Mixing
Possible general solution of second order ODE given solutions $1,x$ and $x^2$?
$begingroup$
Let $1$,$x$, and $x^2$ be the solution of a second order linear non-homogeneous differential equation on $-1<x<1$. Then its general solution, can be be: (see image)
I tried solving it using the theorem :
Theorem: Let $y1,y2$ be the solution to differential equation $(D^2+p(x)D+q(x))y=0$ $(text{where}; D=frac{d}{dx})$ over an interval $J=[a,b]$, then the wronskian of $y1, y2, W(y1,y2)$ is either 0 on J or doesn't vanish at any point of J.
However, I don't have a closed interval as required for the theorem.
I took $y1 $ and $y2$ as the functions whose coefficients are $C1 $and $C2$ respectively.
I got the following wronskians $W(y1,y2)$
- Option $(a)$: $(x-1)^2$ : Does not vanish on $(-1,1)$
- Option $(b)$: $x^2$ : Vanish at $x=0$ but not at $x=0.4$ so this option is ruled out
- Option $(c)$: $x^2+2x-1$ : Vanish at $x=-1+sqrt(2)in (-1,1)$ but not at $x=0.1$, this is also ruled out.
- Option $(d)$: $W(y1,y2)=1 $, does not vanish anywhere.
Now how do I decide between $(a)$ or $(d)$?
ordinary-differential-equations
asked Jan 25 at 18:30
AbhayAbhay
3789
$endgroup$
add a comment |
$begingroup$
Let $1$,$x$, and $x^2$ be the solution of a second order linear non-homogeneous differential equation on $-1<x<1$. Then its general solution, can be be: (see image)
I tried solving it using the theorem :
Theorem: Let $y1,y2$ be the solution to differential equation $(D^2+p(x)D+q(x))y=0$ $(text{where}; D=frac{d}{dx})$ over an interval $J=[a,b]$, then the wronskian of $y1, y2, W(y1,y2)$ is either 0 on J or doesn't vanish at any point of J.
However, I don't have a closed interval as required for the theorem.
I took $y1 $ and $y2$ as the functions whose coefficients are $C1 $and $C2$ respectively.
I got the following wronskians $W(y1,y2)$
- Option $(a)$: $(x-1)^2$ : Does not vanish on $(-1,1)$
- Option $(b)$: $x^2$ : Vanish at $x=0$ but not at $x=0.4$ so this option is ruled out
- Option $(c)$: $x^2+2x-1$ : Vanish at $x=-1+sqrt(2)in (-1,1)$ but not at $x=0.1$, this is also ruled out.
- Option $(d)$: $W(y1,y2)=1 $, does not vanish anywhere.
Now how do I decide between $(a)$ or $(d)$?
ordinary-differential-equations
asked Jan 25 at 18:30
AbhayAbhay
3789
$endgroup$
$begingroup$
Can variant d) deliver the solution $y=1$?
$endgroup$
– LutzL
Jan 25 at 18:56
add a comment |
$begingroup$
Let $1$,$x$, and $x^2$ be the solution of a second order linear non-homogeneous differential equation on $-1<x<1$. Then its general solution, can be be: (see image)
I tried solving it using the theorem :
Theorem: Let $y1,y2$ be the solution to differential equation $(D^2+p(x)D+q(x))y=0$ $(text{where}; D=frac{d}{dx})$ over an interval $J=[a,b]$, then the wronskian of $y1, y2, W(y1,y2)$ is either 0 on J or doesn't vanish at any point of J.
However, I don't have a closed interval as required for the theorem.
I took $y1 $ and $y2$ as the functions whose coefficients are $C1 $and $C2$ respectively.
I got the following wronskians $W(y1,y2)$
- Option $(a)$: $(x-1)^2$ : Does not vanish on $(-1,1)$
- Option $(b)$: $x^2$ : Vanish at $x=0$ but not at $x=0.4$ so this option is ruled out
- Option $(c)$: $x^2+2x-1$ : Vanish at $x=-1+sqrt(2)in (-1,1)$ but not at $x=0.1$, this is also ruled out.
