Mixing
How can I get $Acos(alpha t)+sin(alpha t)$ from $A'e^{ialpha t}+B'e^{-ialpha t}$ (solution of ODE with...
$begingroup$
Let for example the ODE $y''(t)+alpha ^2y(t)=0$. The caracteristic equation is $r^2+t^2=0$ and thus $r=pm it$. I know that the general solution can be written as $$Acos(alpha t)+Bsin(alpha t).$$
But since $e^{pm alpha t}$ solve the ODE, $$y(t)=Ae^{it}+Be^{-it}$$ solve the equation as well. So I should be able to find $$A'cos(alpha t)+B'sin(alpha t),$$
form $$A'e^{ialpha t}+B'e^{-ialpha t} ?$$
I tried as follow : $$A'e^{ialpha t}+B'e^{-ialpha t}=frac{A'}{sqrt{A'^2+B'^2}} e^{ialpha t}+frac{B'}{sqrt{A'^2+B'^2}}e^{-ialpha t}=cos(varphi )e^{ialpha t}+sin(varphi )e^{-ialpha t}=(sin(varphi )+sin(varphi ))cos(alpha t)+i(cos(varphi)-sin(varphi ) )sin(alpha t),$$
but I still have complex coefficient... may be there is an other way ? Because I can't conclude.
ordinary-differential-equations
asked Jan 14 at 9:12
user623855user623855
1507
$endgroup$
add a comment |
$begingroup$
Let for example the ODE $y''(t)+alpha ^2y(t)=0$. The caracteristic equation is $r^2+t^2=0$ and thus $r=pm it$. I know that the general solution can be written as $$Acos(alpha t)+Bsin(alpha t).$$
But since $e^{pm alpha t}$ solve the ODE, $$y(t)=Ae^{it}+Be^{-it}$$ solve the equation as well. So I should be able to find $$A'cos(alpha t)+B'sin(alpha t),$$
form $$A'e^{ialpha t}+B'e^{-ialpha t} ?$$
I tried as follow : $$A'e^{ialpha t}+B'e^{-ialpha t}=frac{A'}{sqrt{A'^2+B'^2}} e^{ialpha t}+frac{B'}{sqrt{A'^2+B'^2}}e^{-ialpha t}=cos(varphi )e^{ialpha t}+sin(varphi )e^{-ialpha t}=(sin(varphi )+sin(varphi ))cos(alpha t)+i(cos(varphi)-sin(varphi ) )sin(alpha t),$$
but I still have complex coefficient... may be there is an other way ? Because I can't conclude.
ordinary-differential-equations
asked Jan 14 at 9:12
user623855user623855
1507
$endgroup$
add a comment |
$begingroup$
Let for example the ODE $y''(t)+alpha ^2y(t)=0$. The caracteristic equation is $r^2+t^2=0$ and thus $r=pm it$. I know that the general solution can be written as $$Acos(alpha t)+Bsin(alpha t).$$
But since $e^{pm alpha t}$ solve the ODE, $$y(t)=Ae^{it}+Be^{-it}$$ solve the equation as well. So I should be able to find $$A'cos(alpha t)+B'sin(alpha t),$$
form $$A'e^{ialpha t}+B'e^{-ialpha t} ?$$
I tried as follow : $$A'e^{ialpha t}+B'e^{-ialpha t}=frac{A'}{sqrt{A'^2+B'^2}} e^{ialpha t}+frac{B'}{sqrt{A'^2+B'^2}}e^{-ialpha t}=cos(varphi )e^{ialpha t}+sin(varphi )e^{-ialpha t}=(sin(varphi )+sin(varphi ))cos(alpha t)+i(cos(varphi)-sin(varphi ) )sin(alpha t),$$
but I still have complex coefficient... may be there is an other way ? Because I can't conclude.
