Solving a simple homogeneous ODE
$begingroup$
$y^{(4)}-2y^{(3)}+5y''(x)=0$
Ansatz: $y(x)=e^{lambda x}$
We find: $P(lambda)=lambda^4-2lambda^3+5lambda^2=lambda^2(lambda^2-2lambda+5)=0 quad Rightarrow lambda_{1,2}=0, lambda_{3,4}=1+-2i$
So we find:
$$y_H(x)=A+Bx+e^x[Ccos(2x)+Dsin(2x)]tag{1}$$
Now, the solution they gave me is this:
$$y_h(x)=k_1+k_2x+frac{1}{25}{k_3(4cos(2x)+3sin(2x))+k_4(3cos(2x)-4sin(2x)-4sin(2x))}e^xtag{2}$$
Now I have no idea how they obtained that. When I plot them I can see they are the same but anyone has any idea how to obtain (2)?
ordinary-differential-equations
asked Jan 9 at 13:53
xotixxotix
291411
$endgroup$
add a comment |
$begingroup$
$y^{(4)}-2y^{(3)}+5y''(x)=0$
Ansatz: $y(x)=e^{lambda x}$
We find: $P(lambda)=lambda^4-2lambda^3+5lambda^2=lambda^2(lambda^2-2lambda+5)=0 quad Rightarrow lambda_{1,2}=0, lambda_{3,4}=1+-2i$
So we find:
$$y_H(x)=A+Bx+e^x[Ccos(2x)+Dsin(2x)]tag{1}$$
Now, the solution they gave me is this:
$$y_h(x)=k_1+k_2x+frac{1}{25}{k_3(4cos(2x)+3sin(2x))+k_4(3cos(2x)-4sin(2x)-4sin(2x))}e^xtag{2}$$
Now I have no idea how they obtained that. When I plot them I can see they are the same but anyone has any idea how to obtain (2)?
ordinary-differential-equations
asked Jan 9 at 13:53
xotixxotix
291411
$endgroup$
$begingroup$
Constants we need find from initial conditions: $y(x_0)=y_0$, $y'(x_0)=y_1$, $y''(x_0)=y_2$, $y'''(x_0)=y_3$,
$endgroup$
– Aleksas Domarkas
Jan 9 at 14:04
$begingroup$
Not entirely sure who "they" is, but the two solutions are equivalent. If you collect all the things multiplying $e^xcos(2x),$ you'll see that it amounts to an arbitrary constant. The same goes for everything multiplying $e^xsin(2x)$. If "they" is a CAS like Mathematica, then the answer is you'd have to dive into its documentation.
$endgroup$
– Adrian Keister
Jan 9 at 14:33
$begingroup$
yeah I also though like that and I also confirmed that the solutions are the same - but I am very curious to know how that other solution was derived. (Can't ask them atm). They being the people who made the exersice sheet.
$endgroup$
– xotix
Jan 9 at 15:03
$begingroup$
Are $k_1,k_2,k_3,k_4$ the initial conditions?
$endgroup$
– Dylan
Jan 11 at 12:56
$begingroup$
Yeah - to be determined by some initial values.
$endgroup$
– xotix
Jan 12 at 17:14
add a comment |
$begingroup$
$y^{(4)}-2y^{(3)}+5y''(x)=0$
Ansatz: $y(x)=e^{lambda x}$
We find: $P(lambda)=lambda^4-2lambda^3+5lambda^2=lambda^2(lambda^2-2lambda+5)=0 quad Rightarrow lambda_{1,2}=0, lambda_{3,4}=1+-2i$
So we find:
$$y_H(x)=A+Bx+e^x[Ccos(2x)+Dsin(2x)]tag{1}$$
Now, the solution they gave me is this:
$$y_h(x)=k_1+k_2x+frac{1}{25}{k_3(4cos(2x)+3sin(2x))+k_4(3cos(2x)-4sin(2x)-4sin(2x))}e^xtag{2}$$
Now I have no idea how they obtained that. When I plot them I can see they are the same but anyone has any idea how to obtain (2)?
ordinary-differential-equations
asked Jan 9 at 13:53
xotixxotix
291411
$endgroup$
$y^{(4)}-2y^{(3)}+5y''(x)=0$
Ansatz: $y(x)=e^{lambda x}$
We find: $P(lambda)=lambda^4-2lambda^3+5lambda^2=lambda^2(lambda^2-2lambda+5)=0 quad Rightarrow lambda_{1,2}=0, lambda_{3,4}=1+-2i$
So we find:
$$y_H(x)=A+Bx+e^x[Ccos(2x)+Dsin(2x)]tag{1}$$
Now, the solution they gave me is this:
$$y_h(x)=k_1+k_2x+frac{1}{25}{k_3(4cos(2x)+3sin(2x))+k_4(3cos(2x)-4sin(2x)-4sin(2x))}e^xtag{2}$$
Now I have no idea how they obtained that. When I plot them I can see they are the same but anyone has any idea how to obtain (2)?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 9 at 13:53
xotixxotix
291411
asked Jan 9 at 13:53
xotixxotix
291411
asked Jan 9 at 13:53
xotixxotix
291411
asked Jan 9 at 13:53
xotixxotix
291411
asked Jan 9 at 13:53
xotixxotix
291411
291411
$begingroup$
Constants we need find from initial conditions: $y(x_0)=y_0$, $y'(x_0)=y_1$, $y''(x_0)=y_2$, $y'''(x_0)=y_3$,
$endgroup$
– Aleksas Domarkas
Jan 9 at 14:04
$begingroup$
Not entirely sure who "they" is, but the two solutions are equivalent. If you collect all the things multiplying $e^xcos(2x),$ you'll see that it amounts to an arbitrary constant. The same goes for everything multiplying $e^xsin(2x)$. If "they" is a CAS like Mathematica, then the answer is you'd have to dive into its documentation.
