Modification of a differential equation
$begingroup$
I have been given a second-order differential equation of the form:
begin{equation}
frac{d^2z}{dt^2}=x_0beta ,frac{dz}{dt},e^{-beta z/gamma} - gamma ,frac{dz}{dt}
end{equation}
where, $ x_{0}, beta, gamma $ are constants.
By introducing the function
begin{equation}
u=e^{-beta z/gamma}
end{equation}
and substituting it into the equation, how can one achieve the form:
begin{equation}
ufrac{d^{2}u}{dt^{2}}-bigg(frac{du}{dt}bigg)^{2}+bigg(gamma-x_{0}beta ubigg)ufrac{du}{dt} = 0?
end{equation}
ordinary-differential-equations
edited Jan 9 at 14:30
Adrian Keister
5,26971933
asked Jan 9 at 14:21
nipohc88nipohc88
163
$endgroup$
add a comment |
$begingroup$
I have been given a second-order differential equation of the form:
begin{equation}
frac{d^2z}{dt^2}=x_0beta ,frac{dz}{dt},e^{-beta z/gamma} - gamma ,frac{dz}{dt}
end{equation}
where, $ x_{0}, beta, gamma $ are constants.
By introducing the function
begin{equation}
u=e^{-beta z/gamma}
end{equation}
and substituting it into the equation, how can one achieve the form:
begin{equation}
ufrac{d^{2}u}{dt^{2}}-bigg(frac{du}{dt}bigg)^{2}+bigg(gamma-x_{0}beta ubigg)ufrac{du}{dt} = 0?
end{equation}
ordinary-differential-equations
edited Jan 9 at 14:30
Adrian Keister
5,26971933
asked Jan 9 at 14:21
nipohc88nipohc88
163
$endgroup$
$begingroup$
Does $'$ denote differentiation with respect to $t$?
$endgroup$
– John Doe
Jan 9 at 14:24
$begingroup$
Sorry for not clarifying, yes indeed everything is differentiated with respect to $t$
$endgroup$
– nipohc88
Jan 9 at 14:27
add a comment |
$begingroup$
I have been given a second-order differential equation of the form:
begin{equation}
frac{d^2z}{dt^2}=x_0beta ,frac{dz}{dt},e^{-beta z/gamma} - gamma ,frac{dz}{dt}
end{equation}
where, $ x_{0}, beta, gamma $ are constants.
By introducing the function
begin{equation}
u=e^{-beta z/gamma}
end{equation}
and substituting it into the equation, how can one achieve the form:
begin{equation}
ufrac{d^{2}u}{dt^{2}}-bigg(frac{du}{dt}bigg)^{2}+bigg(gamma-x_{0}beta ubigg)ufrac{du}{dt} = 0?
end{equation}
ordinary-differential-equations
edited Jan 9 at 14:30
Adrian Keister
5,26971933
asked Jan 9 at 14:21
nipohc88nipohc88
163
$endgroup$
I have been given a second-order differential equation of the form:
begin{equation}
frac{d^2z}{dt^2}=x_0beta ,frac{dz}{dt},e^{-beta z/gamma} - gamma ,frac{dz}{dt}
end{equation}
where, $ x_{0}, beta, gamma $ are constants.
By introducing the function
begin{equation}
u=e^{-beta z/gamma}
end{equation}
and substituting it into the equation, how can one achieve the form:
begin{equation}
ufrac{d^{2}u}{dt^{2}}-bigg(frac{du}{dt}bigg)^{2}+bigg(gamma-x_{0}beta ubigg)ufrac{du}{dt} = 0?
end{equation}
ordinary-differential-equations
ordinary-differential-equations
edited Jan 9 at 14:30
Adrian Keister
5,26971933
asked Jan 9 at 14:21
nipohc88nipohc88
163
edited Jan 9 at 14:30
Adrian Keister
5,26971933
asked Jan 9 at 14:21
nipohc88nipohc88
163
edited Jan 9 at 14:30
Adrian Keister
5,26971933
edited Jan 9 at 14:30
Adrian Keister
5,26971933
edited Jan 9 at 14:30
Adrian Keister
5,26971933
5,26971933
asked Jan 9 at 14:21
nipohc88nipohc88
163
asked Jan 9 at 14:21
nipohc88nipohc88
163
asked Jan 9 at 14:21
nipohc88nipohc88
163
163
$begingroup$
Does $'$ denote differentiation with respect to $t$?
