ODE of second order, proving that polynomials at $t_0$ are zero












5














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?










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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
















5














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?










share|cite|improve this question


















  • 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














5












5








5







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?










share|cite|improve this question













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






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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














  • 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










1 Answer
1






active

oldest

votes


















3














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?






share|cite|improve this answer





















  • Thanks! This helped me solve it
    – Gabi G
    Dec 26 '18 at 21:47











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














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?






share|cite|improve this answer





















  • Thanks! This helped me solve it
    – Gabi G
    Dec 26 '18 at 21:47
















3














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?






share|cite|improve this answer





















  • Thanks! This helped me solve it
    – Gabi G
    Dec 26 '18 at 21:47














3












3








3






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?






share|cite|improve this answer












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?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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