$ tx''+ x' - tx + f(t) = 0 $ with $displaystyle f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big)$












1












$begingroup$


A solution of this ODE is $displaystyleint_0^{pi/2}sec{(theta)}e^{-tsec{(theta)}}dtheta$ and I don't know how to start with the following ODE:
$$ tx''+ x' - tx + f(t) = 0 $$
with $displaystyle f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big)$. Is this ODE possible to solve it?




Note: I don't know what is the value of $displaystylelim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big)$, I am going to change this for the value when I know it.






ordinary-differential-equations limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    The homogeneous linear ODE $tx_0'' + x_0' - tx_0 = 0$ can be written as the modified Bessel's equation by multiplying by $t$: $t^2 x_0'' + t x_0' - t^2 x_0 = 0$. Thus the general solution is $x_0(t) = C_1 I_0(t) + C_2 K_0(t)$, with the zeroth-order modified Bessel functions $I_0, K_0$.
    $endgroup$
    – Christoph
    Feb 3 at 9:20


















1












$begingroup$


A solution of this ODE is $displaystyleint_0^{pi/2}sec{(theta)}e^{-tsec{(theta)}}dtheta$ and I don't know how to start with the following ODE:
$$ tx''+ x' - tx + f(t) = 0 $$
with $displaystyle f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big)$. Is this ODE possible to solve it?




Note: I don't know what is the value of $displaystylelim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big)$, I am going to change this for the value when I know it.






ordinary-differential-equations limits







share|cite|improve this question









$endgroup$












  • $begingroup$
    The homogeneous linear ODE $tx_0'' + x_0' - tx_0 = 0$ can be written as the modified Bessel's equation by multiplying by $t$: $t^2 x_0'' + t x_0' - t^2 x_0 = 0$. Thus the general solution is $x_0(t) = C_1 I_0(t) + C_2 K_0(t)$, with the zeroth-order modified Bessel functions $I_0, K_0$.
    $endgroup$
    – Christoph
    Feb 3 at 9:20
















1












1








1





$begingroup$


A solution of this ODE is $displaystyleint_0^{pi/2}sec{(theta)}e^{-tsec{(theta)}}dtheta$ and I don't know how to start with the following ODE:
$$ tx''+ x' - tx + f(t) = 0 $$
with $displaystyle f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big)$. Is this ODE possible to solve it?




Note: I don't know what is the value of $displaystylelim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big)$, I am going to change this for the value when I know it.






ordinary-differential-equations limits







share|cite|improve this question









$endgroup$




A solution of this ODE is $displaystyleint_0^{pi/2}sec{(theta)}e^{-tsec{(theta)}}dtheta$ and I don't know how to start with the following ODE:
$$ tx''+ x' - tx + f(t) = 0 $$
with $displaystyle f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big)$. Is this ODE possible to solve it?




Note: I don't know what is the value of $displaystylelim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big)$, I am going to change this for the value when I know it.






ordinary-differential-equations limits




ordinary-differential-equations limits






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 3 at 8:50









El boritoEl borito

664216




664216












  • $begingroup$
    The homogeneous linear ODE $tx_0'' + x_0' - tx_0 = 0$ can be written as the modified Bessel's equation by multiplying by $t$: $t^2 x_0'' + t x_0' - t^2 x_0 = 0$. Thus the general solution is $x_0(t) = C_1 I_0(t) + C_2 K_0(t)$, with the zeroth-order modified Bessel functions $I_0, K_0$.
    $endgroup$
    – Christoph
    Feb 3 at 9:20




















  • $begingroup$
    The homogeneous linear ODE $tx_0'' + x_0' - tx_0 = 0$ can be written as the modified Bessel's equation by multiplying by $t$: $t^2 x_0'' + t x_0' - t^2 x_0 = 0$. Thus the general solution is $x_0(t) = C_1 I_0(t) + C_2 K_0(t)$, with the zeroth-order modified Bessel functions $I_0, K_0$.
    $endgroup$
    – Christoph
    Feb 3 at 9:20


















$begingroup$
The homogeneous linear ODE $tx_0'' + x_0' - tx_0 = 0$ can be written as the modified Bessel's equation by multiplying by $t$: $t^2 x_0'' + t x_0' - t^2 x_0 = 0$. Thus the general solution is $x_0(t) = C_1 I_0(t) + C_2 K_0(t)$, with the zeroth-order modified Bessel functions $I_0, K_0$.
$endgroup$
– Christoph
Feb 3 at 9:20






$begingroup$
The homogeneous linear ODE $tx_0'' + x_0' - tx_0 = 0$ can be written as the modified Bessel's equation by multiplying by $t$: $t^2 x_0'' + t x_0' - t^2 x_0 = 0$. Thus the general solution is $x_0(t) = C_1 I_0(t) + C_2 K_0(t)$, with the zeroth-order modified Bessel functions $I_0, K_0$.
$endgroup$
– Christoph
Feb 3 at 9:20












