How to solve a system of ODE's using laplace transform?
$begingroup$
System of ODE's:
$$y_1^"=16y_2$$
$$y_2^"=16y_1$$
Initial conditions : $$y_1(0)=2,space y_1^{'}(0)=12$$
$$y_2(0)=6, space y_2^{'}(0)=4$$
The equations i get once i do the laplace transform are:
$$ s^2 mathbf Y_1(s)-12s-2=16 mathbf Y_2(s)$$
$$ s^2 mathbf Y_2(s)-4s-6=16 mathbf Y_1(s)$$
Then I put $mathbf Y_2(s)$ in terms of $mathbf Y_1(s)$
This gave me $$ mathbf Y_2(s)=frac{4s^3+6s^2+192s+32}{s^4-256}$$
I've tried to do partial fractions but keep getting the wrong answer, is there another way?
ordinary-differential-equations laplace-transform
asked Dec 10 '18 at 16:13
Elliot SilverElliot Silver
406
$endgroup$
add a comment |
$begingroup$
System of ODE's:
$$y_1^"=16y_2$$
$$y_2^"=16y_1$$
Initial conditions : $$y_1(0)=2,space y_1^{'}(0)=12$$
$$y_2(0)=6, space y_2^{'}(0)=4$$
The equations i get once i do the laplace transform are:
$$ s^2 mathbf Y_1(s)-12s-2=16 mathbf Y_2(s)$$
$$ s^2 mathbf Y_2(s)-4s-6=16 mathbf Y_1(s)$$
Then I put $mathbf Y_2(s)$ in terms of $mathbf Y_1(s)$
This gave me $$ mathbf Y_2(s)=frac{4s^3+6s^2+192s+32}{s^4-256}$$
I've tried to do partial fractions but keep getting the wrong answer, is there another way?
ordinary-differential-equations laplace-transform
asked Dec 10 '18 at 16:13
Elliot SilverElliot Silver
406
$endgroup$
$begingroup$
You can consider the equations for the sum and difference of $y_1$ and $y_2$, these are decoupled second order equations.
$endgroup$
– LutzL
Dec 10 '18 at 16:20
$begingroup$
I have to use Laplace transform, I meant is there a way other than putting Y2 in terms of Y1
$endgroup$
– Elliot Silver
Dec 10 '18 at 16:22
$begingroup$
@ElliotSilver You can still use Laplace transform after using LutzL's hint
$endgroup$
– Federico
Dec 10 '18 at 16:43
$begingroup$
Or compute the sums and differences after the Laplace transform.
$endgroup$
– LutzL
Dec 11 '18 at 17:53
add a comment |
$begingroup$
System of ODE's:
$$y_1^"=16y_2$$
$$y_2^"=16y_1$$
Initial conditions : $$y_1(0)=2,space y_1^{'}(0)=12$$
$$y_2(0)=6, space y_2^{'}(0)=4$$
The equations i get once i do the laplace transform are:
$$ s^2 mathbf Y_1(s)-12s-2=16 mathbf Y_2(s)$$
$$ s^2 mathbf Y_2(s)-4s-6=16 mathbf Y_1(s)$$
Then I put $mathbf Y_2(s)$ in terms of $mathbf Y_1(s)$
This gave me $$ mathbf Y_2(s)=frac{4s^3+6s^2+192s+32}{s^4-256}$$
I've tried to do partial fractions but keep getting the wrong answer, is there another way?
ordinary-differential-equations laplace-transform
asked Dec 10 '18 at 16:13
Elliot SilverElliot Silver
406
$endgroup$
System of ODE's:
$$y_1^"=16y_2$$
$$y_2^"=16y_1$$
Initial conditions : $$y_1(0)=2,space y_1^{'}(0)=12$$
$$y_2(0)=6, space y_2^{'}(0)=4$$
The equations i get once i do the laplace transform are:
$$ s^2 mathbf Y_1(s)-12s-2=16 mathbf Y_2(s)$$
$$ s^2 mathbf Y_2(s)-4s-6=16 mathbf Y_1(s)$$
Then I put $mathbf Y_2(s)$ in terms of $mathbf Y_1(s)$
This gave me $$ mathbf Y_2(s)=frac{4s^3+6s^2+192s+32}{s^4-256}$$
I've tried to do partial fractions but keep getting the wrong answer, is there another way?
ordinary-differential-equations laplace-transform
ordinary-differential-equations laplace-transform
asked Dec 10 '18 at 16:13
Elliot SilverElliot Silver
406
asked Dec 10 '18 at 16:13
Elliot SilverElliot Silver
406
asked Dec 10 '18 at 16:13
Elliot SilverElliot Silver
406
asked Dec 10 '18 at 16:13
Elliot SilverElliot Silver
406
asked Dec 10 '18 at 16:13
Elliot SilverElliot Silver
406
406
$begingroup$
You can consider the equations for the sum and difference of $y_1$ and $y_2$, these are decoupled second order equations.