- Option $(d)$: $W(y1,y2)=1 $, does not vanish anywhere.
Now how do I decide between $(a)$ or $(d)$?
ordinary-differential-equations
asked Jan 25 at 18:30
AbhayAbhay
3789
$endgroup$
Let $1$,$x$, and $x^2$ be the solution of a second order linear non-homogeneous differential equation on $-1<x<1$. Then its general solution, can be be: (see image)
I tried solving it using the theorem :
Theorem: Let $y1,y2$ be the solution to differential equation $(D^2+p(x)D+q(x))y=0$ $(text{where}; D=frac{d}{dx})$ over an interval $J=[a,b]$, then the wronskian of $y1, y2, W(y1,y2)$ is either 0 on J or doesn't vanish at any point of J.
However, I don't have a closed interval as required for the theorem.
I took $y1 $ and $y2$ as the functions whose coefficients are $C1 $and $C2$ respectively.
I got the following wronskians $W(y1,y2)$
- Option $(a)$: $(x-1)^2$ : Does not vanish on $(-1,1)$
- Option $(b)$: $x^2$ : Vanish at $x=0$ but not at $x=0.4$ so this option is ruled out
- Option $(c)$: $x^2+2x-1$ : Vanish at $x=-1+sqrt(2)in (-1,1)$ but not at $x=0.1$, this is also ruled out.
- Option $(d)$: $W(y1,y2)=1 $, does not vanish anywhere.
Now how do I decide between $(a)$ or $(d)$?
ordinary-differential-equations
ordinary-differential-equations
asked Jan 25 at 18:30
AbhayAbhay
3789
asked Jan 25 at 18:30
AbhayAbhay
3789
asked Jan 25 at 18:30
AbhayAbhay
3789
asked Jan 25 at 18:30
AbhayAbhay
3789
asked Jan 25 at 18:30
AbhayAbhay
3789
3789
$begingroup$
Can variant d) deliver the solution $y=1$?
$endgroup$
– LutzL
Jan 25 at 18:56
add a comment |
$begingroup$
Can variant d) deliver the solution $y=1$?
$endgroup$
– LutzL
Jan 25 at 18:56
$begingroup$
Can variant d) deliver the solution $y=1$?
$endgroup$
– LutzL
Jan 25 at 18:56
$begingroup$
Can variant d) deliver the solution $y=1$?
$endgroup$
– LutzL
Jan 25 at 18:56
add a comment |
1 Answer
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$begingroup$
If $y_1,y_2,y_3$ are solutions of the inhomogeneous equation, then $y_1-y_2,y_2-y_3,y_1-y_3$ are solutions of the homogeneous equation, and you can pick any two of them and one of the inhomogeneous solutions to compose the solution formula. Only the first variant has that pattern.
With 3 functions that are solutions of a second order inhomogeneous linear ODE any other solution satisfies
$$
0=detpmatrix{1&1&1&1\y_1&y_2&y_3&y\y_1'&y_2'&y_3'&y'\y_1''&y_2''&y_3''&y''}
$$
so for the given system you get
$$
0=detpmatrix{1&1&1&1\1&x&x^2&y\0&1&2x&y'\0&0&2&y''}
=2-2y+2(x-1)y'-(x-1)^2y''
$$
where by Euler-Cauchy you get in return the solution formula $y-1=A(x-1)+B(x-1)^2$.
answered Jan 25 at 18:54
LutzLLutzL
59.7k42057
$endgroup$
add a comment |
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1 Answer
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$begingroup$
If $y_1,y_2,y_3$ are solutions of the inhomogeneous equation, then $y_1-y_2,y_2-y_3,y_1-y_3$ are solutions of the homogeneous equation, and you can pick any two of them and one of the inhomogeneous solutions to compose the solution formula. Only the first variant has that pattern.
