Mixing
$ tx''+ x' - tx + f(t) = 0 $ with $displaystyle f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big)$
$begingroup$
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
asked Feb 3 at 8:50
El boritoEl borito
664216
$endgroup$
add a comment |
$begingroup$
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
asked Feb 3 at 8:50
El boritoEl borito
664216
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Feb 3 at 8:50
El boritoEl borito
664216
$endgroup$
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
asked Feb 3 at 8:50
El boritoEl borito
664216
asked Feb 3 at 8:50
El boritoEl borito
664216
asked Feb 3 at 8:50
El boritoEl borito
664216
asked Feb 3 at 8:50
El boritoEl borito
664216
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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$begingroup$
$$e^{-tsec(theta)}tan(theta)=e^{-t/sin(epsilon)}frac{cos(epsilon)}{sin(epsilon)}quad;quad theta=frac{pi}{2}-epsilon$$
$thetato(pi/2)^- quad;quad epsilon>0to 0^+$
$$e^{-tsec(theta)}tan(theta)sim frac{e^{-t/epsilon}}{epsilon}$$
$f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big) =
begin{cases}
0qquad text{if}quad t>0 \
inftyqquad text{if}quad tleq 0
end{cases}$
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.
answered Feb 3 at 10:16
JJacquelinJJacquelin
45.7k21858
$endgroup$
$begingroup$
This is a very interesant answer, thank you!!
$endgroup$
– El borito
Feb 4 at 4:58
add a comment |
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1 Answer
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$begingroup$
$$e^{-tsec(theta)}tan(theta)=e^{-t/sin(epsilon)}frac{cos(epsilon)}{sin(epsilon)}quad;quad theta=frac{pi}{2}-epsilon$$
$thetato(pi/2)^- quad;quad epsilon>0to 0^+$
$$e^{-tsec(theta)}tan(theta)sim frac{e^{-t/epsilon}}{epsilon}$$
$f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big) =
begin{cases}
0qquad text{if}quad t>0 \
inftyqquad text{if}quad tleq 0
end{cases}$
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.
answered Feb 3 at 10:16
JJacquelinJJacquelin
45.7k21858
$endgroup$
$begingroup$
This is a very interesant answer, thank you!!
$endgroup$
– El borito
Feb 4 at 4:58
add a comment |
$begingroup$
$$e^{-tsec(theta)}tan(theta)=e^{-t/sin(epsilon)}frac{cos(epsilon)}{sin(epsilon)}quad;quad theta=frac{pi}{2}-epsilon$$
$thetato(pi/2)^- quad;quad epsilon>0to 0^+$
$$e^{-tsec(theta)}tan(theta)sim frac{e^{-t/epsilon}}{epsilon}$$
$f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big) =
begin{cases}
0qquad text{if}quad t>0 \
inftyqquad text{if}quad tleq 0
end{cases}$
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.
answered Feb 3 at 10:16
JJacquelinJJacquelin
45.7k21858
$endgroup$
$begingroup$
This is a very interesant answer, thank you!!
$endgroup$
– El borito
Feb 4 at 4:58
add a comment |
$begingroup$
$$e^{-tsec(theta)}tan(theta)=e^{-t/sin(epsilon)}frac{cos(epsilon)}{sin(epsilon)}quad;quad theta=frac{pi}{2}-epsilon$$
$thetato(pi/2)^- quad;quad epsilon>0to 0^+$
$$e^{-tsec(theta)}tan(theta)sim frac{e^{-t/epsilon}}{epsilon}$$
$f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big) =
begin{cases}
0qquad text{if}quad t>0 \
inftyqquad text{if}quad tleq 0
end{cases}$
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.
answered Feb 3 at 10:16
JJacquelinJJacquelin
45.7k21858
$endgroup$
$$e^{-tsec(theta)}tan(theta)=e^{-t/sin(epsilon)}frac{cos(epsilon)}{sin(epsilon)}quad;quad theta=frac{pi}{2}-epsilon$$
$thetato(pi/2)^- quad;quad epsilon>0to 0^+$
$$e^{-tsec(theta)}tan(theta)sim frac{e^{-t/epsilon}}{epsilon}$$
$f(t)=lim_{thetatopi/2^-}Big(e^{-t(sec{theta})}tan{theta}Big) =
begin{cases}
0qquad text{if}quad t>0 \
inftyqquad text{if}quad tleq 0
end{cases}$
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.
answered Feb 3 at 10:16
JJacquelinJJacquelin
45.7k21858
answered Feb 3 at 10:16
JJacquelinJJacquelin
45.7k21858
answered Feb 3 at 10:16
JJacquelinJJacquelin
45.7k21858
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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