$endgroup$
– LutzL
Dec 10 '18 at 16:20
$begingroup$
I have to use Laplace transform, I meant is there a way other than putting Y2 in terms of Y1
$endgroup$
– Elliot Silver
Dec 10 '18 at 16:22
$begingroup$
@ElliotSilver You can still use Laplace transform after using LutzL's hint
$endgroup$
– Federico
Dec 10 '18 at 16:43
$begingroup$
Or compute the sums and differences after the Laplace transform.
$endgroup$
– LutzL
Dec 11 '18 at 17:53
add a comment |
$begingroup$
You can consider the equations for the sum and difference of $y_1$ and $y_2$, these are decoupled second order equations.
$endgroup$
– LutzL
Dec 10 '18 at 16:20
$begingroup$
I have to use Laplace transform, I meant is there a way other than putting Y2 in terms of Y1
$endgroup$
– Elliot Silver
Dec 10 '18 at 16:22
$begingroup$
@ElliotSilver You can still use Laplace transform after using LutzL's hint
$endgroup$
– Federico
Dec 10 '18 at 16:43
$begingroup$
Or compute the sums and differences after the Laplace transform.
$endgroup$
– LutzL
Dec 11 '18 at 17:53
$begingroup$
You can consider the equations for the sum and difference of $y_1$ and $y_2$, these are decoupled second order equations.
$endgroup$
– LutzL
Dec 10 '18 at 16:20
$begingroup$
You can consider the equations for the sum and difference of $y_1$ and $y_2$, these are decoupled second order equations.
$endgroup$
– LutzL
Dec 10 '18 at 16:20
$begingroup$
I have to use Laplace transform, I meant is there a way other than putting Y2 in terms of Y1
$endgroup$
– Elliot Silver
Dec 10 '18 at 16:22
$begingroup$
I have to use Laplace transform, I meant is there a way other than putting Y2 in terms of Y1
$endgroup$
– Elliot Silver
Dec 10 '18 at 16:22
$begingroup$
@ElliotSilver You can still use Laplace transform after using LutzL's hint
$endgroup$
– Federico
Dec 10 '18 at 16:43
$begingroup$
@ElliotSilver You can still use Laplace transform after using LutzL's hint
$endgroup$
– Federico
Dec 10 '18 at 16:43
$begingroup$
Or compute the sums and differences after the Laplace transform.
$endgroup$
– LutzL
Dec 11 '18 at 17:53
$begingroup$
Or compute the sums and differences after the Laplace transform.
$endgroup$
– LutzL
Dec 11 '18 at 17:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you do partial fractions this is what you should get
$$
{bf Y}_2(x) = frac{9}{2}frac{1}{s - 4} + frac{7}{2}frac{1}{s + 4} - 2frac{2 s - 1}{s^2 + 4^2}
$$
Use the transform for each term
$$
y_2(t) = frac{9}{2}e^{4t} + frac{7}{2}e^{-4t} - frac{1}{2}sin 4t + 4cos 4t
$$
EDIT Note that $s^4 - 256 = (s^2 - 4^2)(s^2 + 4^2) = (s - 4)(s + 4)(s^2 + 16)$ so that
begin{eqnarray}
frac{4s^3 + 6s^2 + 192 s + 32}{s^4 - 256} &=& frac{A}{s - 4} + frac{B}{s + 4} + frac{Cs + D}{s^2 + 16}\
&=& frac{A(s + 4)(s^2 + 16) + B(s - 4)(s^2 + 16) + (s - 4)(s + 4)(Cs + D)}{(s - 4)(s + 4)(s^2 + 16)} \
&=& frac{A(s^3 + 16s + 4s^2 + 64) + B(s^3 + 16s - 4s^2 - 64)}{s^4 - 256} \
&& + frac{Cs^3 + Ds^2 -16Cs - 16D}{s^4 - 256} \
&=& frac{s^3(A + B + C) + s^2(4A - 4B + D) + s(16A + 16B - 16C)}{s^4 - 256} \
&& + frac{64A - 64B - 16D}{s^4 - 256}
end{eqnarray}
And from here you conclude
begin{eqnarray}
A + B + C &=& 4 \
4A - 4B + D &=& 6\
16A + 16B - 16C &=& 192 \
64A - 64B - 16D &=& 32
end{eqnarray}
And from here is easy to get $A = 9/2$, $B = 7/2$, $C = -4$ and $D = -2$
answered Dec 10 '18 at 16:36
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Could you show how you got those numbers in the partial fractions, I keep doing it and keep getting different numbers, thanks.
