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Solution to ODE from Newton's Second Law
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I have attempted to explore Newton's second law (F = ma) further into its many differential forms. I am not very familiar with differential equations and was searching for the steps and methods to solve the first order equation, second order equation, and the impulse-acceleration relation (that I so called).
For a Impulse-Velocity equation, I think I have found a valid solution.
$$fracdPdt = m fracdvdt$$
$$int fracdPdt dt = m int fracdvdt dt$$
$$int dP = m int dv$$
$$P = mv$$
I think this is correct since this is the momentum equation we already know. However, I also think that this is a trivial solution and there are other important solutions. My question here is, what are the methods one can use to solve differential equations like:
$$F = m fracd^2xdt^2$$
$$F = m fracdvdt$$
and
$$fracdPdt = m fracd^2xdt^2$$
I understand this may be too exhaustive, but if anyone can direct me to the methods only, that would be very useful. I am attempting to learn differential equations with a "head-first" approach of just diving into it. I have a good grounding in calculus as well.
Thank you.
differential-equations physics applications
asked Jul 28 at 22:47
PartialDifferentials
153
add a comment |Â
up vote
1
down vote
favorite
I have attempted to explore Newton's second law (F = ma) further into its many differential forms. I am not very familiar with differential equations and was searching for the steps and methods to solve the first order equation, second order equation, and the impulse-acceleration relation (that I so called).
For a Impulse-Velocity equation, I think I have found a valid solution.
$$fracdPdt = m fracdvdt$$
$$int fracdPdt dt = m int fracdvdt dt$$
$$int dP = m int dv$$
$$P = mv$$
I think this is correct since this is the momentum equation we already know. However, I also think that this is a trivial solution and there are other important solutions. My question here is, what are the methods one can use to solve differential equations like:
$$F = m fracd^2xdt^2$$
$$F = m fracdvdt$$
and
$$fracdPdt = m fracd^2xdt^2$$
I understand this may be too exhaustive, but if anyone can direct me to the methods only, that would be very useful. I am attempting to learn differential equations with a "head-first" approach of just diving into it. I have a good grounding in calculus as well.
Thank you.
differential-equations physics applications
asked Jul 28 at 22:47
PartialDifferentials
153
OP, @Simon gave great answer to the main question. Are you interest in an example where Newton law(using other laws) gave very non-trivial result by solving an ODE(ordinal differential equation)?
– Holo
Jul 29 at 4:29
@Holo yes, his answer is very useful. If you can show me how that's done, it'd be great.
– PartialDifferentials
Jul 29 at 15:13
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have attempted to explore Newton's second law (F = ma) further into its many differential forms. I am not very familiar with differential equations and was searching for the steps and methods to solve the first order equation, second order equation, and the impulse-acceleration relation (that I so called).
For a Impulse-Velocity equation, I think I have found a valid solution.
$$fracdPdt = m fracdvdt$$
$$int fracdPdt dt = m int fracdvdt dt$$
$$int dP = m int dv$$
$$P = mv$$
I think this is correct since this is the momentum equation we already know. However, I also think that this is a trivial solution and there are other important solutions. My question here is, what are the methods one can use to solve differential equations like:
$$F = m fracd^2xdt^2$$
$$F = m fracdvdt$$
and
$$fracdPdt = m fracd^2xdt^2$$
I understand this may be too exhaustive, but if anyone can direct me to the methods only, that would be very useful. I am attempting to learn differential equations with a "head-first" approach of just diving into it. I have a good grounding in calculus as well.
Thank you.
differential-equations physics applications
asked Jul 28 at 22:47
PartialDifferentials
153
I have attempted to explore Newton's second law (F = ma) further into its many differential forms. I am not very familiar with differential equations and was searching for the steps and methods to solve the first order equation, second order equation, and the impulse-acceleration relation (that I so called).
For a Impulse-Velocity equation, I think I have found a valid solution.
$$fracdPdt = m fracdvdt$$
$$int fracdPdt dt = m int fracdvdt dt$$
$$int dP = m int dv$$
$$P = mv$$
I think this is correct since this is the momentum equation we already know. However, I also think that this is a trivial solution and there are other important solutions. My question here is, what are the methods one can use to solve differential equations like:
$$F = m fracd^2xdt^2$$
$$F = m fracdvdt$$
and
$$fracdPdt = m fracd^2xdt^2$$
I understand this may be too exhaustive, but if anyone can direct me to the methods only, that would be very useful. I am attempting to learn differential equations with a "head-first" approach of just diving into it. I have a good grounding in calculus as well.