1 Answer
1






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oldest

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3












$begingroup$

$$e^{-tsec(theta)}tan(theta)=e^{-t/sin(epsilon)}frac{cos(epsilon)}{sin(epsilon)}quad;quad theta=frac{pi}{2}-epsilon$$
$thetato(pi/2)^- quad;quad epsilon>0to 0^+$
$$e^{-tsec(theta)}tan(theta)sim frac{e^{-t/epsilon}}{epsilon}$$
$f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big) =
begin{cases}
0qquad text{if}quad t>0 \
inftyqquad text{if}quad tleq 0
end{cases}$



Thus the case $tleq 0$ is excluded.



$$tx''+x'-tx=0qquad t>0$$
This is an ODE of Bessel kind.
$$x(t)=c_1I_0(t)+c_2K_0(t)$$
Modified Bessel function of first and second kind.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a very interesant answer, thank you!!
    $endgroup$
    – El borito
    Feb 4 at 4:58












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1 Answer
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active

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active

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3












$begingroup$

$$e^{-tsec(theta)}tan(theta)=e^{-t/sin(epsilon)}frac{cos(epsilon)}{sin(epsilon)}quad;quad theta=frac{pi}{2}-epsilon$$
$thetato(pi/2)^- quad;quad epsilon>0to 0^+$
$$e^{-tsec(theta)}tan(theta)sim frac{e^{-t/epsilon}}{epsilon}$$
$f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big) =
begin{cases}
0qquad text{if}quad t>0 \
inftyqquad text{if}quad tleq 0
end{cases}$



Thus the case $tleq 0$ is excluded.



$$tx''+x'-tx=0qquad t>0$$
This is an ODE of Bessel kind.
$$x(t)=c_1I_0(t)+c_2K_0(t)$$
Modified Bessel function of first and second kind.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a very interesant answer, thank you!!
    $endgroup$
    – El borito
    Feb 4 at 4:58
















3












$begingroup$

$$e^{-tsec(theta)}tan(theta)=e^{-t/sin(epsilon)}frac{cos(epsilon)}{sin(epsilon)}quad;quad theta=frac{pi}{2}-epsilon$$
$thetato(pi/2)^- quad;quad epsilon>0to 0^+$
$$e^{-tsec(theta)}tan(theta)sim frac{e^{-t/epsilon}}{epsilon}$$
$f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big) =
begin{cases}
0qquad text{if}quad t>0 \
inftyqquad text{if}quad tleq 0
end{cases}$



Thus the case $tleq 0$ is excluded.



$$tx''+x'-tx=0qquad t>0$$
This is an ODE of Bessel kind.
$$x(t)=c_1I_0(t)+c_2K_0(t)$$
Modified Bessel function of first and second kind.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a very interesant answer, thank you!!
    $endgroup$
    – El borito
    Feb 4 at 4:58














3












3








3





$begingroup$

$$e^{-tsec(theta)}tan(theta)=e^{-t/sin(epsilon)}frac{cos(epsilon)}{sin(epsilon)}quad;quad theta=frac{pi}{2}-epsilon$$
$thetato(pi/2)^- quad;quad epsilon>0to 0^+$
$$e^{-tsec(theta)}tan(theta)sim frac{e^{-t/epsilon}}{epsilon}$$
$f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big) =
begin{cases}
0qquad text{if}quad t>0 \
inftyqquad text{if}quad tleq 0
end{cases}$



Thus the case $tleq 0$ is excluded.



$$tx''+x'-tx=0qquad t>0$$
This is an ODE of Bessel kind.
$$x(t)=c_1I_0(t)+c_2K_0(t)$$
Modified Bessel function of first and second kind.






share|cite|improve this answer









$endgroup$



$$e^{-tsec(theta)}tan(theta)=e^{-t/sin(epsilon)}frac{cos(epsilon)}{sin(epsilon)}quad;quad theta=frac{pi}{2}-epsilon$$
$thetato(pi/2)^- quad;quad epsilon>0to 0^+$
$$e^{-tsec(theta)}tan(theta)sim frac{e^{-t/epsilon}}{epsilon}$$
$f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big) =
begin{cases}
0qquad text{if}quad t>0 \
inftyqquad text{if}quad tleq 0
end{cases}$



Thus the case $tleq 0$ is excluded.



$$tx''+x'-tx=0qquad t>0$$
This is an ODE of Bessel kind.
$$x(t)=c_1I_0(t)+c_2K_0(t)$$
Modified Bessel function of first and second kind.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Feb 3 at 10:16









JJacquelinJJacquelin

45.7k21858




45.7k21858












  • $begingroup$
    This is a very interesant answer, thank you!!
    $endgroup$
    – El borito
    Feb 4 at 4:58


















  • $begingroup$
    This is a very interesant answer, thank you!!
    $endgroup$
    – El borito
    Feb 4 at 4:58
















$begingroup$
This is a very interesant answer, thank you!!
$endgroup$
– El borito
Feb 4 at 4:58




$begingroup$
This is a very interesant answer, thank you!!
$endgroup$
– El borito
Feb 4 at 4:58


















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