Mixing
How to solve these simultaneous differential equations
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I want to solve the following equations for $x_1$, $x_2$, and $x_3$
$$ K_0, x_1(t) + K_1, frac{dx_1(t)}{dt} + M_0, frac{dx_2(t)}{dt} + x_3(t) =A_0,sin (alpha_0 t) tag{1} $$
$$ M_0, frac{dx_1(t)}{dt} + L_0,x_2(t) + K_1, frac{dx_2(t)}{dt}+L_1,int x_2(t), dt =0 tag{2} $$
$$ x_1(t) -alpha_1,B_1,cos(alpha_1,t + phi) , x_3(t)- big[B_0+B_1,sin(alpha_1,t + phi) big] frac{dx_3(t)}{dt} =0 tag{3} $$
Here is my try using Laplace transform:
I can assume $x_1(0)=0$, and define $x_2(0)=x_{20}$, and $x_3(0)=x_{30}$
$$(1) Rightarrow K_0, X_1(s) + K_1, sX_1(s)+ M_0, sX_2(s)+ X_3(s) + M_0, x_{20} = frac{A_0alpha_0}{s^2+alpha_0^2} tag{4} $$
$$(2) Rightarrow M_0, X_1(s) + L_0, X_2(s) +K_1, sX_2(s)+ L_1, frac{X_2(s)}{s}+ X_3(s)+ K_1, x_{20} =0 tag{5} $$
Next step brings me the trouble. I cannot simply take the Laplace transform of (3)
Update 1 (as suggested by @Cesareo)
Using the Euler's expansion of sine and cosine (3) can be transformed to
$$ X_1(s) - B_0 s X_3(s) -\
frac{B_1 (alpha_1 +1) s}{2} big(e^{i phi} X_3(s-i alpha_1) + e^{-i phi} X_3(s+i alpha_1)big)+x_{30} (B_0 - B_1 sin phi)=0 tag{6}$$
Now, my question is:
How can I move forward from here? How can I handle $X_3(s pm i alpha_1)$ ifI want to solve for $X_3(s)$?
differential-equations
asked 2 days ago
Pojj
33
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Pojj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
I want to solve the following equations for $x_1$, $x_2$, and $x_3$
$$ K_0, x_1(t) + K_1, frac{dx_1(t)}{dt} + M_0, frac{dx_2(t)}{dt} + x_3(t) =A_0,sin (alpha_0 t) tag{1} $$
$$ M_0, frac{dx_1(t)}{dt} + L_0,x_2(t) + K_1, frac{dx_2(t)}{dt}+L_1,int x_2(t), dt =0 tag{2} $$
$$ x_1(t) -alpha_1,B_1,cos(alpha_1,t + phi) , x_3(t)- big[B_0+B_1,sin(alpha_1,t + phi) big] frac{dx_3(t)}{dt} =0 tag{3} $$
Here is my try using Laplace transform:
I can assume $x_1(0)=0$, and define $x_2(0)=x_{20}$, and $x_3(0)=x_{30}$
$$(1) Rightarrow K_0, X_1(s) + K_1, sX_1(s)+ M_0, sX_2(s)+ X_3(s) + M_0, x_{20} = frac{A_0alpha_0}{s^2+alpha_0^2} tag{4} $$
$$(2) Rightarrow M_0, X_1(s) + L_0, X_2(s) +K_1, sX_2(s)+ L_1, frac{X_2(s)}{s}+ X_3(s)+ K_1, x_{20} =0 tag{5} $$
Next step brings me the trouble. I cannot simply take the Laplace transform of (3)
Update 1 (as suggested by @Cesareo)
Using the Euler's expansion of sine and cosine (3) can be transformed to
$$ X_1(s) - B_0 s X_3(s) -\
frac{B_1 (alpha_1 +1) s}{2} big(e^{i phi} X_3(s-i alpha_1) + e^{-i phi} X_3(s+i alpha_1)big)+x_{30} (B_0 - B_1 sin phi)=0 tag{6}$$
Now, my question is:
How can I move forward from here? How can I handle $X_3(s pm i alpha_1)$ ifI want to solve for $X_3(s)$?
