mathreadler
Dec 8 '18 at 16:19
$begingroup$
Use Picard's rule.
$endgroup$
– Cesareo
Dec 8 '18 at 16:27
2
$begingroup$
The problem is that you translated each identity $$2nleft(n+1right)C_{n+1}+nC_n = 0$$ as $$2left(n+1right)C_{n+1}+C_n = 0$$ This holds true for every $nne0$ but not for $n=0$.
$endgroup$
– Did
Dec 8 '18 at 16:27
3
3
$begingroup$
If you substitute $y(x) = C_0 + C_1 x + C_2 x^2 + ldots$ then the equation reads $2(2C_2 + 3C_3 x + ldots) + (C_1 + 2C_2 x + ldots) = 0$. The recurssion you get only holds from $n=1$ so $C_0$ can be anything so and $C_n$ is determined by $C_1$ giving you a solution on the form $Ae^{-x/2} + B$.
$endgroup$
– Winther
Dec 8 '18 at 16:05
$begingroup$
If you substitute $y(x) = C_0 + C_1 x + C_2 x^2 + ldots$ then the equation reads $2(2C_2 + 3C_3 x + ldots) + (C_1 + 2C_2 x + ldots) = 0$. The recurssion you get only holds from $n=1$ so $C_0$ can be anything so and $C_n$ is determined by $C_1$ giving you a solution on the form $Ae^{-x/2} + B$.
$endgroup$
– Winther
Dec 8 '18 at 16:05
$begingroup$
You can fix the underdetermination with regularization.
$endgroup$
– mathreadler
Dec 8 '18 at 16:11
$begingroup$
You can fix the underdetermination with regularization.
$endgroup$
– mathreadler
Dec 8 '18 at 16:11
1
1
$begingroup$
Also don't omit the question in the text-body even if you have it in title.
$endgroup$
– mathreadler
Dec 8 '18 at 16:19
$begingroup$
Also don't omit the question in the text-body even if you have it in title.
$endgroup$
– mathreadler
Dec 8 '18 at 16:19
$begingroup$
Use Picard's rule.
$endgroup$
– Cesareo
Dec 8 '18 at 16:27
$begingroup$
Use Picard's rule.
$endgroup$
– Cesareo
Dec 8 '18 at 16:27
2
2
$begingroup$
The problem is that you translated each identity $$2nleft(n+1right)C_{n+1}+nC_n = 0$$ as $$2left(n+1right)C_{n+1}+C_n = 0$$ This holds true for every $nne0$ but not for $n=0$.
$endgroup$
– Did
Dec 8 '18 at 16:27
$begingroup$
The problem is that you translated each identity $$2nleft(n+1right)C_{n+1}+nC_n = 0$$ as $$2left(n+1right)C_{n+1}+C_n = 0$$ This holds true for every $nne0$ but not for $n=0$.
$endgroup$
– Did
Dec 8 '18 at 16:27
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
$$ frac{mathrm{d^2}y }{mathrm{d} x^2}=-frac{1}{2} frac{mathrm{d} y}{mathrm{d} x} \
frac{mathrm{d} }{mathrm{d}x}left ( frac{mathrm{d}y }{mathrm{d} x} right ) = -frac{1}{2} frac{mathrm{d} y}{mathrm{d} x} \
ln{frac{mathrm{d}y}{mathrm{d}x}} = frac{-x}{2}+c \
frac{mathrm{d}y}{mathrm{d}x} = c_{1}e^{-frac{x}{2}}
\ y=c_{1}e^{-frac{x}{2}}+c_{2}
$$
answered Dec 8 '18 at 16:17
Lakshya SinhaLakshya Sinha
724
$endgroup$
1
$begingroup$
The intent is to use power series.
$endgroup$
– Sean Roberson
Dec 15 '18 at 6:06
add a comment |
Your Answer
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1 Answer
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$begingroup$
$$ frac{mathrm{d^2}y }{mathrm{d} x^2}=-frac{1}{2} frac{mathrm{d} y}{mathrm{d} x} \
frac{mathrm{d} }{mathrm{d}x}left ( frac{mathrm{d}y }{mathrm{d} x} right ) = -frac{1}{2} frac{mathrm{d} y}{mathrm{d} x} \
ln{frac{mathrm{d}y}{mathrm{d}x}} = frac{-x}{2}+c \
frac{mathrm{d}y}{mathrm{d}x} = c_{1}e^{-frac{x}{2}}
\ y=c_{1}e^{-frac{x}{2}}+c_{2}
$$
answered Dec 8 '18 at 16:17
Lakshya SinhaLakshya Sinha
724
$endgroup$
1
$begingroup$
The intent is to use power series.