With 3 functions that are solutions of a second order inhomogeneous linear ODE any other solution satisfies
$$
0=detpmatrix{1&1&1&1\y_1&y_2&y_3&y\y_1'&y_2'&y_3'&y'\y_1''&y_2''&y_3''&y''}
$$
so for the given system you get
$$
0=detpmatrix{1&1&1&1\1&x&x^2&y\0&1&2x&y'\0&0&2&y''}
=2-2y+2(x-1)y'-(x-1)^2y''
$$
where by Euler-Cauchy you get in return the solution formula $y-1=A(x-1)+B(x-1)^2$.
answered Jan 25 at 18:54
LutzLLutzL
59.7k42057
$endgroup$
add a comment |
$begingroup$
If $y_1,y_2,y_3$ are solutions of the inhomogeneous equation, then $y_1-y_2,y_2-y_3,y_1-y_3$ are solutions of the homogeneous equation, and you can pick any two of them and one of the inhomogeneous solutions to compose the solution formula. Only the first variant has that pattern.
With 3 functions that are solutions of a second order inhomogeneous linear ODE any other solution satisfies
$$
0=detpmatrix{1&1&1&1\y_1&y_2&y_3&y\y_1'&y_2'&y_3'&y'\y_1''&y_2''&y_3''&y''}
$$
so for the given system you get
$$
0=detpmatrix{1&1&1&1\1&x&x^2&y\0&1&2x&y'\0&0&2&y''}
=2-2y+2(x-1)y'-(x-1)^2y''
$$
where by Euler-Cauchy you get in return the solution formula $y-1=A(x-1)+B(x-1)^2$.
answered Jan 25 at 18:54
LutzLLutzL
59.7k42057
$endgroup$
add a comment |
$begingroup$
If $y_1,y_2,y_3$ are solutions of the inhomogeneous equation, then $y_1-y_2,y_2-y_3,y_1-y_3$ are solutions of the homogeneous equation, and you can pick any two of them and one of the inhomogeneous solutions to compose the solution formula. Only the first variant has that pattern.
With 3 functions that are solutions of a second order inhomogeneous linear ODE any other solution satisfies
$$
0=detpmatrix{1&1&1&1\y_1&y_2&y_3&y\y_1'&y_2'&y_3'&y'\y_1''&y_2''&y_3''&y''}
$$
so for the given system you get
$$
0=detpmatrix{1&1&1&1\1&x&x^2&y\0&1&2x&y'\0&0&2&y''}
=2-2y+2(x-1)y'-(x-1)^2y''
$$
where by Euler-Cauchy you get in return the solution formula $y-1=A(x-1)+B(x-1)^2$.
answered Jan 25 at 18:54
LutzLLutzL
59.7k42057
$endgroup$
If $y_1,y_2,y_3$ are solutions of the inhomogeneous equation, then $y_1-y_2,y_2-y_3,y_1-y_3$ are solutions of the homogeneous equation, and you can pick any two of them and one of the inhomogeneous solutions to compose the solution formula. Only the first variant has that pattern.
With 3 functions that are solutions of a second order inhomogeneous linear ODE any other solution satisfies
$$
0=detpmatrix{1&1&1&1\y_1&y_2&y_3&y\y_1'&y_2'&y_3'&y'\y_1''&y_2''&y_3''&y''}
$$
so for the given system you get
$$
0=detpmatrix{1&1&1&1\1&x&x^2&y\0&1&2x&y'\0&0&2&y''}
=2-2y+2(x-1)y'-(x-1)^2y''
$$
where by Euler-Cauchy you get in return the solution formula $y-1=A(x-1)+B(x-1)^2$.
answered Jan 25 at 18:54
LutzLLutzL
59.7k42057
answered Jan 25 at 18:54
LutzLLutzL
59.7k42057
answered Jan 25 at 18:54
LutzLLutzL
59.7k42057
answered Jan 25 at 18:54
LutzLLutzL
59.7k42057
59.7k42057
add a comment |
add a comment |
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$begingroup$
Can variant d) deliver the solution $y=1$?
$endgroup$
– LutzL
Jan 25 at 18:56