Fundamentals of differential equations by Nagle Saff Snider, 8th edition, No. 26, Page 445
$begingroup$
find at least the first four nonzero
terms in a power series expansion about for the
solution to the given initial value problem.
$$
(x^2-x+1)y''-y'-y=0;\~\
y(0)=0, ~~ y'(0)=1
$$
I confused how to manipulate..
$$sum_2^infty n(n-1)a_nx^n$$
ordinary-differential-equations power-series
edited Jan 19 at 8:13
LutzL
60.9k42157
asked Jan 19 at 4:41
Nurul AbidahNurul Abidah
12
$endgroup$
add a comment |
$begingroup$
find at least the first four nonzero
terms in a power series expansion about for the
solution to the given initial value problem.
$$
(x^2-x+1)y''-y'-y=0;\~\
y(0)=0, ~~ y'(0)=1
$$
I confused how to manipulate..
$$sum_2^infty n(n-1)a_nx^n$$
ordinary-differential-equations power-series
edited Jan 19 at 8:13
LutzL
60.9k42157
asked Jan 19 at 4:41
Nurul AbidahNurul Abidah
12
$endgroup$
$begingroup$
Welcome to MSE! Please try to include what you have tried so far, and what exactly you are having trouble with, as your not doing so implies the whole "Here's my homework, do it for me vibe". We are here to help, but we need to see some effort on your side first. Thanks!.
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– Aniruddh Venkatesan
Jan 19 at 4:46
add a comment |
$begingroup$
find at least the first four nonzero
terms in a power series expansion about for the
solution to the given initial value problem.
$$
(x^2-x+1)y''-y'-y=0;\~\
y(0)=0, ~~ y'(0)=1
$$
I confused how to manipulate..
$$sum_2^infty n(n-1)a_nx^n$$
ordinary-differential-equations power-series
edited Jan 19 at 8:13
LutzL
60.9k42157
asked Jan 19 at 4:41
Nurul AbidahNurul Abidah
12
$endgroup$
find at least the first four nonzero
terms in a power series expansion about for the
solution to the given initial value problem.
$$
(x^2-x+1)y''-y'-y=0;\~\
y(0)=0, ~~ y'(0)=1
$$
I confused how to manipulate..
$$sum_2^infty n(n-1)a_nx^n$$
ordinary-differential-equations power-series
ordinary-differential-equations power-series
edited Jan 19 at 8:13
LutzL
60.9k42157
asked Jan 19 at 4:41
Nurul AbidahNurul Abidah
12
edited Jan 19 at 8:13
LutzL
60.9k42157
asked Jan 19 at 4:41
Nurul AbidahNurul Abidah
12
edited Jan 19 at 8:13
LutzL
60.9k42157
edited Jan 19 at 8:13
LutzL
60.9k42157
edited Jan 19 at 8:13
LutzL
60.9k42157
60.9k42157
asked Jan 19 at 4:41
Nurul AbidahNurul Abidah
12
asked Jan 19 at 4:41
Nurul AbidahNurul Abidah
12
asked Jan 19 at 4:41
Nurul AbidahNurul Abidah
12
12
$begingroup$
Welcome to MSE! Please try to include what you have tried so far, and what exactly you are having trouble with, as your not doing so implies the whole "Here's my homework, do it for me vibe". We are here to help, but we need to see some effort on your side first. Thanks!.
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:46
add a comment |
$begingroup$
Welcome to MSE! Please try to include what you have tried so far, and what exactly you are having trouble with, as your not doing so implies the whole "Here's my homework, do it for me vibe". We are here to help, but we need to see some effort on your side first. Thanks!.
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:46
$begingroup$
Welcome to MSE! Please try to include what you have tried so far, and what exactly you are having trouble with, as your not doing so implies the whole "Here's my homework, do it for me vibe". We are here to help, but we need to see some effort on your side first. Thanks!.
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:46
$begingroup$
Welcome to MSE! Please try to include what you have tried so far, and what exactly you are having trouble with, as your not doing so implies the whole "Here's my homework, do it for me vibe". We are here to help, but we need to see some effort on your side first. Thanks!.
$endgroup$
– Aniruddh Venkatesan
Jan 19 at 4:46
add a comment |
1 Answer
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I assume your differential equation is $(x^2-x+1)y'' - y'- y = 0$ with the initial conditions $y(0)=0, y'(0) = 1$
The power series (Taylor/Maclaurin) of $y(x) = frac{y(0)}{0!}x^0+frac{y'(0)}{1!}x^1+frac{y''(0)}{2!}x^2+...+frac{y^{(r)}(0)}{r!}x^r+...$
They asked for the first four non-zero terms. Since $y(0) = 0$, that doesn't count. If every derivative up to and including $y^{(4)}$ is non-zero, you're done.
