ODE of second order, proving that polynomials at $t_0$ are zero
The following ODE is given: $$y''(t) + p(t)y'(t) + q(t)y(t)=0$$
When $p(t), q(t)$ are continuous functions.
We are given two linear independent solutions $y_1(t), y_2(t)$ and also $y_1''(t_0) = y_2''(t_0) = 0$.
I need to prove that $p(t_0) = q(t_0) = 0$.
What I've tried is just placing zero in the second derivative for each function in the ODE, and working with the Wronskian. However I end up with $$p(t)(y_1'(t_0) - y_2'(t_0)) + q(t)(y_1(t_0) - y_2(t_0))$$ which is not the Wronskian.
Any help?
differential-equations
asked Dec 26 '18 at 20:36
Gabi G
36218
add a comment |
The following ODE is given: $$y''(t) + p(t)y'(t) + q(t)y(t)=0$$
When $p(t), q(t)$ are continuous functions.
We are given two linear independent solutions $y_1(t), y_2(t)$ and also $y_1''(t_0) = y_2''(t_0) = 0$.
I need to prove that $p(t_0) = q(t_0) = 0$.
What I've tried is just placing zero in the second derivative for each function in the ODE, and working with the Wronskian. However I end up with $$p(t)(y_1'(t_0) - y_2'(t_0)) + q(t)(y_1(t_0) - y_2(t_0))$$ which is not the Wronskian.
Any help?
differential-equations
asked Dec 26 '18 at 20:36
Gabi G
36218
1
Use the two equations to eliminate one of $p(t_0)$ or $q(t_0)$ and consider the remaining terms in view of the Wronskian.
– LutzL
Dec 26 '18 at 20:53
add a comment |
The following ODE is given: $$y''(t) + p(t)y'(t) + q(t)y(t)=0$$
When $p(t), q(t)$ are continuous functions.
We are given two linear independent solutions $y_1(t), y_2(t)$ and also $y_1''(t_0) = y_2''(t_0) = 0$.
I need to prove that $p(t_0) = q(t_0) = 0$.
What I've tried is just placing zero in the second derivative for each function in the ODE, and working with the Wronskian. However I end up with $$p(t)(y_1'(t_0) - y_2'(t_0)) + q(t)(y_1(t_0) - y_2(t_0))$$ which is not the Wronskian.
Any help?
differential-equations
asked Dec 26 '18 at 20:36
Gabi G
36218
The following ODE is given: $$y''(t) + p(t)y'(t) + q(t)y(t)=0$$
When $p(t), q(t)$ are continuous functions.
We are given two linear independent solutions $y_1(t), y_2(t)$ and also $y_1''(t_0) = y_2''(t_0) = 0$.
I need to prove that $p(t_0) = q(t_0) = 0$.
What I've tried is just placing zero in the second derivative for each function in the ODE, and working with the Wronskian. However I end up with $$p(t)(y_1'(t_0) - y_2'(t_0)) + q(t)(y_1(t_0) - y_2(t_0))$$ which is not the Wronskian.
Any help?
differential-equations
differential-equations
asked Dec 26 '18 at 20:36
Gabi G
36218
asked Dec 26 '18 at 20:36
Gabi G
36218
asked Dec 26 '18 at 20:36
Gabi G
36218
asked Dec 26 '18 at 20:36
Gabi G
36218
asked Dec 26 '18 at 20:36
Gabi G
36218
36218
1
Use the two equations to eliminate one of $p(t_0)$ or $q(t_0)$ and consider the remaining terms in view of the Wronskian.
– LutzL
Dec 26 '18 at 20:53
add a comment |
1
Use the two equations to eliminate one of $p(t_0)$ or $q(t_0)$ and consider the remaining terms in view of the Wronskian.
– LutzL
Dec 26 '18 at 20:53
1
1
Use the two equations to eliminate one of $p(t_0)$ or $q(t_0)$ and consider the remaining terms in view of the Wronskian.
– LutzL
Dec 26 '18 at 20:53
Use the two equations to eliminate one of $p(t_0)$ or $q(t_0)$ and consider the remaining terms in view of the Wronskian.
– LutzL
Dec 26 '18 at 20:53
add a comment |
1 Answer
1
active
oldest
votes
From the assumptions you get
$$
begin{aligned}
0 + p(t_0) y_1'(t_0) + q(t_0) y_1(t_0) &= 0
,
\
0 + p(t_0) y_2'(t_0) + q(t_0) y_2(t_0) &= 0
.
end{aligned}
$$
That's a $2 times 2$ linear system for $p(t_0)$ and $q(t_0)$. Can you take it from here?
answered Dec 26 '18 at 20:59
Hans Lundmark
35.1k564113
Thanks! This helped me solve it
– Gabi G
Dec 26 '18 at 21:47
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
From the assumptions you get
$$
begin{aligned}
0 + p(t_0) y_1'(t_0) + q(t_0) y_1(t_0) &= 0
,
\
0 + p(t_0) y_2'(t_0) + q(t_0) y_2(t_0) &= 0
.
end{aligned}
$$
That's a $2 times 2$ linear system for $p(t_0)$ and $q(t_0)$. Can you take it from here?
answered Dec 26 '18 at 20:59
Hans Lundmark
35.1k564113
Thanks! This helped me solve it
– Gabi G
Dec 26 '18 at 21:47
add a comment |
From the assumptions you get
$$
begin{aligned}
0 + p(t_0) y_1'(t_0) + q(t_0) y_1(t_0) &= 0
,
\
0 + p(t_0) y_2'(t_0) + q(t_0) y_2(t_0) &= 0
.
end{aligned}
$$
That's a $2 times 2$ linear system for $p(t_0)$ and $q(t_0)$. Can you take it from here?
answered Dec 26 '18 at 20:59
Hans Lundmark
35.1k564113
Thanks! This helped me solve it
– Gabi G
Dec 26 '18 at 21:47
add a comment |
From the assumptions you get
$$
begin{aligned}
0 + p(t_0) y_1'(t_0) + q(t_0) y_1(t_0) &= 0
,
\
0 + p(t_0) y_2'(t_0) + q(t_0) y_2(t_0) &= 0
.
end{aligned}
$$
That's a $2 times 2$ linear system for $p(t_0)$ and $q(t_0)$. Can you take it from here?
answered Dec 26 '18 at 20:59
Hans Lundmark
35.1k564113
From the assumptions you get
$$
begin{aligned}
0 + p(t_0) y_1'(t_0) + q(t_0) y_1(t_0) &= 0
,
\
0 + p(t_0) y_2'(t_0) + q(t_0) y_2(t_0) &= 0
.
end{aligned}
$$
That's a $2 times 2$ linear system for $p(t_0)$ and $q(t_0)$. Can you take it from here?
answered Dec 26 '18 at 20:59
Hans Lundmark
35.1k564113
answered Dec 26 '18 at 20:59
Hans Lundmark
35.1k564113
answered Dec 26 '18 at 20:59
Hans Lundmark
35.1k564113
answered Dec 26 '18 at 20:59
Hans Lundmark
35.1k564113
35.1k564113
Thanks! This helped me solve it
– Gabi G
Dec 26 '18 at 21:47
add a comment |
Thanks! This helped me solve it
– Gabi G
Dec 26 '18 at 21:47
Thanks! This helped me solve it
– Gabi G
Dec 26 '18 at 21:47
Thanks! This helped me solve it
– Gabi G
Dec 26 '18 at 21:47
add a comment |
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1
Use the two equations to eliminate one of $p(t_0)$ or $q(t_0)$ and consider the remaining terms in view of the Wronskian.
– LutzL
Dec 26 '18 at 20:53