Solution of the differential equation $ddot{x}=e^{-x}$
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I would like to know how to get the solution of the following differential equation:
$$frac{d^2x}{dt^2}(t)=e^{-x(t)}$$
A differential equation of similar form appears in a paper (Rapp & Kassal 1968) that I'm studying. The authors show us its solution (it involves an hyperbolic secant function), but don't tell us how did they get it. So, at least I know that this differential equation has an analytical solution.
ordinary-differential-equations
asked Jan 7 at 16:15
Élio PereiraÉlio Pereira
402514
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add a comment |
$begingroup$
I would like to know how to get the solution of the following differential equation:
$$frac{d^2x}{dt^2}(t)=e^{-x(t)}$$
A differential equation of similar form appears in a paper (Rapp & Kassal 1968) that I'm studying. The authors show us its solution (it involves an hyperbolic secant function), but don't tell us how did they get it. So, at least I know that this differential equation has an analytical solution.
ordinary-differential-equations
asked Jan 7 at 16:15
Élio PereiraÉlio Pereira
402514
$endgroup$
add a comment |
$begingroup$
I would like to know how to get the solution of the following differential equation:
$$frac{d^2x}{dt^2}(t)=e^{-x(t)}$$
A differential equation of similar form appears in a paper (Rapp & Kassal 1968) that I'm studying. The authors show us its solution (it involves an hyperbolic secant function), but don't tell us how did they get it. So, at least I know that this differential equation has an analytical solution.
ordinary-differential-equations
asked Jan 7 at 16:15
Élio PereiraÉlio Pereira
402514
$endgroup$
I would like to know how to get the solution of the following differential equation:
$$frac{d^2x}{dt^2}(t)=e^{-x(t)}$$
A differential equation of similar form appears in a paper (Rapp & Kassal 1968) that I'm studying. The authors show us its solution (it involves an hyperbolic secant function), but don't tell us how did they get it. So, at least I know that this differential equation has an analytical solution.
ordinary-differential-equations
ordinary-differential-equations
asked Jan 7 at 16:15
Élio PereiraÉlio Pereira
402514
asked Jan 7 at 16:15
Élio PereiraÉlio Pereira
402514
asked Jan 7 at 16:15
Élio PereiraÉlio Pereira
402514
asked Jan 7 at 16:15
Élio PereiraÉlio Pereira
402514
asked Jan 7 at 16:15
Élio PereiraÉlio Pereira
402514
402514
add a comment |
add a comment |
1 Answer
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We have $$ddot x=e^{-x}\dot xddot x=dot x e^{-x}$$Integrate both sides with respect to $t$.
$$frac12 dot x^2=C_1-e^{-x}\intfrac{mathrm dx}{sqrt{2(C_1-e^{-x})}}=intmathrm dt$$ This can then be integrated using the substitution $u=e^{-x}$.
Edit: On second thoughts, the substitution $u=sqrt{C_1-e^{-x}}$ works pretty well, giving $$frac1{sqrt2}intfrac{mathrm du}{C_1-u^2}=frac1{sqrt{2C_1}}tanh^{-1}left(frac{u}{sqrt{C_1}}right)=frac1{sqrt{2C_1}}tanh^{-1}left({sqrt{1-frac{1}{C_1e^x}}}right)$$
answered Jan 7 at 16:25
John DoeJohn Doe
11.1k11238
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2
$begingroup$
You can also use $u=e^{x/2}$. Lots of possibilities!
$endgroup$
– Dylan
Jan 7 at 16:47
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Thank you very much for the help!
$endgroup$
– Élio Pereira
Jan 7 at 16:49
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
We have $$ddot x=e^{-x}\dot xddot x=dot x e^{-x}$$Integrate both sides with respect to $t$.
$$frac12 dot x^2=C_1-e^{-x}\intfrac{mathrm dx}{sqrt{2(C_1-e^{-x})}}=intmathrm dt$$ This can then be integrated using the substitution $u=e^{-x}$.
Edit: On second thoughts, the substitution $u=sqrt{C_1-e^{-x}}$ works pretty well, giving $$frac1{sqrt2}intfrac{mathrm du}{C_1-u^2}=frac1{sqrt{2C_1}}tanh^{-1}left(frac{u}{sqrt{C_1}}right)=frac1{sqrt{2C_1}}tanh^{-1}left({sqrt{1-frac{1}{C_1e^x}}}right)$$
answered Jan 7 at 16:25
John DoeJohn Doe
11.1k11238
$endgroup$
2
$begingroup$
You can also use $u=e^{x/2}$. Lots of possibilities!