You can easily compute the second derivative directly from the equation by setting $x=0$ and using the given conditions.
You can then differentiate the original equation, and then differentiate again using product rule carefully, and set $x=0$ each time to work out the value of third and fourth derivatives at $x=0$. That should be sufficient. Then apply the Taylor expansion.
answered Jan 19 at 5:56
DeepakDeepak
18.2k11641
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
I assume your differential equation is $(x^2-x+1)y'' - y'- y = 0$ with the initial conditions $y(0)=0, y'(0) = 1$
The power series (Taylor/Maclaurin) of $y(x) = frac{y(0)}{0!}x^0+frac{y'(0)}{1!}x^1+frac{y''(0)}{2!}x^2+...+frac{y^{(r)}(0)}{r!}x^r+...$
They asked for the first four non-zero terms. Since $y(0) = 0$, that doesn't count. If every derivative up to and including $y^{(4)}$ is non-zero, you're done.
You can easily compute the second derivative directly from the equation by setting $x=0$ and using the given conditions.
You can then differentiate the original equation, and then differentiate again using product rule carefully, and set $x=0$ each time to work out the value of third and fourth derivatives at $x=0$. That should be sufficient. Then apply the Taylor expansion.
answered Jan 19 at 5:56
DeepakDeepak
18.2k11641
$endgroup$
add a comment |
$begingroup$
I assume your differential equation is $(x^2-x+1)y'' - y'- y = 0$ with the initial conditions $y(0)=0, y'(0) = 1$
The power series (Taylor/Maclaurin) of $y(x) = frac{y(0)}{0!}x^0+frac{y'(0)}{1!}x^1+frac{y''(0)}{2!}x^2+...+frac{y^{(r)}(0)}{r!}x^r+...$
They asked for the first four non-zero terms. Since $y(0) = 0$, that doesn't count. If every derivative up to and including $y^{(4)}$ is non-zero, you're done.
You can easily compute the second derivative directly from the equation by setting $x=0$ and using the given conditions.
You can then differentiate the original equation, and then differentiate again using product rule carefully, and set $x=0$ each time to work out the value of third and fourth derivatives at $x=0$. That should be sufficient. Then apply the Taylor expansion.
answered Jan 19 at 5:56
DeepakDeepak
18.2k11641
$endgroup$
add a comment |
$begingroup$
I assume your differential equation is $(x^2-x+1)y'' - y'- y = 0$ with the initial conditions $y(0)=0, y'(0) = 1$
The power series (Taylor/Maclaurin) of $y(x) = frac{y(0)}{0!}x^0+frac{y'(0)}{1!}x^1+frac{y''(0)}{2!}x^2+...+frac{y^{(r)}(0)}{r!}x^r+...$
They asked for the first four non-zero terms. Since $y(0) = 0$, that doesn't count. If every derivative up to and including $y^{(4)}$ is non-zero, you're done.
You can easily compute the second derivative directly from the equation by setting $x=0$ and using the given conditions.
You can then differentiate the original equation, and then differentiate again using product rule carefully, and set $x=0$ each time to work out the value of third and fourth derivatives at $x=0$. That should be sufficient. Then apply the Taylor expansion.
answered Jan 19 at 5:56
DeepakDeepak
18.2k11641
$endgroup$
I assume your differential equation is $(x^2-x+1)y'' - y'- y = 0$ with the initial conditions $y(0)=0, y'(0) = 1$
The power series (Taylor/Maclaurin) of $y(x) = frac{y(0)}{0!}x^0+frac{y'(0)}{1!}x^1+frac{y''(0)}{2!}x^2+...+frac{y^{(r)}(0)}{r!}x^r+...$
They asked for the first four non-zero terms. Since $y(0) = 0$, that doesn't count. If every derivative up to and including $y^{(4)}$ is non-zero, you're done.
You can easily compute the second derivative directly from the equation by setting $x=0$ and using the given conditions.
You can then differentiate the original equation, and then differentiate again using product rule carefully, and set $x=0$ each time to work out the value of third and fourth derivatives at $x=0$. That should be sufficient. Then apply the Taylor expansion.
answered Jan 19 at 5:56
DeepakDeepak
18.2k11641
answered Jan 19 at 5:56
DeepakDeepak
18.2k11641
answered Jan 19 at 5:56
DeepakDeepak
18.2k11641
answered Jan 19 at 5:56
DeepakDeepak
18.2k11641
18.2k11641
add a comment |
add a comment |
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Welcome to MSE! Please try to include what you have tried so far, and what exactly you are having trouble with, as your not doing so implies the whole "Here's my homework, do it for me vibe". We are here to help, but we need to see some effort on your side first. Thanks!.
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– Aniruddh Venkatesan
Jan 19 at 4:46