$endgroup$
– Sean Roberson
Dec 15 '18 at 6:06
add a comment |
$begingroup$
$$ frac{mathrm{d^2}y }{mathrm{d} x^2}=-frac{1}{2} frac{mathrm{d} y}{mathrm{d} x} \
frac{mathrm{d} }{mathrm{d}x}left ( frac{mathrm{d}y }{mathrm{d} x} right ) = -frac{1}{2} frac{mathrm{d} y}{mathrm{d} x} \
ln{frac{mathrm{d}y}{mathrm{d}x}} = frac{-x}{2}+c \
frac{mathrm{d}y}{mathrm{d}x} = c_{1}e^{-frac{x}{2}}
\ y=c_{1}e^{-frac{x}{2}}+c_{2}
$$
answered Dec 8 '18 at 16:17
Lakshya SinhaLakshya Sinha
724
$endgroup$
1
$begingroup$
The intent is to use power series.
$endgroup$
– Sean Roberson
Dec 15 '18 at 6:06
add a comment |
$begingroup$
$$ frac{mathrm{d^2}y }{mathrm{d} x^2}=-frac{1}{2} frac{mathrm{d} y}{mathrm{d} x} \
frac{mathrm{d} }{mathrm{d}x}left ( frac{mathrm{d}y }{mathrm{d} x} right ) = -frac{1}{2} frac{mathrm{d} y}{mathrm{d} x} \
ln{frac{mathrm{d}y}{mathrm{d}x}} = frac{-x}{2}+c \
frac{mathrm{d}y}{mathrm{d}x} = c_{1}e^{-frac{x}{2}}
\ y=c_{1}e^{-frac{x}{2}}+c_{2}
$$
answered Dec 8 '18 at 16:17
Lakshya SinhaLakshya Sinha
724
$endgroup$
$$ frac{mathrm{d^2}y }{mathrm{d} x^2}=-frac{1}{2} frac{mathrm{d} y}{mathrm{d} x} \
frac{mathrm{d} }{mathrm{d}x}left ( frac{mathrm{d}y }{mathrm{d} x} right ) = -frac{1}{2} frac{mathrm{d} y}{mathrm{d} x} \
ln{frac{mathrm{d}y}{mathrm{d}x}} = frac{-x}{2}+c \
frac{mathrm{d}y}{mathrm{d}x} = c_{1}e^{-frac{x}{2}}
\ y=c_{1}e^{-frac{x}{2}}+c_{2}
$$
answered Dec 8 '18 at 16:17
Lakshya SinhaLakshya Sinha
724
answered Dec 8 '18 at 16:17
Lakshya SinhaLakshya Sinha
724
answered Dec 8 '18 at 16:17
Lakshya SinhaLakshya Sinha
724
answered Dec 8 '18 at 16:17
Lakshya SinhaLakshya Sinha
724
724
1
$begingroup$
The intent is to use power series.
$endgroup$
– Sean Roberson
Dec 15 '18 at 6:06
add a comment |
1
$begingroup$
The intent is to use power series.
$endgroup$
– Sean Roberson
Dec 15 '18 at 6:06
1
1
$begingroup$
The intent is to use power series.
$endgroup$
– Sean Roberson
Dec 15 '18 at 6:06
$begingroup$
The intent is to use power series.
$endgroup$
– Sean Roberson
Dec 15 '18 at 6:06
add a comment |
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$begingroup$
If you substitute $y(x) = C_0 + C_1 x + C_2 x^2 + ldots$ then the equation reads $2(2C_2 + 3C_3 x + ldots) + (C_1 + 2C_2 x + ldots) = 0$. The recurssion you get only holds from $n=1$ so $C_0$ can be anything so and $C_n$ is determined by $C_1$ giving you a solution on the form $Ae^{-x/2} + B$.
$endgroup$
– Winther
Dec 8 '18 at 16:05
$begingroup$
You can fix the underdetermination with regularization.
$endgroup$
– mathreadler
Dec 8 '18 at 16:11
1
$begingroup$
Also don't omit the question in the text-body even if you have it in title.
$endgroup$
– mathreadler
Dec 8 '18 at 16:19
$begingroup$
Use Picard's rule.
$endgroup$
– Cesareo
Dec 8 '18 at 16:27
2
$begingroup$
The problem is that you translated each identity $$2nleft(n+1right)C_{n+1}+nC_n = 0$$ as $$2left(n+1right)C_{n+1}+C_n = 0$$ This holds true for every $nne0$ but not for $n=0$.
$endgroup$
– Did
Dec 8 '18 at 16:27