ordinary-differential-equations
asked Jan 14 at 9:12
user623855user623855
1507
$endgroup$
Let for example the ODE $y''(t)+alpha ^2y(t)=0$. The caracteristic equation is $r^2+t^2=0$ and thus $r=pm it$. I know that the general solution can be written as $$Acos(alpha t)+Bsin(alpha t).$$
But since $e^{pm alpha t}$ solve the ODE, $$y(t)=Ae^{it}+Be^{-it}$$ solve the equation as well. So I should be able to find $$A'cos(alpha t)+B'sin(alpha t),$$
form $$A'e^{ialpha t}+B'e^{-ialpha t} ?$$
I tried as follow : $$A'e^{ialpha t}+B'e^{-ialpha t}=frac{A'}{sqrt{A'^2+B'^2}} e^{ialpha t}+frac{B'}{sqrt{A'^2+B'^2}}e^{-ialpha t}=cos(varphi )e^{ialpha t}+sin(varphi )e^{-ialpha t}=(sin(varphi )+sin(varphi ))cos(alpha t)+i(cos(varphi)-sin(varphi ) )sin(alpha t),$$
but I still have complex coefficient... may be there is an other way ? Because I can't conclude.
ordinary-differential-equations
ordinary-differential-equations
asked Jan 14 at 9:12
user623855user623855
1507
asked Jan 14 at 9:12
user623855user623855
1507
asked Jan 14 at 9:12
user623855user623855
1507
asked Jan 14 at 9:12
user623855user623855
1507
asked Jan 14 at 9:12
user623855user623855
1507
1507
add a comment |
add a comment |
1 Answer
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$begingroup$
Maybe you want to use different symbols for the coefficients,
begin{eqnarray}
A' e^{ialpha t} + B' e^{-ialpha t} &=& A' left[cos alpha t + i sin alpha t right] + B' left[cos alpha t - i sin alpha t right] \
&=& underbrace{(A' + B')}_{A''}cos alpha t + underbrace{i(A' - B')}_{B''}sin alpha t \
&=& A'' cos alpha t + B''sin alpha t
end{eqnarray}
The thing here is that $A'$ or $B'$ maybe complex themselves, so $A'$ and $B'$ will be real.
answered Jan 14 at 9:34
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
You mean $A''$ and $B''$ will be real no ? How can we prove the fact that $A''$ and $B''$ will be real ?
$endgroup$
– user623855
Jan 14 at 11:00
$begingroup$
@user623855 Yes I mean $A''$ and $B''$, sorry, typo. If you require the coefficients $A''$ and $B''$ to be real you work it backwards: $Im A' = - Im B'$ and $Re A' = Re B' $, these are the conditions $A', B'$ must follow so that both $A''$ and $B''$ are real
$endgroup$
– caverac
Jan 14 at 11:13
add a comment |
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1 Answer
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active
oldest
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1 Answer
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$begingroup$
Maybe you want to use different symbols for the coefficients,
begin{eqnarray}
A' e^{ialpha t} + B' e^{-ialpha t} &=& A' left[cos alpha t + i sin alpha t right] + B' left[cos alpha t - i sin alpha t right] \
&=& underbrace{(A' + B')}_{A''}cos alpha t + underbrace{i(A' - B')}_{B''}sin alpha t \
&=& A'' cos alpha t + B''sin alpha t
end{eqnarray}
The thing here is that $A'$ or $B'$ maybe complex themselves, so $A'$ and $B'$ will be real.
answered Jan 14 at 9:34
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
You mean $A''$ and $B''$ will be real no ? How can we prove the fact that $A''$ and $B''$ will be real ?
$endgroup$
– user623855
Jan 14 at 11:00
$begingroup$
@user623855 Yes I mean $A''$ and $B''$, sorry, typo. If you require the coefficients $A''$ and $B''$ to be real you work it backwards: $Im A' = - Im B'$ and $Re A' = Re B' $, these are the conditions $A', B'$ must follow so that both $A''$ and $B''$ are real
$endgroup$
– caverac
Jan 14 at 11:13
add a comment |
$begingroup$
Maybe you want to use different symbols for the coefficients,
begin{eqnarray}
A' e^{ialpha t} + B' e^{-ialpha t} &=& A' left[cos alpha t + i sin alpha t right] + B' left[cos alpha t - i sin alpha t right] \
&=& underbrace{(A' + B')}_{A''}cos alpha t + underbrace{i(A' - B')}_{B''}sin alpha t \
&=& A'' cos alpha t + B''sin alpha t
end{eqnarray}
The thing here is that $A'$ or $B'$ maybe complex themselves, so $A'$ and $B'$ will be real.
answered Jan 14 at 9:34
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
You mean $A''$ and $B''$ will be real no ? How can we prove the fact that $A''$ and $B''$ will be real ?
