Why are the coefficients of the complementary function of a second order differential equation with complex...
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Once I have the complementary function of a second order differential equation with complex roots; $y=Ae^{(a+bi)x} + Be^{(a-bi)x}$
I expanded this out to $e^{ax} ((A+B)cos(bx) + i(A-B)sin(bx))$
Apparently $A+B$ can be written as $C$ and $i(A-B)$ can be written as $D$ where $C$ and $D$ are real numbers as $A$ and $B$ are conjugate pairs. The textbook explains why they are conjugate by using an example question but I wanted to know more generally why the coefficients $A$ and $B$ are conjugate pairs. Thanks for any help.
differential-equations complex-numbers
edited Nov 13 at 13:04
LutzL
53.6k41953
asked Nov 11 at 11:39
James Rodriguez
1
add a comment |
up vote
0
down vote
favorite
Once I have the complementary function of a second order differential equation with complex roots; $y=Ae^{(a+bi)x} + Be^{(a-bi)x}$
I expanded this out to $e^{ax} ((A+B)cos(bx) + i(A-B)sin(bx))$
Apparently $A+B$ can be written as $C$ and $i(A-B)$ can be written as $D$ where $C$ and $D$ are real numbers as $A$ and $B$ are conjugate pairs. The textbook explains why they are conjugate by using an example question but I wanted to know more generally why the coefficients $A$ and $B$ are conjugate pairs. Thanks for any help.
differential-equations complex-numbers
edited Nov 13 at 13:04
LutzL
53.6k41953
asked Nov 11 at 11:39
James Rodriguez
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 11 at 11:41
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Once I have the complementary function of a second order differential equation with complex roots; $y=Ae^{(a+bi)x} + Be^{(a-bi)x}$
I expanded this out to $e^{ax} ((A+B)cos(bx) + i(A-B)sin(bx))$
Apparently $A+B$ can be written as $C$ and $i(A-B)$ can be written as $D$ where $C$ and $D$ are real numbers as $A$ and $B$ are conjugate pairs. The textbook explains why they are conjugate by using an example question but I wanted to know more generally why the coefficients $A$ and $B$ are conjugate pairs. Thanks for any help.
differential-equations complex-numbers
edited Nov 13 at 13:04
LutzL
53.6k41953
asked Nov 11 at 11:39
James Rodriguez
1
Once I have the complementary function of a second order differential equation with complex roots; $y=Ae^{(a+bi)x} + Be^{(a-bi)x}$
I expanded this out to $e^{ax} ((A+B)cos(bx) + i(A-B)sin(bx))$
Apparently $A+B$ can be written as $C$ and $i(A-B)$ can be written as $D$ where $C$ and $D$ are real numbers as $A$ and $B$ are conjugate pairs. The textbook explains why they are conjugate by using an example question but I wanted to know more generally why the coefficients $A$ and $B$ are conjugate pairs. Thanks for any help.
differential-equations complex-numbers
differential-equations complex-numbers
edited Nov 13 at 13:04
LutzL
53.6k41953
asked Nov 11 at 11:39
James Rodriguez
1
edited Nov 13 at 13:04
LutzL
53.6k41953
asked Nov 11 at 11:39
James Rodriguez
1
edited Nov 13 at 13:04
LutzL
53.6k41953
edited Nov 13 at 13:04
LutzL
53.6k41953
edited Nov 13 at 13:04
LutzL
53.6k41953
53.6k41953
asked Nov 11 at 11:39
James Rodriguez
1
asked Nov 11 at 11:39
James Rodriguez
1
asked Nov 11 at 11:39
James Rodriguez
1
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 11 at 11:41
add a comment |
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 11 at 11:41
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 11 at 11:41
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 11 at 11:41
add a comment |
1 Answer
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$y$ is a real function, $y=bar y$. Comparing coefficients in
$$
Ae^{(a+bi)x}+Be^{(a-bi)x}=y(x)=overline{y(x)}=bar Ae^{(a-bi)x}+bar Be^{(a+bi)x}
$$
gives directly $B=bar A$.
answered Nov 11 at 11:51
LutzL
53.6k41953
Hi, thanks for the reply. I am not sure what the bar above the y means, sorry. This is probably because I am an A level student. Is there a simpler explanation?
– James Rodriguez
Nov 11 at 12:05
It is one notation for the complex conjugate. In more operator oriented fields the hermitean adjoint is noted as $A^*$, for matrices that is $bar A^T$, and that is sometimes also used for scalars, so that $bar y=y^*$.