Thank you.
differential-equations physics applications
asked Jul 28 at 22:47
PartialDifferentials
153
asked Jul 28 at 22:47
PartialDifferentials
153
asked Jul 28 at 22:47
PartialDifferentials
153
asked Jul 28 at 22:47
PartialDifferentials
153
153
OP, @Simon gave great answer to the main question. Are you interest in an example where Newton law(using other laws) gave very non-trivial result by solving an ODE(ordinal differential equation)?
– Holo
Jul 29 at 4:29
@Holo yes, his answer is very useful. If you can show me how that's done, it'd be great.
– PartialDifferentials
Jul 29 at 15:13
add a comment |Â
OP, @Simon gave great answer to the main question. Are you interest in an example where Newton law(using other laws) gave very non-trivial result by solving an ODE(ordinal differential equation)?
– Holo
Jul 29 at 4:29
@Holo yes, his answer is very useful. If you can show me how that's done, it'd be great.
– PartialDifferentials
Jul 29 at 15:13
OP, @Simon gave great answer to the main question. Are you interest in an example where Newton law(using other laws) gave very non-trivial result by solving an ODE(ordinal differential equation)?
– Holo
Jul 29 at 4:29
OP, @Simon gave great answer to the main question. Are you interest in an example where Newton law(using other laws) gave very non-trivial result by solving an ODE(ordinal differential equation)?
– Holo
Jul 29 at 4:29
@Holo yes, his answer is very useful. If you can show me how that's done, it'd be great.
– PartialDifferentials
Jul 29 at 15:13
@Holo yes, his answer is very useful. If you can show me how that's done, it'd be great.
– PartialDifferentials
Jul 29 at 15:13
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
For the first differential equation (DE), there are two things I want to point out.
The equation is what is called a separable equation, for which your method of direct integration suffices.
Be careful. Without specifying boundary/initial conditions, there can be infinite solutions to the DE. When you performed the integration, you forgot to include the constant of integration. The particular value of that constant is determined by the boundary/initial conditions, which arise from physical conditions that are imposed on the system.
For first order equations (where the highest derivative is the first derivative), there is the method of integrating factors. The idea behind this method is to turn the left-hand side of the DE into the chain rule. Then, you can perform integration on both sides and solve for the solution.
For second order equations (where the highest derivative is the second derivative), there are many methods. Here is a list of some of the methods. Which method you use, will ultimately come down to the structure of your problem.
For higher-order differential equations, your best bet is to turn the DE into a system of first-order equations, after which you may apply linear-algebra tools to solve the problem. That is explained here.
Some alternative methods which apply to any order of DE are the power series method, and the Laplace transform. These are more advanced methods that are sometimes used, but I suggest you get down first-order and second-order equations first. The main idea behind the power series method is to assume the solution can be expressed as an infinite series, and then derive a relationship between the coefficients in the series. The main idea behind the Laplace transform is to transform your problem into an algebraic one, after which you would use a table of Laplace transforms to find the solution.
answered Jul 28 at 23:16
Simon
380211
add a comment |Â
up vote
1
down vote
Here are 2 simple examples where ODEs are used in physics:
1
Remember Newton's Second law:$$F=ma=mfracd^2xdt^2$$Also remember Hooke’s Law:
Given a spring, the power on a spring is linearly proportionally to the distance:$$F=kx$$Hence the power on the body connected to a spring is exactly $$F=-kx$$($k$ is the stiffness constant of a string, all you need to know is that it is positive number)
Now let's assume that we have a body on a spring with no forces apart from the spring, combining the 2 laws we get $$mfracd^2xdt^2=-kxiff mfracd^2xdt^2+kx=0$$This ODE is in pretty standard form, I won't add here the way because in this answered I just want to show you examples but the final solution will be $$x(t)=C_1cosleft(tsqrtfrackmright)+C_2sinleft(tsqrtfrackmright)$$Where $C_1,C_2$ are some numbers(we can determine them if given initial conditions). This is simple harmonic motion.
2
In resistor-capacitor circuit(RC) we are dealing with a system where we have the charge($q(t)$), the voltage($V(t)$) and the current($I(t)$) depends on time.
