Mixing
Exercise: second-order linear differential equation
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I have the system
$$begin{cases} y'_{1} = dfrac{1}{2};y_{1} - dfrac{3}{2};y_{2} \ y'_{2} = - dfrac{3}{2};y_{1} + dfrac{1}{2};y_{2}end{cases}$$
Retrieving $y_{2}$ from the first one
$$y_{2} = - dfrac{2}{3};y'_{1} + dfrac{1}{3};y_{1}$$
Now, the result I should get by replacing $y_{2}$ in the second one is
$$y''_{1} - y'_{1} + 2y_{1} = 0$$
Yes I know how to resolv a second-order differential equation. What I would like to know, is how to get this equation by replacing in the second one
Thank you in advances
ordinary-differential-equations
asked Jan 22 at 14:56
user3204810user3204810
1947
$endgroup$
add a comment |
$begingroup$
I have the system
$$begin{cases} y'_{1} = dfrac{1}{2};y_{1} - dfrac{3}{2};y_{2} \ y'_{2} = - dfrac{3}{2};y_{1} + dfrac{1}{2};y_{2}end{cases}$$
Retrieving $y_{2}$ from the first one
$$y_{2} = - dfrac{2}{3};y'_{1} + dfrac{1}{3};y_{1}$$
Now, the result I should get by replacing $y_{2}$ in the second one is
$$y''_{1} - y'_{1} + 2y_{1} = 0$$
Yes I know how to resolv a second-order differential equation. What I would like to know, is how to get this equation by replacing in the second one
Thank you in advances
ordinary-differential-equations
asked Jan 22 at 14:56
user3204810user3204810
1947
$endgroup$
$begingroup$
Make the ansatz $$y_1=e^{lambda t}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 14:58
$begingroup$
Yes I know how to resolv a second-order differential equation. What I would like to know, is how to get this equation
$endgroup$
– user3204810
Jan 22 at 15:03
1
$begingroup$
Just plug it in?
$endgroup$
– Klaus
Jan 22 at 15:06
1
$begingroup$
Take the derivative of your expression for $y_2$ and plug it in the second equation.
$endgroup$
– WarreG
Jan 22 at 15:24
$begingroup$
@WarreG this is the solution! Thanks!
$endgroup$
– user3204810
Jan 22 at 15:39
add a comment |
$begingroup$
I have the system
$$begin{cases} y'_{1} = dfrac{1}{2};y_{1} - dfrac{3}{2};y_{2} \ y'_{2} = - dfrac{3}{2};y_{1} + dfrac{1}{2};y_{2}end{cases}$$
Retrieving $y_{2}$ from the first one
$$y_{2} = - dfrac{2}{3};y'_{1} + dfrac{1}{3};y_{1}$$
Now, the result I should get by replacing $y_{2}$ in the second one is
$$y''_{1} - y'_{1} + 2y_{1} = 0$$
Yes I know how to resolv a second-order differential equation. What I would like to know, is how to get this equation by replacing in the second one
Thank you in advances
ordinary-differential-equations
asked Jan 22 at 14:56
user3204810user3204810
1947
$endgroup$
I have the system
$$begin{cases} y'_{1} = dfrac{1}{2};y_{1} - dfrac{3}{2};y_{2} \ y'_{2} = - dfrac{3}{2};y_{1} + dfrac{1}{2};y_{2}end{cases}$$
Retrieving $y_{2}$ from the first one
$$y_{2} = - dfrac{2}{3};y'_{1} + dfrac{1}{3};y_{1}$$
Now, the result I should get by replacing $y_{2}$ in the second one is
$$y''_{1} - y'_{1} + 2y_{1} = 0$$
Yes I know how to resolv a second-order differential equation. What I would like to know, is how to get this equation by replacing in the second one
Thank you in advances
ordinary-differential-equations
ordinary-differential-equations
asked Jan 22 at 14:56
user3204810user3204810
1947
asked Jan 22 at 14:56
user3204810user3204810
1947
edited Jan 22 at 15:22
user3204810
asked Jan 22 at 14:56
user3204810user3204810
1947
asked Jan 22 at 14:56
user3204810user3204810
1947
asked Jan 22 at 14:56
user3204810user3204810
1947
1947
$begingroup$
Make the ansatz $$y_1=e^{lambda t}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 14:58
$begingroup$
Yes I know how to resolv a second-order differential equation. What I would like to know, is how to get this equation
$endgroup$
– user3204810
Jan 22 at 15:03
1
$begingroup$
Just plug it in?
$endgroup$
– Klaus
Jan 22 at 15:06
1
$begingroup$
Take the derivative of your expression for $y_2$ and plug it in the second equation.
$endgroup$
– WarreG
Jan 22 at 15:24
$begingroup$
@WarreG this is the solution! Thanks!
$endgroup$
– user3204810
Jan 22 at 15:39
add a comment |
$begingroup$
Make the ansatz $$y_1=e^{lambda t}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 14:58
$begingroup$
Yes I know how to resolv a second-order differential equation. What I would like to know, is how to get this equation
$endgroup$
– user3204810
Jan 22 at 15:03
1
$begingroup$
Just plug it in?
$endgroup$
– Klaus
Jan 22 at 15:06
1
$begingroup$
Take the derivative of your expression for $y_2$ and plug it in the second equation.
$endgroup$
– WarreG
Jan 22 at 15:24
$begingroup$
@WarreG this is the solution! Thanks!
$endgroup$
– user3204810
Jan 22 at 15:39
$begingroup$
Make the ansatz $$y_1=e^{lambda t}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 14:58
$begingroup$
Make the ansatz $$y_1=e^{lambda t}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 14:58
$begingroup$
Yes I know how to resolv a second-order differential equation. What I would like to know, is how to get this equation
$endgroup$
– user3204810
Jan 22 at 15:03
$begingroup$
Yes I know how to resolv a second-order differential equation. What I would like to know, is how to get this equation
$endgroup$
– user3204810
Jan 22 at 15:03
1
1
$begingroup$
Just plug it in?
$endgroup$
– Klaus
Jan 22 at 15:06
$begingroup$
Just plug it in?
$endgroup$
– Klaus
Jan 22 at 15:06
1
1
$begingroup$
Take the derivative of your expression for $y_2$ and plug it in the second equation.
$endgroup$
– WarreG
Jan 22 at 15:24
$begingroup$
Take the derivative of your expression for $y_2$ and plug it in the second equation.
$endgroup$
– WarreG
Jan 22 at 15:24
$begingroup$
@WarreG this is the solution! Thanks!
$endgroup$
– user3204810
Jan 22 at 15:39
$begingroup$
@WarreG this is the solution! Thanks!
$endgroup$
– user3204810
Jan 22 at 15:39
add a comment |
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$begingroup$
Make the ansatz $$y_1=e^{lambda t}$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 14:58
$begingroup$
Yes I know how to resolv a second-order differential equation. What I would like to know, is how to get this equation
$endgroup$
– user3204810
Jan 22 at 15:03
1
$begingroup$
Just plug it in?
$endgroup$
– Klaus
Jan 22 at 15:06
1
$begingroup$
Take the derivative of your expression for $y_2$ and plug it in the second equation.
$endgroup$
– WarreG
Jan 22 at 15:24
$begingroup$
@WarreG this is the solution! Thanks!
$endgroup$
– user3204810
Jan 22 at 15:39