$endgroup$
– Elliot Silver
Dec 11 '18 at 13:02
$begingroup$
@ElliotSilver Updated answer
$endgroup$
– caverac
Dec 11 '18 at 17:37
$begingroup$
Thanks a lot, I see where i went wrong now. Appreciate the help :)
$endgroup$
– Elliot Silver
Dec 12 '18 at 11:40
$begingroup$
@ElliotSilver Happy to help
$endgroup$
– caverac
Dec 12 '18 at 12:14
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If you do partial fractions this is what you should get
$$
{bf Y}_2(x) = frac{9}{2}frac{1}{s - 4} + frac{7}{2}frac{1}{s + 4} - 2frac{2 s - 1}{s^2 + 4^2}
$$
Use the transform for each term
$$
y_2(t) = frac{9}{2}e^{4t} + frac{7}{2}e^{-4t} - frac{1}{2}sin 4t + 4cos 4t
$$
EDIT Note that $s^4 - 256 = (s^2 - 4^2)(s^2 + 4^2) = (s - 4)(s + 4)(s^2 + 16)$ so that
begin{eqnarray}
frac{4s^3 + 6s^2 + 192 s + 32}{s^4 - 256} &=& frac{A}{s - 4} + frac{B}{s + 4} + frac{Cs + D}{s^2 + 16}\
&=& frac{A(s + 4)(s^2 + 16) + B(s - 4)(s^2 + 16) + (s - 4)(s + 4)(Cs + D)}{(s - 4)(s + 4)(s^2 + 16)} \
&=& frac{A(s^3 + 16s + 4s^2 + 64) + B(s^3 + 16s - 4s^2 - 64)}{s^4 - 256} \
&& + frac{Cs^3 + Ds^2 -16Cs - 16D}{s^4 - 256} \
&=& frac{s^3(A + B + C) + s^2(4A - 4B + D) + s(16A + 16B - 16C)}{s^4 - 256} \
&& + frac{64A - 64B - 16D}{s^4 - 256}
end{eqnarray}
And from here you conclude
begin{eqnarray}
A + B + C &=& 4 \
4A - 4B + D &=& 6\
16A + 16B - 16C &=& 192 \
64A - 64B - 16D &=& 32
end{eqnarray}
And from here is easy to get $A = 9/2$, $B = 7/2$, $C = -4$ and $D = -2$
answered Dec 10 '18 at 16:36
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Could you show how you got those numbers in the partial fractions, I keep doing it and keep getting different numbers, thanks.
$endgroup$
– Elliot Silver
Dec 11 '18 at 13:02
$begingroup$
@ElliotSilver Updated answer
$endgroup$
– caverac
Dec 11 '18 at 17:37
$begingroup$
Thanks a lot, I see where i went wrong now. Appreciate the help :)
$endgroup$
– Elliot Silver
Dec 12 '18 at 11:40
$begingroup$
@ElliotSilver Happy to help
$endgroup$
– caverac
Dec 12 '18 at 12:14
add a comment |
$begingroup$
If you do partial fractions this is what you should get
$$
{bf Y}_2(x) = frac{9}{2}frac{1}{s - 4} + frac{7}{2}frac{1}{s + 4} - 2frac{2 s - 1}{s^2 + 4^2}
$$
Use the transform for each term
$$
y_2(t) = frac{9}{2}e^{4t} + frac{7}{2}e^{-4t} - frac{1}{2}sin 4t + 4cos 4t
$$
EDIT Note that $s^4 - 256 = (s^2 - 4^2)(s^2 + 4^2) = (s - 4)(s + 4)(s^2 + 16)$ so that
begin{eqnarray}
frac{4s^3 + 6s^2 + 192 s + 32}{s^4 - 256} &=& frac{A}{s - 4} + frac{B}{s + 4} + frac{Cs + D}{s^2 + 16}\
&=& frac{A(s + 4)(s^2 + 16) + B(s - 4)(s^2 + 16) + (s - 4)(s + 4)(Cs + D)}{(s - 4)(s + 4)(s^2 + 16)} \
&=& frac{A(s^3 + 16s + 4s^2 + 64) + B(s^3 + 16s - 4s^2 - 64)}{s^4 - 256} \
&& + frac{Cs^3 + Ds^2 -16Cs - 16D}{s^4 - 256} \
&=& frac{s^3(A + B + C) + s^2(4A - 4B + D) + s(16A + 16B - 16C)}{s^4 - 256} \
&& + frac{64A - 64B - 16D}{s^4 - 256}
end{eqnarray}
And from here you conclude
begin{eqnarray}
A + B + C &=& 4 \
4A - 4B + D &=& 6\
16A + 16B - 16C &=& 192 \
64A - 64B - 16D &=& 32
end{eqnarray}
And from here is easy to get $A = 9/2$, $B = 7/2$, $C = -4$ and $D = -2$
answered Dec 10 '18 at 16:36
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
Could you show how you got those numbers in the partial fractions, I keep doing it and keep getting different numbers, thanks.
