Solving an Unwieldy Differential Equation
I am at the end of a very long math problem and am left with this equation.
I know. It's monstrous.
This equation is equal to $frac{dtheta}{dt}$, and I would like to solve it as a differential equation. The only way I've found that might work is the separation of variables strategy:
$$frac{dtheta}{dt}=f(t)g(theta)$$
$$frac{1}{g(theta)}dtheta=f(t)dt$$
$$intfrac{1}{g(theta)}dtheta=int f(t)dt$$
$$G(theta)+c_1=F(t)+c_2$$
$$G(theta)-F(t)=C$$
The issue is that teasing apart the variables in this equation seems like it will take a very, very long time, not to mention taking the integral of what is left. It feels like this is the wrong way to do it.
So, my question is this: how do I solve this differential equation?
Thank you for all the help you can give.
calculus differential-equations
asked Dec 4 '18 at 16:27
Levi Buckwalter
1
add a comment |
I am at the end of a very long math problem and am left with this equation.
I know. It's monstrous.
This equation is equal to $frac{dtheta}{dt}$, and I would like to solve it as a differential equation. The only way I've found that might work is the separation of variables strategy:
$$frac{dtheta}{dt}=f(t)g(theta)$$
$$frac{1}{g(theta)}dtheta=f(t)dt$$
$$intfrac{1}{g(theta)}dtheta=int f(t)dt$$
$$G(theta)+c_1=F(t)+c_2$$
$$G(theta)-F(t)=C$$
The issue is that teasing apart the variables in this equation seems like it will take a very, very long time, not to mention taking the integral of what is left. It feels like this is the wrong way to do it.
So, my question is this: how do I solve this differential equation?
Thank you for all the help you can give.
calculus differential-equations
asked Dec 4 '18 at 16:27
Levi Buckwalter
1
add a comment |
I am at the end of a very long math problem and am left with this equation.
I know. It's monstrous.
This equation is equal to $frac{dtheta}{dt}$, and I would like to solve it as a differential equation. The only way I've found that might work is the separation of variables strategy:
$$frac{dtheta}{dt}=f(t)g(theta)$$
$$frac{1}{g(theta)}dtheta=f(t)dt$$
$$intfrac{1}{g(theta)}dtheta=int f(t)dt$$
$$G(theta)+c_1=F(t)+c_2$$
$$G(theta)-F(t)=C$$
The issue is that teasing apart the variables in this equation seems like it will take a very, very long time, not to mention taking the integral of what is left. It feels like this is the wrong way to do it.
So, my question is this: how do I solve this differential equation?
Thank you for all the help you can give.
calculus differential-equations
asked Dec 4 '18 at 16:27
Levi Buckwalter
1
I am at the end of a very long math problem and am left with this equation.
I know. It's monstrous.
This equation is equal to $frac{dtheta}{dt}$, and I would like to solve it as a differential equation. The only way I've found that might work is the separation of variables strategy:
$$frac{dtheta}{dt}=f(t)g(theta)$$
$$frac{1}{g(theta)}dtheta=f(t)dt$$
$$intfrac{1}{g(theta)}dtheta=int f(t)dt$$
$$G(theta)+c_1=F(t)+c_2$$
$$G(theta)-F(t)=C$$
The issue is that teasing apart the variables in this equation seems like it will take a very, very long time, not to mention taking the integral of what is left. It feels like this is the wrong way to do it.
So, my question is this: how do I solve this differential equation?
Thank you for all the help you can give.
calculus differential-equations
calculus differential-equations
asked Dec 4 '18 at 16:27
Levi Buckwalter
1
asked Dec 4 '18 at 16:27
Levi Buckwalter
1
asked Dec 4 '18 at 16:27
Levi Buckwalter
1
asked Dec 4 '18 at 16:27
Levi Buckwalter
1
asked Dec 4 '18 at 16:27
Levi Buckwalter
1
1
add a comment |
add a comment |
1 Answer
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HINT
For starters, simplify the denominator noting that
$$
l^2 sin^2 theta + l^2 cos^2 theta = l^2,
$$
and cancel one of the $l$ factor with the numerator. Finally, I would try to simplify the massive trig function on the RHS using various trig summation of sines and cosines to compact it down.
answered Dec 4 '18 at 16:35
gt6989b
33.1k22452
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT
For starters, simplify the denominator noting that
$$
l^2 sin^2 theta + l^2 cos^2 theta = l^2,
$$
and cancel one of the $l$ factor with the numerator. Finally, I would try to simplify the massive trig function on the RHS using various trig summation of sines and cosines to compact it down.
answered Dec 4 '18 at 16:35
gt6989b
33.1k22452
add a comment |
HINT
For starters, simplify the denominator noting that
$$
l^2 sin^2 theta + l^2 cos^2 theta = l^2,
$$
and cancel one of the $l$ factor with the numerator. Finally, I would try to simplify the massive trig function on the RHS using various trig summation of sines and cosines to compact it down.
answered Dec 4 '18 at 16:35
gt6989b
33.1k22452
add a comment |
HINT
For starters, simplify the denominator noting that
$$
l^2 sin^2 theta + l^2 cos^2 theta = l^2,
$$
and cancel one of the $l$ factor with the numerator. Finally, I would try to simplify the massive trig function on the RHS using various trig summation of sines and cosines to compact it down.
answered Dec 4 '18 at 16:35
gt6989b
33.1k22452
HINT
For starters, simplify the denominator noting that
$$
l^2 sin^2 theta + l^2 cos^2 theta = l^2,
$$
and cancel one of the $l$ factor with the numerator. Finally, I would try to simplify the massive trig function on the RHS using various trig summation of sines and cosines to compact it down.
answered Dec 4 '18 at 16:35
gt6989b
33.1k22452
answered Dec 4 '18 at 16:35
gt6989b
33.1k22452
answered Dec 4 '18 at 16:35
gt6989b
33.1k22452
answered Dec 4 '18 at 16:35
gt6989b
33.1k22452
33.1k22452
add a comment |
add a comment |
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