Mixing
Quadratic second-order differential equation involving tan x
$begingroup$
Is there a way to exactly solve the following differential equation:
$$left(frac{dx}{dt}right)^2 = -tan x frac{d^2x}{dt^2} $$
ordinary-differential-equations
edited Jan 31 at 11:14
Dylan
14.2k31127
asked Jan 29 at 1:00
user153388user153388
514
$endgroup$
add a comment |
$begingroup$
Is there a way to exactly solve the following differential equation:
$$left(frac{dx}{dt}right)^2 = -tan x frac{d^2x}{dt^2} $$
ordinary-differential-equations
edited Jan 31 at 11:14
Dylan
14.2k31127
asked Jan 29 at 1:00
user153388user153388
514
$endgroup$
$begingroup$
The equation would be easier to read if you used MathJax
$endgroup$
– J. W. Tanner
Jan 29 at 1:42
add a comment |
$begingroup$
Is there a way to exactly solve the following differential equation:
$$left(frac{dx}{dt}right)^2 = -tan x frac{d^2x}{dt^2} $$
ordinary-differential-equations
edited Jan 31 at 11:14
Dylan
14.2k31127
asked Jan 29 at 1:00
user153388user153388
514
$endgroup$
Is there a way to exactly solve the following differential equation:
$$left(frac{dx}{dt}right)^2 = -tan x frac{d^2x}{dt^2} $$
ordinary-differential-equations
ordinary-differential-equations
edited Jan 31 at 11:14
Dylan
14.2k31127
asked Jan 29 at 1:00
user153388user153388
514
edited Jan 31 at 11:14
Dylan
14.2k31127
asked Jan 29 at 1:00
user153388user153388
514
edited Jan 31 at 11:14
Dylan
14.2k31127
edited Jan 31 at 11:14
Dylan
14.2k31127
edited Jan 31 at 11:14
Dylan
14.2k31127
14.2k31127
asked Jan 29 at 1:00
user153388user153388
514
asked Jan 29 at 1:00
user153388user153388
514
asked Jan 29 at 1:00
user153388user153388
514
514
$begingroup$
The equation would be easier to read if you used MathJax
$endgroup$
– J. W. Tanner
Jan 29 at 1:42
add a comment |
$begingroup$
The equation would be easier to read if you used MathJax
$endgroup$
– J. W. Tanner
Jan 29 at 1:42
$begingroup$
The equation would be easier to read if you used MathJax
$endgroup$
– J. W. Tanner
Jan 29 at 1:42
$begingroup$
The equation would be easier to read if you used MathJax
$endgroup$
– J. W. Tanner
Jan 29 at 1:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint
The equation being
$$(x')^2+tan(x) x''=0$$ make
$$cos(x)=z implies x=cos ^{-1}(z)implies x'=-frac{z'}{sqrt{1-z^2}}implies x''=frac{left(z^2-1right) z''-(z')^2}{left(1-z^2right)^{3/2}}$$ as well as $tan(x)=frac{sqrt{1-z^2}}{z}$.
You should get something very simple.
answered Jan 29 at 3:02
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Rewrite
$$ -(cot x)x' = frac{x''}{x'} $$
Integrate both sides
$$ ln(x') = -ln(sin x) + C $$
$$ implies x' = frac{A}{sin x} $$
Separate and integrate again
$$ (sin x)x' = A $$
$$ cos x = B-At $$
Final solution
$$ x(t) = arccos(B-At) $$
answered Jan 31 at 11:19
DylanDylan
14.2k31127
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
The equation being
$$(x')^2+tan(x) x''=0$$ make
$$cos(x)=z implies x=cos ^{-1}(z)implies x'=-frac{z'}{sqrt{1-z^2}}implies x''=frac{left(z^2-1right) z''-(z')^2}{left(1-z^2right)^{3/2}}$$ as well as $tan(x)=frac{sqrt{1-z^2}}{z}$.
You should get something very simple.
answered Jan 29 at 3:02
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Hint
The equation being
$$(x')^2+tan(x) x''=0$$ make
$$cos(x)=z implies x=cos ^{-1}(z)implies x'=-frac{z'}{sqrt{1-z^2}}implies x''=frac{left(z^2-1right) z''-(z')^2}{left(1-z^2right)^{3/2}}$$ as well as $tan(x)=frac{sqrt{1-z^2}}{z}$.
You should get something very simple.
answered Jan 29 at 3:02
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Hint
The equation being
$$(x')^2+tan(x) x''=0$$ make
$$cos(x)=z implies x=cos ^{-1}(z)implies x'=-frac{z'}{sqrt{1-z^2}}implies x''=frac{left(z^2-1right) z''-(z')^2}{left(1-z^2right)^{3/2}}$$ as well as $tan(x)=frac{sqrt{1-z^2}}{z}$.
You should get something very simple.
answered Jan 29 at 3:02
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
Hint
The equation being
$$(x')^2+tan(x) x''=0$$ make
$$cos(x)=z implies x=cos ^{-1}(z)implies x'=-frac{z'}{sqrt{1-z^2}}implies x''=frac{left(z^2-1right) z''-(z')^2}{left(1-z^2right)^{3/2}}$$ as well as $tan(x)=frac{sqrt{1-z^2}}{z}$.
You should get something very simple.
answered Jan 29 at 3:02
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 29 at 3:02
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 29 at 3:02
Claude LeiboviciClaude Leibovici
125k1158135
answered Jan 29 at 3:02
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
$begingroup$
Rewrite
$$ -(cot x)x' = frac{x''}{x'} $$
Integrate both sides
$$ ln(x') = -ln(sin x) + C $$
$$ implies x' = frac{A}{sin x} $$
Separate and integrate again
$$ (sin x)x' = A $$
$$ cos x = B-At $$
Final solution
$$ x(t) = arccos(B-At) $$
answered Jan 31 at 11:19
DylanDylan
14.2k31127
$endgroup$
add a comment |
$begingroup$
Rewrite
$$ -(cot x)x' = frac{x''}{x'} $$
Integrate both sides
$$ ln(x') = -ln(sin x) + C $$
$$ implies x' = frac{A}{sin x} $$
Separate and integrate again
$$ (sin x)x' = A $$
$$ cos x = B-At $$
Final solution
$$ x(t) = arccos(B-At) $$
answered Jan 31 at 11:19
DylanDylan
14.2k31127
$endgroup$
add a comment |
$begingroup$
Rewrite
$$ -(cot x)x' = frac{x''}{x'} $$
Integrate both sides
$$ ln(x') = -ln(sin x) + C $$
$$ implies x' = frac{A}{sin x} $$
Separate and integrate again
$$ (sin x)x' = A $$
$$ cos x = B-At $$
Final solution
$$ x(t) = arccos(B-At) $$
answered Jan 31 at 11:19
DylanDylan
14.2k31127
$endgroup$
Rewrite
$$ -(cot x)x' = frac{x''}{x'} $$
Integrate both sides
$$ ln(x') = -ln(sin x) + C $$
$$ implies x' = frac{A}{sin x} $$
Separate and integrate again
$$ (sin x)x' = A $$
$$ cos x = B-At $$
Final solution
$$ x(t) = arccos(B-At) $$
answered Jan 31 at 11:19
DylanDylan
14.2k31127
answered Jan 31 at 11:19
DylanDylan
14.2k31127
answered Jan 31 at 11:19
DylanDylan
14.2k31127
answered Jan 31 at 11:19
DylanDylan
14.2k31127
14.2k31127
add a comment |
add a comment |
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$begingroup$
The equation would be easier to read if you used MathJax
$endgroup$
– J. W. Tanner
Jan 29 at 1:42