$endgroup$
– user623855
Jan 14 at 11:00
$begingroup$
@user623855 Yes I mean $A''$ and $B''$, sorry, typo. If you require the coefficients $A''$ and $B''$ to be real you work it backwards: $Im A' = - Im B'$ and $Re A' = Re B' $, these are the conditions $A', B'$ must follow so that both $A''$ and $B''$ are real
$endgroup$
– caverac
Jan 14 at 11:13
add a comment |
$begingroup$
Maybe you want to use different symbols for the coefficients,
begin{eqnarray}
A' e^{ialpha t} + B' e^{-ialpha t} &=& A' left[cos alpha t + i sin alpha t right] + B' left[cos alpha t - i sin alpha t right] \
&=& underbrace{(A' + B')}_{A''}cos alpha t + underbrace{i(A' - B')}_{B''}sin alpha t \
&=& A'' cos alpha t + B''sin alpha t
end{eqnarray}
The thing here is that $A'$ or $B'$ maybe complex themselves, so $A'$ and $B'$ will be real.
answered Jan 14 at 9:34
caveraccaverac
14.6k31130
$endgroup$
Maybe you want to use different symbols for the coefficients,
begin{eqnarray}
A' e^{ialpha t} + B' e^{-ialpha t} &=& A' left[cos alpha t + i sin alpha t right] + B' left[cos alpha t - i sin alpha t right] \
&=& underbrace{(A' + B')}_{A''}cos alpha t + underbrace{i(A' - B')}_{B''}sin alpha t \
&=& A'' cos alpha t + B''sin alpha t
end{eqnarray}
The thing here is that $A'$ or $B'$ maybe complex themselves, so $A'$ and $B'$ will be real.
answered Jan 14 at 9:34
caveraccaverac
14.6k31130
answered Jan 14 at 9:34
caveraccaverac
14.6k31130
answered Jan 14 at 9:34
caveraccaverac
14.6k31130
answered Jan 14 at 9:34
caveraccaverac
14.6k31130
14.6k31130
$begingroup$
You mean $A''$ and $B''$ will be real no ? How can we prove the fact that $A''$ and $B''$ will be real ?
$endgroup$
– user623855
Jan 14 at 11:00
$begingroup$
@user623855 Yes I mean $A''$ and $B''$, sorry, typo. If you require the coefficients $A''$ and $B''$ to be real you work it backwards: $Im A' = - Im B'$ and $Re A' = Re B' $, these are the conditions $A', B'$ must follow so that both $A''$ and $B''$ are real
$endgroup$
– caverac
Jan 14 at 11:13
add a comment |
$begingroup$
You mean $A''$ and $B''$ will be real no ? How can we prove the fact that $A''$ and $B''$ will be real ?
$endgroup$
– user623855
Jan 14 at 11:00
$begingroup$
@user623855 Yes I mean $A''$ and $B''$, sorry, typo. If you require the coefficients $A''$ and $B''$ to be real you work it backwards: $Im A' = - Im B'$ and $Re A' = Re B' $, these are the conditions $A', B'$ must follow so that both $A''$ and $B''$ are real
$endgroup$
– caverac
Jan 14 at 11:13
$begingroup$
You mean $A''$ and $B''$ will be real no ? How can we prove the fact that $A''$ and $B''$ will be real ?
$endgroup$
– user623855
Jan 14 at 11:00
$begingroup$
You mean $A''$ and $B''$ will be real no ? How can we prove the fact that $A''$ and $B''$ will be real ?
$endgroup$
– user623855
Jan 14 at 11:00
$begingroup$
@user623855 Yes I mean $A''$ and $B''$, sorry, typo. If you require the coefficients $A''$ and $B''$ to be real you work it backwards: $Im A' = - Im B'$ and $Re A' = Re B' $, these are the conditions $A', B'$ must follow so that both $A''$ and $B''$ are real
$endgroup$
– caverac
Jan 14 at 11:13
$begingroup$
@user623855 Yes I mean $A''$ and $B''$, sorry, typo. If you require the coefficients $A''$ and $B''$ to be real you work it backwards: $Im A' = - Im B'$ and $Re A' = Re B' $, these are the conditions $A', B'$ must follow so that both $A''$ and $B''$ are real
$endgroup$
– caverac
Jan 14 at 11:13
add a comment |
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