differential-equations
asked 2 days ago
Pojj
33
New contributor
Pojj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to solve the following equations for $x_1$, $x_2$, and $x_3$
$$ K_0, x_1(t) + K_1, frac{dx_1(t)}{dt} + M_0, frac{dx_2(t)}{dt} + x_3(t) =A_0,sin (alpha_0 t) tag{1} $$
$$ M_0, frac{dx_1(t)}{dt} + L_0,x_2(t) + K_1, frac{dx_2(t)}{dt}+L_1,int x_2(t), dt =0 tag{2} $$
$$ x_1(t) -alpha_1,B_1,cos(alpha_1,t + phi) , x_3(t)- big[B_0+B_1,sin(alpha_1,t + phi) big] frac{dx_3(t)}{dt} =0 tag{3} $$
Here is my try using Laplace transform:
I can assume $x_1(0)=0$, and define $x_2(0)=x_{20}$, and $x_3(0)=x_{30}$
$$(1) Rightarrow K_0, X_1(s) + K_1, sX_1(s)+ M_0, sX_2(s)+ X_3(s) + M_0, x_{20} = frac{A_0alpha_0}{s^2+alpha_0^2} tag{4} $$
$$(2) Rightarrow M_0, X_1(s) + L_0, X_2(s) +K_1, sX_2(s)+ L_1, frac{X_2(s)}{s}+ X_3(s)+ K_1, x_{20} =0 tag{5} $$
Next step brings me the trouble. I cannot simply take the Laplace transform of (3)
Update 1 (as suggested by @Cesareo)
Using the Euler's expansion of sine and cosine (3) can be transformed to
$$ X_1(s) - B_0 s X_3(s) -\
frac{B_1 (alpha_1 +1) s}{2} big(e^{i phi} X_3(s-i alpha_1) + e^{-i phi} X_3(s+i alpha_1)big)+x_{30} (B_0 - B_1 sin phi)=0 tag{6}$$
Now, my question is:
How can I move forward from here? How can I handle $X_3(s pm i alpha_1)$ ifI want to solve for $X_3(s)$?
differential-equations
asked 2 days ago
Pojj
33
New contributor
Pojj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I want to solve the following equations for $x_1$, $x_2$, and $x_3$
$$ K_0, x_1(t) + K_1, frac{dx_1(t)}{dt} + M_0, frac{dx_2(t)}{dt} + x_3(t) =A_0,sin (alpha_0 t) tag{1} $$
$$ M_0, frac{dx_1(t)}{dt} + L_0,x_2(t) + K_1, frac{dx_2(t)}{dt}+L_1,int x_2(t), dt =0 tag{2} $$
$$ x_1(t) -alpha_1,B_1,cos(alpha_1,t + phi) , x_3(t)- big[B_0+B_1,sin(alpha_1,t + phi) big] frac{dx_3(t)}{dt} =0 tag{3} $$
Here is my try using Laplace transform:
I can assume $x_1(0)=0$, and define $x_2(0)=x_{20}$, and $x_3(0)=x_{30}$
$$(1) Rightarrow K_0, X_1(s) + K_1, sX_1(s)+ M_0, sX_2(s)+ X_3(s) + M_0, x_{20} = frac{A_0alpha_0}{s^2+alpha_0^2} tag{4} $$
$$(2) Rightarrow M_0, X_1(s) + L_0, X_2(s) +K_1, sX_2(s)+ L_1, frac{X_2(s)}{s}+ X_3(s)+ K_1, x_{20} =0 tag{5} $$
Next step brings me the trouble. I cannot simply take the Laplace transform of (3)
Update 1 (as suggested by @Cesareo)
Using the Euler's expansion of sine and cosine (3) can be transformed to
$$ X_1(s) - B_0 s X_3(s) -\
frac{B_1 (alpha_1 +1) s}{2} big(e^{i phi} X_3(s-i alpha_1) + e^{-i phi} X_3(s+i alpha_1)big)+x_{30} (B_0 - B_1 sin phi)=0 tag{6}$$
Now, my question is:
How can I move forward from here? How can I handle $X_3(s pm i alpha_1)$ ifI want to solve for $X_3(s)$?
differential-equations
differential-equations
asked 2 days ago
Pojj
33
New contributor
Pojj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Pojj
33
New contributor
Pojj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
asked 2 days ago
Pojj
33
New contributor
Pojj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Pojj
33
asked 2 days ago
Pojj
33
33
New contributor
Pojj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Pojj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Pojj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Hint.