We know that the charge is nothing but $CV(t)$($C$ is a constant depends on the capacitor in the system).
Also the rate of change in the charge is just the current, so $$fracdqdt=CfracdVdt=I(t)$$Ohm's law gives us that $$I(t)=fracV(t)R$$In the resistor. Where $R$ is the resistance(another constant)
Now we can't compare the $2$ yet, because the first equation is about the capacitor and the second is about the resistor.
Now we look for help from kirchoff's law, which says that the sum of the currents must be equal $0$, so we get the ODE$$CfracdVdt+fracV(t)R=0$$Which solving it will result with $$V(t)=V(0)e^-frac tRC$$
This is just 2 examples of non-trivial results of ODEs in physics
answered Jul 29 at 17:56
Holo
4,1072528
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For the first differential equation (DE), there are two things I want to point out.
The equation is what is called a separable equation, for which your method of direct integration suffices.
Be careful. Without specifying boundary/initial conditions, there can be infinite solutions to the DE. When you performed the integration, you forgot to include the constant of integration. The particular value of that constant is determined by the boundary/initial conditions, which arise from physical conditions that are imposed on the system.
For first order equations (where the highest derivative is the first derivative), there is the method of integrating factors. The idea behind this method is to turn the left-hand side of the DE into the chain rule. Then, you can perform integration on both sides and solve for the solution.
For second order equations (where the highest derivative is the second derivative), there are many methods. Here is a list of some of the methods. Which method you use, will ultimately come down to the structure of your problem.
For higher-order differential equations, your best bet is to turn the DE into a system of first-order equations, after which you may apply linear-algebra tools to solve the problem. That is explained here.
Some alternative methods which apply to any order of DE are the power series method, and the Laplace transform. These are more advanced methods that are sometimes used, but I suggest you get down first-order and second-order equations first. The main idea behind the power series method is to assume the solution can be expressed as an infinite series, and then derive a relationship between the coefficients in the series. The main idea behind the Laplace transform is to transform your problem into an algebraic one, after which you would use a table of Laplace transforms to find the solution.
answered Jul 28 at 23:16
Simon
380211
add a comment |Â
up vote
2
down vote
accepted
For the first differential equation (DE), there are two things I want to point out.
The equation is what is called a separable equation, for which your method of direct integration suffices.
Be careful. Without specifying boundary/initial conditions, there can be infinite solutions to the DE. When you performed the integration, you forgot to include the constant of integration. The particular value of that constant is determined by the boundary/initial conditions, which arise from physical conditions that are imposed on the system.
For first order equations (where the highest derivative is the first derivative), there is the method of integrating factors. The idea behind this method is to turn the left-hand side of the DE into the chain rule. Then, you can perform integration on both sides and solve for the solution.
For second order equations (where the highest derivative is the second derivative), there are many methods. Here is a list of some of the methods. Which method you use, will ultimately come down to the structure of your problem.
For higher-order differential equations, your best bet is to turn the DE into a system of first-order equations, after which you may apply linear-algebra tools to solve the problem. That is explained here.
Some alternative methods which apply to any order of DE are the power series method, and the Laplace transform. These are more advanced methods that are sometimes used, but I suggest you get down first-order and second-order equations first. The main idea behind the power series method is to assume the solution can be expressed as an infinite series, and then derive a relationship between the coefficients in the series. The main idea behind the Laplace transform is to transform your problem into an algebraic one, after which you would use a table of Laplace transforms to find the solution.
answered Jul 28 at 23:16
Simon
380211
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For the first differential equation (DE), there are two things I want to point out.
The equation is what is called a separable equation, for which your method of direct integration suffices.
Be careful. Without specifying boundary/initial conditions, there can be infinite solutions to the DE. When you performed the integration, you forgot to include the constant of integration. The particular value of that constant is determined by the boundary/initial conditions, which arise from physical conditions that are imposed on the system.
For first order equations (where the highest derivative is the first derivative), there is the method of integrating factors. The idea behind this method is to turn the left-hand side of the DE into the chain rule. Then, you can perform integration on both sides and solve for the solution.
For second order equations (where the highest derivative is the second derivative), there are many methods. Here is a list of some of the methods. Which method you use, will ultimately come down to the structure of your problem.
