Second order linear ode with variable coefficient and imaginary roots
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Given the ode:$$t^2y''+ty'+y=0$$
The solution should be of the form $$y=t^r$$
So:
$t^2(t^r)''+t(t^r)'+t^r=0$
$$...$$
$t^r(r^2+1)=0$
$r_1=i$ and $r_2=-i$
$y_1=c_1e^{it}$ and $y_2=c_2e^{-it}$
Using Eulers Formula we get
$y_1=c_1(cost+isint)$ and $y_2=c_2(cost-isint)$
$y(t)=y_1(t)+y_2(t)$
$y(t)=c_1(cost+isint)+c_2(cost-isint)$
$y(t)=cost(c_1+c_2)+isint(c_1-c_2)$
Is this correct? If not where am I mistaken?
differential-equations
asked Dec 1 at 2:59
VakiPitsi
1637
add a comment |
up vote
0
down vote
favorite
Given the ode:$$t^2y''+ty'+y=0$$
The solution should be of the form $$y=t^r$$
So:
$t^2(t^r)''+t(t^r)'+t^r=0$
$$...$$
$t^r(r^2+1)=0$
$r_1=i$ and $r_2=-i$
$y_1=c_1e^{it}$ and $y_2=c_2e^{-it}$
Using Eulers Formula we get
$y_1=c_1(cost+isint)$ and $y_2=c_2(cost-isint)$
$y(t)=y_1(t)+y_2(t)$
$y(t)=c_1(cost+isint)+c_2(cost-isint)$
$y(t)=cost(c_1+c_2)+isint(c_1-c_2)$
Is this correct? If not where am I mistaken?
differential-equations
asked Dec 1 at 2:59
VakiPitsi
1637
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given the ode:$$t^2y''+ty'+y=0$$
The solution should be of the form $$y=t^r$$
So:
$t^2(t^r)''+t(t^r)'+t^r=0$
$$...$$
$t^r(r^2+1)=0$
$r_1=i$ and $r_2=-i$
$y_1=c_1e^{it}$ and $y_2=c_2e^{-it}$
Using Eulers Formula we get
$y_1=c_1(cost+isint)$ and $y_2=c_2(cost-isint)$
$y(t)=y_1(t)+y_2(t)$
$y(t)=c_1(cost+isint)+c_2(cost-isint)$
$y(t)=cost(c_1+c_2)+isint(c_1-c_2)$
Is this correct? If not where am I mistaken?
differential-equations
asked Dec 1 at 2:59
VakiPitsi
1637
Given the ode:$$t^2y''+ty'+y=0$$
The solution should be of the form $$y=t^r$$
So:
$t^2(t^r)''+t(t^r)'+t^r=0$
$$...$$
$t^r(r^2+1)=0$
$r_1=i$ and $r_2=-i$
$y_1=c_1e^{it}$ and $y_2=c_2e^{-it}$
Using Eulers Formula we get
$y_1=c_1(cost+isint)$ and $y_2=c_2(cost-isint)$
$y(t)=y_1(t)+y_2(t)$
$y(t)=c_1(cost+isint)+c_2(cost-isint)$
$y(t)=cost(c_1+c_2)+isint(c_1-c_2)$
Is this correct? If not where am I mistaken?
differential-equations
differential-equations
asked Dec 1 at 2:59
VakiPitsi
1637
asked Dec 1 at 2:59
VakiPitsi
1637
asked Dec 1 at 2:59
VakiPitsi
1637
asked Dec 1 at 2:59
VakiPitsi
1637
asked Dec 1 at 2:59
VakiPitsi
1637
1637
add a comment |
add a comment |
1 Answer
1
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oldest
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up vote
2
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accepted
You started off assuming that the solution has the form $t^r$ for some $r$ and correctly obtained $r=pm i$.
But then you wrote $y_1=c_1e^{it}$ instead of $y_1=c_1t^i$, plugging back into the wrong form.
Instead, go on with $$t^i=left(e^{ln(t)}right)^i=e^{iln(t)}=cos(ln(t))+isin(ln(t))$$
and you should be able to finish the problem.
answered Dec 1 at 3:24
obscurans
52117
Ohh I see thanks a lot!
– VakiPitsi
Dec 1 at 3:34
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You started off assuming that the solution has the form $t^r$ for some $r$ and correctly obtained $r=pm i$.
But then you wrote $y_1=c_1e^{it}$ instead of $y_1=c_1t^i$, plugging back into the wrong form.
Instead, go on with $$t^i=left(e^{ln(t)}right)^i=e^{iln(t)}=cos(ln(t))+isin(ln(t))$$
and you should be able to finish the problem.
answered Dec 1 at 3:24
obscurans
52117
Ohh I see thanks a lot!
– VakiPitsi
Dec 1 at 3:34
add a comment |
up vote
2
down vote
accepted
You started off assuming that the solution has the form $t^r$ for some $r$ and correctly obtained $r=pm i$.
But then you wrote $y_1=c_1e^{it}$ instead of $y_1=c_1t^i$, plugging back into the wrong form.
Instead, go on with $$t^i=left(e^{ln(t)}right)^i=e^{iln(t)}=cos(ln(t))+isin(ln(t))$$
and you should be able to finish the problem.
answered Dec 1 at 3:24
obscurans
52117
Ohh I see thanks a lot!
– VakiPitsi
Dec 1 at 3:34
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You started off assuming that the solution has the form $t^r$ for some $r$ and correctly obtained $r=pm i$.
But then you wrote $y_1=c_1e^{it}$ instead of $y_1=c_1t^i$, plugging back into the wrong form.
Instead, go on with $$t^i=left(e^{ln(t)}right)^i=e^{iln(t)}=cos(ln(t))+isin(ln(t))$$
and you should be able to finish the problem.
answered Dec 1 at 3:24
obscurans
52117
You started off assuming that the solution has the form $t^r$ for some $r$ and correctly obtained $r=pm i$.
But then you wrote $y_1=c_1e^{it}$ instead of $y_1=c_1t^i$, plugging back into the wrong form.
Instead, go on with $$t^i=left(e^{ln(t)}right)^i=e^{iln(t)}=cos(ln(t))+isin(ln(t))$$
and you should be able to finish the problem.
answered Dec 1 at 3:24
obscurans
52117
answered Dec 1 at 3:24
obscurans
52117
answered Dec 1 at 3:24
obscurans
52117
answered Dec 1 at 3:24
obscurans
52117
52117
Ohh I see thanks a lot!
– VakiPitsi
Dec 1 at 3:34
add a comment |
Ohh I see thanks a lot!
– VakiPitsi
Dec 1 at 3:34
Ohh I see thanks a lot!
– VakiPitsi
Dec 1 at 3:34
Ohh I see thanks a lot!
– VakiPitsi
Dec 1 at 3:34
add a comment |
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