$endgroup$
– Elliot Silver
Dec 11 '18 at 13:02
$begingroup$
@ElliotSilver Updated answer
$endgroup$
– caverac
Dec 11 '18 at 17:37
$begingroup$
Thanks a lot, I see where i went wrong now. Appreciate the help :)
$endgroup$
– Elliot Silver
Dec 12 '18 at 11:40
$begingroup$
@ElliotSilver Happy to help
$endgroup$
– caverac
Dec 12 '18 at 12:14
add a comment |
$begingroup$
If you do partial fractions this is what you should get
$$
{bf Y}_2(x) = frac{9}{2}frac{1}{s - 4} + frac{7}{2}frac{1}{s + 4} - 2frac{2 s - 1}{s^2 + 4^2}
$$
Use the transform for each term
$$
y_2(t) = frac{9}{2}e^{4t} + frac{7}{2}e^{-4t} - frac{1}{2}sin 4t + 4cos 4t
$$
EDIT Note that $s^4 - 256 = (s^2 - 4^2)(s^2 + 4^2) = (s - 4)(s + 4)(s^2 + 16)$ so that
begin{eqnarray}
frac{4s^3 + 6s^2 + 192 s + 32}{s^4 - 256} &=& frac{A}{s - 4} + frac{B}{s + 4} + frac{Cs + D}{s^2 + 16}\
&=& frac{A(s + 4)(s^2 + 16) + B(s - 4)(s^2 + 16) + (s - 4)(s + 4)(Cs + D)}{(s - 4)(s + 4)(s^2 + 16)} \
&=& frac{A(s^3 + 16s + 4s^2 + 64) + B(s^3 + 16s - 4s^2 - 64)}{s^4 - 256} \
&& + frac{Cs^3 + Ds^2 -16Cs - 16D}{s^4 - 256} \
&=& frac{s^3(A + B + C) + s^2(4A - 4B + D) + s(16A + 16B - 16C)}{s^4 - 256} \
&& + frac{64A - 64B - 16D}{s^4 - 256}
end{eqnarray}
And from here you conclude
begin{eqnarray}
A + B + C &=& 4 \
4A - 4B + D &=& 6\
16A + 16B - 16C &=& 192 \
64A - 64B - 16D &=& 32
end{eqnarray}
And from here is easy to get $A = 9/2$, $B = 7/2$, $C = -4$ and $D = -2$
answered Dec 10 '18 at 16:36
caveraccaverac
14.8k31130
$endgroup$
If you do partial fractions this is what you should get
$$
{bf Y}_2(x) = frac{9}{2}frac{1}{s - 4} + frac{7}{2}frac{1}{s + 4} - 2frac{2 s - 1}{s^2 + 4^2}
$$
Use the transform for each term
$$
y_2(t) = frac{9}{2}e^{4t} + frac{7}{2}e^{-4t} - frac{1}{2}sin 4t + 4cos 4t
$$
EDIT Note that $s^4 - 256 = (s^2 - 4^2)(s^2 + 4^2) = (s - 4)(s + 4)(s^2 + 16)$ so that
begin{eqnarray}
frac{4s^3 + 6s^2 + 192 s + 32}{s^4 - 256} &=& frac{A}{s - 4} + frac{B}{s + 4} + frac{Cs + D}{s^2 + 16}\
&=& frac{A(s + 4)(s^2 + 16) + B(s - 4)(s^2 + 16) + (s - 4)(s + 4)(Cs + D)}{(s - 4)(s + 4)(s^2 + 16)} \
&=& frac{A(s^3 + 16s + 4s^2 + 64) + B(s^3 + 16s - 4s^2 - 64)}{s^4 - 256} \
&& + frac{Cs^3 + Ds^2 -16Cs - 16D}{s^4 - 256} \
&=& frac{s^3(A + B + C) + s^2(4A - 4B + D) + s(16A + 16B - 16C)}{s^4 - 256} \
&& + frac{64A - 64B - 16D}{s^4 - 256}
end{eqnarray}
And from here you conclude
begin{eqnarray}
A + B + C &=& 4 \
4A - 4B + D &=& 6\
16A + 16B - 16C &=& 192 \
64A - 64B - 16D &=& 32
end{eqnarray}
And from here is easy to get $A = 9/2$, $B = 7/2$, $C = -4$ and $D = -2$
answered Dec 10 '18 at 16:36
caveraccaverac
14.8k31130
edited Dec 11 '18 at 17:37
answered Dec 10 '18 at 16:36
caveraccaverac
14.8k31130
answered Dec 10 '18 at 16:36
caveraccaverac
14.8k31130
answered Dec 10 '18 at 16:36