For higher-order differential equations, your best bet is to turn the DE into a system of first-order equations, after which you may apply linear-algebra tools to solve the problem. That is explained here.
Some alternative methods which apply to any order of DE are the power series method, and the Laplace transform. These are more advanced methods that are sometimes used, but I suggest you get down first-order and second-order equations first. The main idea behind the power series method is to assume the solution can be expressed as an infinite series, and then derive a relationship between the coefficients in the series. The main idea behind the Laplace transform is to transform your problem into an algebraic one, after which you would use a table of Laplace transforms to find the solution.
answered Jul 28 at 23:16
Simon
380211
For the first differential equation (DE), there are two things I want to point out.
The equation is what is called a separable equation, for which your method of direct integration suffices.
Be careful. Without specifying boundary/initial conditions, there can be infinite solutions to the DE. When you performed the integration, you forgot to include the constant of integration. The particular value of that constant is determined by the boundary/initial conditions, which arise from physical conditions that are imposed on the system.
For first order equations (where the highest derivative is the first derivative), there is the method of integrating factors. The idea behind this method is to turn the left-hand side of the DE into the chain rule. Then, you can perform integration on both sides and solve for the solution.
For second order equations (where the highest derivative is the second derivative), there are many methods. Here is a list of some of the methods. Which method you use, will ultimately come down to the structure of your problem.
For higher-order differential equations, your best bet is to turn the DE into a system of first-order equations, after which you may apply linear-algebra tools to solve the problem. That is explained here.
Some alternative methods which apply to any order of DE are the power series method, and the Laplace transform. These are more advanced methods that are sometimes used, but I suggest you get down first-order and second-order equations first. The main idea behind the power series method is to assume the solution can be expressed as an infinite series, and then derive a relationship between the coefficients in the series. The main idea behind the Laplace transform is to transform your problem into an algebraic one, after which you would use a table of Laplace transforms to find the solution.
answered Jul 28 at 23:16
Simon
380211
answered Jul 28 at 23:16
Simon
380211
answered Jul 28 at 23:16
Simon
380211
answered Jul 28 at 23:16
Simon
380211
380211
add a comment |Â
add a comment |Â
up vote
1
down vote
Here are 2 simple examples where ODEs are used in physics:
1
Remember Newton's Second law:$$F=ma=mfracd^2xdt^2$$Also remember Hooke’s Law:
Given a spring, the power on a spring is linearly proportionally to the distance:$$F=kx$$Hence the power on the body connected to a spring is exactly $$F=-kx$$($k$ is the stiffness constant of a string, all you need to know is that it is positive number)
Now let's assume that we have a body on a spring with no forces apart from the spring, combining the 2 laws we get $$mfracd^2xdt^2=-kxiff mfracd^2xdt^2+kx=0$$This ODE is in pretty standard form, I won't add here the way because in this answered I just want to show you examples but the final solution will be $$x(t)=C_1cosleft(tsqrtfrackmright)+C_2sinleft(tsqrtfrackmright)$$Where $C_1,C_2$ are some numbers(we can determine them if given initial conditions). This is simple harmonic motion.
2
In resistor-capacitor circuit(RC) we are dealing with a system where we have the charge($q(t)$), the voltage($V(t)$) and the current($I(t)$) depends on time.
We know that the charge is nothing but $CV(t)$($C$ is a constant depends on the capacitor in the system).
Also the rate of change in the charge is just the current, so $$fracdqdt=CfracdVdt=I(t)$$Ohm's law gives us that $$I(t)=fracV(t)R$$In the resistor. Where $R$ is the resistance(another constant)
Now we can't compare the $2$ yet, because the first equation is about the capacitor and the second is about the resistor.