$$
mathcal{L}(cos(alpha t)x(t)) = mathcal{L}(frac 12(e^{ialpha t}+e^{-ialpha t})x(t))
$$
and
$$
int_0^{infty}e^{ialpha t}x(t)e^{-s}dt = int_0^{infty}e^{-(s-ialpha)t}x(t) dt = X(s-ialpha)
$$
so after the Laplace transformation you should handle a system of difference equations ...
answered 2 days ago
Cesareo
7,2523416
Thanks for the tip. I will try it and update here. isn't $mathcal{L} big(e^{i alpha t} x(t) big) = X(s-i alpha) $?
– Pojj
yesterday
@Pojj You are right. I had a little imagination left.
– Cesareo
yesterday
any suggestion how I can move forward from here? (Edited the question)
– Pojj
16 hours ago
@Pojj In the same line, eliminating $X_1(s), X_2(s)$ we arrive to a quasi polynomial equation for $X_3(s)$ with the structure $$ f_1(s)X_3(s)+f_2(s)left(e^{iphi}X_3(s-ialpha_1)+e^{-iphi}X_3(s+ialpha_1)right)= C_0 $$ with $f_1(s),f_2(s)$ polynomial fractions ...
– Cesareo
9 hours ago
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
accepted
Hint.
$$
mathcal{L}(cos(alpha t)x(t)) = mathcal{L}(frac 12(e^{ialpha t}+e^{-ialpha t})x(t))
$$
and
$$
int_0^{infty}e^{ialpha t}x(t)e^{-s}dt = int_0^{infty}e^{-(s-ialpha)t}x(t) dt = X(s-ialpha)
$$
so after the Laplace transformation you should handle a system of difference equations ...
answered 2 days ago
Cesareo
7,2523416
Thanks for the tip. I will try it and update here. isn't $mathcal{L} big(e^{i alpha t} x(t) big) = X(s-i alpha) $?
– Pojj
yesterday
@Pojj You are right. I had a little imagination left.
– Cesareo
yesterday
any suggestion how I can move forward from here? (Edited the question)
– Pojj
16 hours ago
@Pojj In the same line, eliminating $X_1(s), X_2(s)$ we arrive to a quasi polynomial equation for $X_3(s)$ with the structure $$ f_1(s)X_3(s)+f_2(s)left(e^{iphi}X_3(s-ialpha_1)+e^{-iphi}X_3(s+ialpha_1)right)= C_0 $$ with $f_1(s),f_2(s)$ polynomial fractions ...
– Cesareo
9 hours ago
add a comment |
up vote
0
down vote
accepted
Hint.
$$
mathcal{L}(cos(alpha t)x(t)) = mathcal{L}(frac 12(e^{ialpha t}+e^{-ialpha t})x(t))
$$
and
$$
int_0^{infty}e^{ialpha t}x(t)e^{-s}dt = int_0^{infty}e^{-(s-ialpha)t}x(t) dt = X(s-ialpha)
$$
so after the Laplace transformation you should handle a system of difference equations ...
answered 2 days ago
Cesareo
7,2523416
Thanks for the tip. I will try it and update here. isn't $mathcal{L} big(e^{i alpha t} x(t) big) = X(s-i alpha) $?
– Pojj
yesterday
@Pojj You are right. I had a little imagination left.
– Cesareo
yesterday
any suggestion how I can move forward from here? (Edited the question)
– Pojj
16 hours ago
@Pojj In the same line, eliminating $X_1(s), X_2(s)$ we arrive to a quasi polynomial equation for $X_3(s)$ with the structure $$ f_1(s)X_3(s)+f_2(s)left(e^{iphi}X_3(s-ialpha_1)+e^{-iphi}X_3(s+ialpha_1)right)= C_0 $$ with $f_1(s),f_2(s)$ polynomial fractions ...