caveraccaverac
14.8k31130
14.8k31130
$begingroup$
Could you show how you got those numbers in the partial fractions, I keep doing it and keep getting different numbers, thanks.
$endgroup$
– Elliot Silver
Dec 11 '18 at 13:02
$begingroup$
@ElliotSilver Updated answer
$endgroup$
– caverac
Dec 11 '18 at 17:37
$begingroup$
Thanks a lot, I see where i went wrong now. Appreciate the help :)
$endgroup$
– Elliot Silver
Dec 12 '18 at 11:40
$begingroup$
@ElliotSilver Happy to help
$endgroup$
– caverac
Dec 12 '18 at 12:14
add a comment |
$begingroup$
Could you show how you got those numbers in the partial fractions, I keep doing it and keep getting different numbers, thanks.
$endgroup$
– Elliot Silver
Dec 11 '18 at 13:02
$begingroup$
@ElliotSilver Updated answer
$endgroup$
– caverac
Dec 11 '18 at 17:37
$begingroup$
Thanks a lot, I see where i went wrong now. Appreciate the help :)
$endgroup$
– Elliot Silver
Dec 12 '18 at 11:40
$begingroup$
@ElliotSilver Happy to help
$endgroup$
– caverac
Dec 12 '18 at 12:14
$begingroup$
Could you show how you got those numbers in the partial fractions, I keep doing it and keep getting different numbers, thanks.
$endgroup$
– Elliot Silver
Dec 11 '18 at 13:02
$begingroup$
Could you show how you got those numbers in the partial fractions, I keep doing it and keep getting different numbers, thanks.
$endgroup$
– Elliot Silver
Dec 11 '18 at 13:02
$begingroup$
@ElliotSilver Updated answer
$endgroup$
– caverac
Dec 11 '18 at 17:37
$begingroup$
@ElliotSilver Updated answer
$endgroup$
– caverac
Dec 11 '18 at 17:37
$begingroup$
Thanks a lot, I see where i went wrong now. Appreciate the help :)
$endgroup$
– Elliot Silver
Dec 12 '18 at 11:40
$begingroup$
Thanks a lot, I see where i went wrong now. Appreciate the help :)
$endgroup$
– Elliot Silver
Dec 12 '18 at 11:40
$begingroup$
@ElliotSilver Happy to help
$endgroup$
– caverac
Dec 12 '18 at 12:14
$begingroup$
@ElliotSilver Happy to help
$endgroup$
– caverac
Dec 12 '18 at 12:14
add a comment |
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$begingroup$
You can consider the equations for the sum and difference of $y_1$ and $y_2$, these are decoupled second order equations.
$endgroup$
– LutzL
Dec 10 '18 at 16:20
$begingroup$
I have to use Laplace transform, I meant is there a way other than putting Y2 in terms of Y1
$endgroup$
– Elliot Silver
Dec 10 '18 at 16:22
$begingroup$
@ElliotSilver You can still use Laplace transform after using LutzL's hint
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– Federico
Dec 10 '18 at 16:43
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Or compute the sums and differences after the Laplace transform.
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– LutzL
Dec 11 '18 at 17:53