$endgroup$
– John Doe
Jan 9 at 14:24
$begingroup$
Sorry for not clarifying, yes indeed everything is differentiated with respect to $t$
$endgroup$
– nipohc88
Jan 9 at 14:27
add a comment |
$begingroup$
Does $'$ denote differentiation with respect to $t$?
$endgroup$
– John Doe
Jan 9 at 14:24
$begingroup$
Sorry for not clarifying, yes indeed everything is differentiated with respect to $t$
$endgroup$
– nipohc88
Jan 9 at 14:27
$begingroup$
Does $'$ denote differentiation with respect to $t$?
$endgroup$
– John Doe
Jan 9 at 14:24
$begingroup$
Does $'$ denote differentiation with respect to $t$?
$endgroup$
– John Doe
Jan 9 at 14:24
$begingroup$
Sorry for not clarifying, yes indeed everything is differentiated with respect to $t$
$endgroup$
– nipohc88
Jan 9 at 14:27
$begingroup$
Sorry for not clarifying, yes indeed everything is differentiated with respect to $t$
$endgroup$
– nipohc88
Jan 9 at 14:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have the equation $$z''=(x_0beta u-gamma)z'tag1$$
Given that substitution, we can compute $z'$ and $z''$ in terms of derivatives of $u$.
$$z=-fracgammabetalog u\z'=-fracgamma{beta u}u'\z''=fracgamma{beta u^2}(u')^2-fracgamma{beta u}u''$$Substitute these results into $(1)$ and multiply through by $-frac{beta u^2}gamma$ to obtain the desired equation.
answered Jan 9 at 14:31
John DoeJohn Doe
11.7k11239
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have the equation $$z''=(x_0beta u-gamma)z'tag1$$
Given that substitution, we can compute $z'$ and $z''$ in terms of derivatives of $u$.
$$z=-fracgammabetalog u\z'=-fracgamma{beta u}u'\z''=fracgamma{beta u^2}(u')^2-fracgamma{beta u}u''$$Substitute these results into $(1)$ and multiply through by $-frac{beta u^2}gamma$ to obtain the desired equation.
answered Jan 9 at 14:31
John DoeJohn Doe
11.7k11239
$endgroup$
add a comment |
$begingroup$
We have the equation $$z''=(x_0beta u-gamma)z'tag1$$
Given that substitution, we can compute $z'$ and $z''$ in terms of derivatives of $u$.
$$z=-fracgammabetalog u\z'=-fracgamma{beta u}u'\z''=fracgamma{beta u^2}(u')^2-fracgamma{beta u}u''$$Substitute these results into $(1)$ and multiply through by $-frac{beta u^2}gamma$ to obtain the desired equation.
answered Jan 9 at 14:31
John DoeJohn Doe
11.7k11239
$endgroup$
add a comment |
$begingroup$
We have the equation $$z''=(x_0beta u-gamma)z'tag1$$
Given that substitution, we can compute $z'$ and $z''$ in terms of derivatives of $u$.
$$z=-fracgammabetalog u\z'=-fracgamma{beta u}u'\z''=fracgamma{beta u^2}(u')^2-fracgamma{beta u}u''$$Substitute these results into $(1)$ and multiply through by $-frac{beta u^2}gamma$ to obtain the desired equation.
answered Jan 9 at 14:31
John DoeJohn Doe
11.7k11239
$endgroup$
We have the equation $$z''=(x_0beta u-gamma)z'tag1$$
Given that substitution, we can compute $z'$ and $z''$ in terms of derivatives of $u$.
$$z=-fracgammabetalog u\z'=-fracgamma{beta u}u'\z''=fracgamma{beta u^2}(u')^2-fracgamma{beta u}u''$$Substitute these results into $(1)$ and multiply through by $-frac{beta u^2}gamma$ to obtain the desired equation.
answered Jan 9 at 14:31
John DoeJohn Doe
11.7k11239
answered Jan 9 at 14:31
John DoeJohn Doe
11.7k11239
answered Jan 9 at 14:31
John DoeJohn Doe
11.7k11239
answered Jan 9 at 14:31
John DoeJohn Doe
11.7k11239
11.7k11239
add a comment |
add a comment |
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$begingroup$
Does $'$ denote differentiation with respect to $t$?
$endgroup$
– John Doe
Jan 9 at 14:24
$begingroup$
Sorry for not clarifying, yes indeed everything is differentiated with respect to $t$
$endgroup$
– nipohc88
Jan 9 at 14:27