$endgroup$
– Dylan
Jan 7 at 16:47
$begingroup$
Thank you very much for the help!
$endgroup$
– Élio Pereira
Jan 7 at 16:49
add a comment |
$begingroup$
We have $$ddot x=e^{-x}\dot xddot x=dot x e^{-x}$$Integrate both sides with respect to $t$.
$$frac12 dot x^2=C_1-e^{-x}\intfrac{mathrm dx}{sqrt{2(C_1-e^{-x})}}=intmathrm dt$$ This can then be integrated using the substitution $u=e^{-x}$.
Edit: On second thoughts, the substitution $u=sqrt{C_1-e^{-x}}$ works pretty well, giving $$frac1{sqrt2}intfrac{mathrm du}{C_1-u^2}=frac1{sqrt{2C_1}}tanh^{-1}left(frac{u}{sqrt{C_1}}right)=frac1{sqrt{2C_1}}tanh^{-1}left({sqrt{1-frac{1}{C_1e^x}}}right)$$
answered Jan 7 at 16:25
John DoeJohn Doe
11.1k11238
$endgroup$
2
$begingroup$
You can also use $u=e^{x/2}$. Lots of possibilities!
$endgroup$
– Dylan
Jan 7 at 16:47
$begingroup$
Thank you very much for the help!
$endgroup$
– Élio Pereira
Jan 7 at 16:49
add a comment |
$begingroup$
We have $$ddot x=e^{-x}\dot xddot x=dot x e^{-x}$$Integrate both sides with respect to $t$.
$$frac12 dot x^2=C_1-e^{-x}\intfrac{mathrm dx}{sqrt{2(C_1-e^{-x})}}=intmathrm dt$$ This can then be integrated using the substitution $u=e^{-x}$.
Edit: On second thoughts, the substitution $u=sqrt{C_1-e^{-x}}$ works pretty well, giving $$frac1{sqrt2}intfrac{mathrm du}{C_1-u^2}=frac1{sqrt{2C_1}}tanh^{-1}left(frac{u}{sqrt{C_1}}right)=frac1{sqrt{2C_1}}tanh^{-1}left({sqrt{1-frac{1}{C_1e^x}}}right)$$
answered Jan 7 at 16:25
John DoeJohn Doe
11.1k11238
$endgroup$
We have $$ddot x=e^{-x}\dot xddot x=dot x e^{-x}$$Integrate both sides with respect to $t$.
$$frac12 dot x^2=C_1-e^{-x}\intfrac{mathrm dx}{sqrt{2(C_1-e^{-x})}}=intmathrm dt$$ This can then be integrated using the substitution $u=e^{-x}$.
Edit: On second thoughts, the substitution $u=sqrt{C_1-e^{-x}}$ works pretty well, giving $$frac1{sqrt2}intfrac{mathrm du}{C_1-u^2}=frac1{sqrt{2C_1}}tanh^{-1}left(frac{u}{sqrt{C_1}}right)=frac1{sqrt{2C_1}}tanh^{-1}left({sqrt{1-frac{1}{C_1e^x}}}right)$$
answered Jan 7 at 16:25
John DoeJohn Doe
11.1k11238
edited Jan 7 at 16:39
answered Jan 7 at 16:25
John DoeJohn Doe
11.1k11238
answered Jan 7 at 16:25
John DoeJohn Doe
11.1k11238
answered Jan 7 at 16:25
John DoeJohn Doe
11.1k11238
11.1k11238
2
$begingroup$
You can also use $u=e^{x/2}$. Lots of possibilities!
$endgroup$
– Dylan
Jan 7 at 16:47
$begingroup$
Thank you very much for the help!
$endgroup$
– Élio Pereira
Jan 7 at 16:49
add a comment |
2
$begingroup$
You can also use $u=e^{x/2}$. Lots of possibilities!
$endgroup$
– Dylan
Jan 7 at 16:47
$begingroup$
Thank you very much for the help!
$endgroup$
– Élio Pereira
Jan 7 at 16:49
2
2
$begingroup$
You can also use $u=e^{x/2}$. Lots of possibilities!
$endgroup$
– Dylan
Jan 7 at 16:47
$begingroup$
You can also use $u=e^{x/2}$. Lots of possibilities!
$endgroup$
– Dylan
Jan 7 at 16:47
$begingroup$
Thank you very much for the help!
$endgroup$
– Élio Pereira
Jan 7 at 16:49
$begingroup$
Thank you very much for the help!
$endgroup$
– Élio Pereira
Jan 7 at 16:49
add a comment |
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