– Cesareo
9 hours ago
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Hint.
$$
mathcal{L}(cos(alpha t)x(t)) = mathcal{L}(frac 12(e^{ialpha t}+e^{-ialpha t})x(t))
$$
and
$$
int_0^{infty}e^{ialpha t}x(t)e^{-s}dt = int_0^{infty}e^{-(s-ialpha)t}x(t) dt = X(s-ialpha)
$$
so after the Laplace transformation you should handle a system of difference equations ...
answered 2 days ago
Cesareo
7,2523416
Hint.
$$
mathcal{L}(cos(alpha t)x(t)) = mathcal{L}(frac 12(e^{ialpha t}+e^{-ialpha t})x(t))
$$
and
$$
int_0^{infty}e^{ialpha t}x(t)e^{-s}dt = int_0^{infty}e^{-(s-ialpha)t}x(t) dt = X(s-ialpha)
$$
so after the Laplace transformation you should handle a system of difference equations ...
answered 2 days ago
Cesareo
7,2523416
edited yesterday
answered 2 days ago
Cesareo
7,2523416
answered 2 days ago
Cesareo
7,2523416
answered 2 days ago
Cesareo
7,2523416
7,2523416
Thanks for the tip. I will try it and update here. isn't $mathcal{L} big(e^{i alpha t} x(t) big) = X(s-i alpha) $?
– Pojj
yesterday
@Pojj You are right. I had a little imagination left.
– Cesareo
yesterday
any suggestion how I can move forward from here? (Edited the question)
– Pojj
16 hours ago
@Pojj In the same line, eliminating $X_1(s), X_2(s)$ we arrive to a quasi polynomial equation for $X_3(s)$ with the structure $$ f_1(s)X_3(s)+f_2(s)left(e^{iphi}X_3(s-ialpha_1)+e^{-iphi}X_3(s+ialpha_1)right)= C_0 $$ with $f_1(s),f_2(s)$ polynomial fractions ...
– Cesareo
9 hours ago
add a comment |
Thanks for the tip. I will try it and update here. isn't $mathcal{L} big(e^{i alpha t} x(t) big) = X(s-i alpha) $?
– Pojj
yesterday
@Pojj You are right. I had a little imagination left.
– Cesareo
yesterday
any suggestion how I can move forward from here? (Edited the question)
– Pojj
16 hours ago
@Pojj In the same line, eliminating $X_1(s), X_2(s)$ we arrive to a quasi polynomial equation for $X_3(s)$ with the structure $$ f_1(s)X_3(s)+f_2(s)left(e^{iphi}X_3(s-ialpha_1)+e^{-iphi}X_3(s+ialpha_1)right)= C_0 $$ with $f_1(s),f_2(s)$ polynomial fractions ...
– Cesareo
9 hours ago
Thanks for the tip. I will try it and update here. isn't $mathcal{L} big(e^{i alpha t} x(t) big) = X(s-i alpha) $?
– Pojj
yesterday
Thanks for the tip. I will try it and update here. isn't $mathcal{L} big(e^{i alpha t} x(t) big) = X(s-i alpha) $?
– Pojj
yesterday
@Pojj You are right. I had a little imagination left.
– Cesareo
yesterday
@Pojj You are right. I had a little imagination left.
– Cesareo
yesterday
any suggestion how I can move forward from here? (Edited the question)
– Pojj
16 hours ago
any suggestion how I can move forward from here? (Edited the question)
– Pojj
16 hours ago
@Pojj In the same line, eliminating $X_1(s), X_2(s)$ we arrive to a quasi polynomial equation for $X_3(s)$ with the structure $$ f_1(s)X_3(s)+f_2(s)left(e^{iphi}X_3(s-ialpha_1)+e^{-iphi}X_3(s+ialpha_1)right)= C_0 $$ with $f_1(s),f_2(s)$ polynomial fractions ...
– Cesareo
9 hours ago
@Pojj In the same line, eliminating $X_1(s), X_2(s)$ we arrive to a quasi polynomial equation for $X_3(s)$ with the structure $$ f_1(s)X_3(s)+f_2(s)left(e^{iphi}X_3(s-ialpha_1)+e^{-iphi}X_3(s+ialpha_1)right)= C_0 $$ with $f_1(s),f_2(s)$ polynomial fractions ...
– Cesareo
9 hours ago
add a comment |
Pojj is a new contributor. Be nice, and check out our Code of Conduct.
Pojj is a new contributor. Be nice, and check out our Code of Conduct.
Pojj is a new contributor. Be nice, and check out our Code of Conduct.
Pojj is a new contributor. Be nice, and check out our Code of Conduct.
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