– LutzL
Nov 11 at 12:19
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$y$ is a real function, $y=bar y$. Comparing coefficients in
$$
Ae^{(a+bi)x}+Be^{(a-bi)x}=y(x)=overline{y(x)}=bar Ae^{(a-bi)x}+bar Be^{(a+bi)x}
$$
gives directly $B=bar A$.
answered Nov 11 at 11:51
LutzL
53.6k41953
Hi, thanks for the reply. I am not sure what the bar above the y means, sorry. This is probably because I am an A level student. Is there a simpler explanation?
– James Rodriguez
Nov 11 at 12:05
It is one notation for the complex conjugate. In more operator oriented fields the hermitean adjoint is noted as $A^*$, for matrices that is $bar A^T$, and that is sometimes also used for scalars, so that $bar y=y^*$.
– LutzL
Nov 11 at 12:19
add a comment |
up vote
0
down vote
$y$ is a real function, $y=bar y$. Comparing coefficients in
$$
Ae^{(a+bi)x}+Be^{(a-bi)x}=y(x)=overline{y(x)}=bar Ae^{(a-bi)x}+bar Be^{(a+bi)x}
$$
gives directly $B=bar A$.
answered Nov 11 at 11:51
LutzL
53.6k41953
Hi, thanks for the reply. I am not sure what the bar above the y means, sorry. This is probably because I am an A level student. Is there a simpler explanation?
– James Rodriguez
Nov 11 at 12:05
It is one notation for the complex conjugate. In more operator oriented fields the hermitean adjoint is noted as $A^*$, for matrices that is $bar A^T$, and that is sometimes also used for scalars, so that $bar y=y^*$.
– LutzL
Nov 11 at 12:19
add a comment |
up vote
0
down vote
up vote
0
down vote
$y$ is a real function, $y=bar y$. Comparing coefficients in
$$
Ae^{(a+bi)x}+Be^{(a-bi)x}=y(x)=overline{y(x)}=bar Ae^{(a-bi)x}+bar Be^{(a+bi)x}
$$
gives directly $B=bar A$.
answered Nov 11 at 11:51
LutzL
53.6k41953
$y$ is a real function, $y=bar y$. Comparing coefficients in
$$
Ae^{(a+bi)x}+Be^{(a-bi)x}=y(x)=overline{y(x)}=bar Ae^{(a-bi)x}+bar Be^{(a+bi)x}
$$
gives directly $B=bar A$.
answered Nov 11 at 11:51
LutzL
53.6k41953
answered Nov 11 at 11:51
LutzL
53.6k41953
answered Nov 11 at 11:51
LutzL
53.6k41953
answered Nov 11 at 11:51
LutzL
53.6k41953
53.6k41953
Hi, thanks for the reply. I am not sure what the bar above the y means, sorry. This is probably because I am an A level student. Is there a simpler explanation?
– James Rodriguez
Nov 11 at 12:05
It is one notation for the complex conjugate. In more operator oriented fields the hermitean adjoint is noted as $A^*$, for matrices that is $bar A^T$, and that is sometimes also used for scalars, so that $bar y=y^*$.
– LutzL
Nov 11 at 12:19
add a comment |
Hi, thanks for the reply. I am not sure what the bar above the y means, sorry. This is probably because I am an A level student. Is there a simpler explanation?
– James Rodriguez
Nov 11 at 12:05
It is one notation for the complex conjugate. In more operator oriented fields the hermitean adjoint is noted as $A^*$, for matrices that is $bar A^T$, and that is sometimes also used for scalars, so that $bar y=y^*$.
– LutzL
Nov 11 at 12:19
Hi, thanks for the reply. I am not sure what the bar above the y means, sorry. This is probably because I am an A level student. Is there a simpler explanation?
– James Rodriguez
Nov 11 at 12:05
Hi, thanks for the reply. I am not sure what the bar above the y means, sorry. This is probably because I am an A level student. Is there a simpler explanation?
– James Rodriguez
Nov 11 at 12:05
It is one notation for the complex conjugate. In more operator oriented fields the hermitean adjoint is noted as $A^*$, for matrices that is $bar A^T$, and that is sometimes also used for scalars, so that $bar y=y^*$.
– LutzL
Nov 11 at 12:19
It is one notation for the complex conjugate. In more operator oriented fields the hermitean adjoint is noted as $A^*$, for matrices that is $bar A^T$, and that is sometimes also used for scalars, so that $bar y=y^*$.
– LutzL
Nov 11 at 12:19
add a comment |
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 11 at 11:41