$endgroup$
– Adrian Keister
Jan 9 at 14:33
$begingroup$
yeah I also though like that and I also confirmed that the solutions are the same - but I am very curious to know how that other solution was derived. (Can't ask them atm). They being the people who made the exersice sheet.
$endgroup$
– xotix
Jan 9 at 15:03
$begingroup$
Are $k_1,k_2,k_3,k_4$ the initial conditions?
$endgroup$
– Dylan
Jan 11 at 12:56
$begingroup$
Yeah - to be determined by some initial values.
$endgroup$
– xotix
Jan 12 at 17:14
add a comment |
$begingroup$
Constants we need find from initial conditions: $y(x_0)=y_0$, $y'(x_0)=y_1$, $y''(x_0)=y_2$, $y'''(x_0)=y_3$,
$endgroup$
– Aleksas Domarkas
Jan 9 at 14:04
$begingroup$
Not entirely sure who "they" is, but the two solutions are equivalent. If you collect all the things multiplying $e^xcos(2x),$ you'll see that it amounts to an arbitrary constant. The same goes for everything multiplying $e^xsin(2x)$. If "they" is a CAS like Mathematica, then the answer is you'd have to dive into its documentation.
$endgroup$
– Adrian Keister
Jan 9 at 14:33
$begingroup$
yeah I also though like that and I also confirmed that the solutions are the same - but I am very curious to know how that other solution was derived. (Can't ask them atm). They being the people who made the exersice sheet.
$endgroup$
– xotix
Jan 9 at 15:03
$begingroup$
Are $k_1,k_2,k_3,k_4$ the initial conditions?
$endgroup$
– Dylan
Jan 11 at 12:56
$begingroup$
Yeah - to be determined by some initial values.
$endgroup$
– xotix
Jan 12 at 17:14
$begingroup$
Constants we need find from initial conditions: $y(x_0)=y_0$, $y'(x_0)=y_1$, $y''(x_0)=y_2$, $y'''(x_0)=y_3$,
$endgroup$
– Aleksas Domarkas
Jan 9 at 14:04
$begingroup$
Constants we need find from initial conditions: $y(x_0)=y_0$, $y'(x_0)=y_1$, $y''(x_0)=y_2$, $y'''(x_0)=y_3$,
$endgroup$
– Aleksas Domarkas
Jan 9 at 14:04
$begingroup$
Not entirely sure who "they" is, but the two solutions are equivalent. If you collect all the things multiplying $e^xcos(2x),$ you'll see that it amounts to an arbitrary constant. The same goes for everything multiplying $e^xsin(2x)$. If "they" is a CAS like Mathematica, then the answer is you'd have to dive into its documentation.
$endgroup$
– Adrian Keister
Jan 9 at 14:33
$begingroup$
Not entirely sure who "they" is, but the two solutions are equivalent. If you collect all the things multiplying $e^xcos(2x),$ you'll see that it amounts to an arbitrary constant. The same goes for everything multiplying $e^xsin(2x)$. If "they" is a CAS like Mathematica, then the answer is you'd have to dive into its documentation.
$endgroup$
– Adrian Keister
Jan 9 at 14:33
$begingroup$
yeah I also though like that and I also confirmed that the solutions are the same - but I am very curious to know how that other solution was derived. (Can't ask them atm). They being the people who made the exersice sheet.
$endgroup$
– xotix
Jan 9 at 15:03
$begingroup$
yeah I also though like that and I also confirmed that the solutions are the same - but I am very curious to know how that other solution was derived. (Can't ask them atm). They being the people who made the exersice sheet.
$endgroup$
– xotix
Jan 9 at 15:03
$begingroup$
Are $k_1,k_2,k_3,k_4$ the initial conditions?
$endgroup$
– Dylan
Jan 11 at 12:56
$begingroup$
Are $k_1,k_2,k_3,k_4$ the initial conditions?
$endgroup$
– Dylan
Jan 11 at 12:56
$begingroup$
Yeah - to be determined by some initial values.
$endgroup$
– xotix
Jan 12 at 17:14
$begingroup$
Yeah - to be determined by some initial values.
$endgroup$
– xotix
Jan 12 at 17:14
add a comment |
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$begingroup$
Constants we need find from initial conditions: $y(x_0)=y_0$, $y'(x_0)=y_1$, $y''(x_0)=y_2$, $y'''(x_0)=y_3$,
$endgroup$
– Aleksas Domarkas
Jan 9 at 14:04
$begingroup$
Not entirely sure who "they" is, but the two solutions are equivalent. If you collect all the things multiplying $e^xcos(2x),$ you'll see that it amounts to an arbitrary constant. The same goes for everything multiplying $e^xsin(2x)$. If "they" is a CAS like Mathematica, then the answer is you'd have to dive into its documentation.
$endgroup$
– Adrian Keister
Jan 9 at 14:33
$begingroup$
yeah I also though like that and I also confirmed that the solutions are the same - but I am very curious to know how that other solution was derived. (Can't ask them atm). They being the people who made the exersice sheet.
$endgroup$
– xotix
Jan 9 at 15:03
$begingroup$
Are $k_1,k_2,k_3,k_4$ the initial conditions?
$endgroup$
– Dylan
Jan 11 at 12:56
$begingroup$
Yeah - to be determined by some initial values.
$endgroup$
– xotix
Jan 12 at 17:14