Now we look for help from kirchoff's law, which says that the sum of the currents must be equal $0$, so we get the ODE$$CfracdVdt+fracV(t)R=0$$Which solving it will result with $$V(t)=V(0)e^-frac tRC$$
This is just 2 examples of non-trivial results of ODEs in physics
answered Jul 29 at 17:56
Holo
4,1072528
add a comment |Â
up vote
1
down vote
Here are 2 simple examples where ODEs are used in physics:
1
Remember Newton's Second law:$$F=ma=mfracd^2xdt^2$$Also remember Hooke’s Law:
Given a spring, the power on a spring is linearly proportionally to the distance:$$F=kx$$Hence the power on the body connected to a spring is exactly $$F=-kx$$($k$ is the stiffness constant of a string, all you need to know is that it is positive number)
Now let's assume that we have a body on a spring with no forces apart from the spring, combining the 2 laws we get $$mfracd^2xdt^2=-kxiff mfracd^2xdt^2+kx=0$$This ODE is in pretty standard form, I won't add here the way because in this answered I just want to show you examples but the final solution will be $$x(t)=C_1cosleft(tsqrtfrackmright)+C_2sinleft(tsqrtfrackmright)$$Where $C_1,C_2$ are some numbers(we can determine them if given initial conditions). This is simple harmonic motion.
2
In resistor-capacitor circuit(RC) we are dealing with a system where we have the charge($q(t)$), the voltage($V(t)$) and the current($I(t)$) depends on time.
We know that the charge is nothing but $CV(t)$($C$ is a constant depends on the capacitor in the system).
Also the rate of change in the charge is just the current, so $$fracdqdt=CfracdVdt=I(t)$$Ohm's law gives us that $$I(t)=fracV(t)R$$In the resistor. Where $R$ is the resistance(another constant)
Now we can't compare the $2$ yet, because the first equation is about the capacitor and the second is about the resistor.
Now we look for help from kirchoff's law, which says that the sum of the currents must be equal $0$, so we get the ODE$$CfracdVdt+fracV(t)R=0$$Which solving it will result with $$V(t)=V(0)e^-frac tRC$$
This is just 2 examples of non-trivial results of ODEs in physics
answered Jul 29 at 17:56
Holo
4,1072528
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here are 2 simple examples where ODEs are used in physics:
1
Remember Newton's Second law:$$F=ma=mfracd^2xdt^2$$Also remember Hooke’s Law:
Given a spring, the power on a spring is linearly proportionally to the distance:$$F=kx$$Hence the power on the body connected to a spring is exactly $$F=-kx$$($k$ is the stiffness constant of a string, all you need to know is that it is positive number)
Now let's assume that we have a body on a spring with no forces apart from the spring, combining the 2 laws we get $$mfracd^2xdt^2=-kxiff mfracd^2xdt^2+kx=0$$This ODE is in pretty standard form, I won't add here the way because in this answered I just want to show you examples but the final solution will be $$x(t)=C_1cosleft(tsqrtfrackmright)+C_2sinleft(tsqrtfrackmright)$$Where $C_1,C_2$ are some numbers(we can determine them if given initial conditions). This is simple harmonic motion.
2
In resistor-capacitor circuit(RC) we are dealing with a system where we have the charge($q(t)$), the voltage($V(t)$) and the current($I(t)$) depends on time.
We know that the charge is nothing but $CV(t)$($C$ is a constant depends on the capacitor in the system).
Also the rate of change in the charge is just the current, so $$fracdqdt=CfracdVdt=I(t)$$Ohm's law gives us that $$I(t)=fracV(t)R$$In the resistor. Where $R$ is the resistance(another constant)
Now we can't compare the $2$ yet, because the first equation is about the capacitor and the second is about the resistor.
Now we look for help from kirchoff's law, which says that the sum of the currents must be equal $0$, so we get the ODE$$CfracdVdt+fracV(t)R=0$$Which solving it will result with $$V(t)=V(0)e^-frac tRC$$
This is just 2 examples of non-trivial results of ODEs in physics
answered Jul 29 at 17:56
Holo
4,1072528
Here are 2 simple examples where ODEs are used in physics:
1
Remember Newton's Second law:$$F=ma=mfracd^2xdt^2$$Also remember Hooke’s Law:
Given a spring, the power on a spring is linearly proportionally to the distance:$$F=kx$$Hence the power on the body connected to a spring is exactly $$F=-kx$$($k$ is the stiffness constant of a string, all you need to know is that it is positive number)
Now let's assume that we have a body on a spring with no forces apart from the spring, combining the 2 laws we get $$mfracd^2xdt^2=-kxiff mfracd^2xdt^2+kx=0$$This ODE is in pretty standard form, I won't add here the way because in this answered I just want to show you examples but the final solution will be $$x(t)=C_1cosleft(tsqrtfrackmright)+C_2sinleft(tsqrtfrackmright)$$Where $C_1,C_2$ are some numbers(we can determine them if given initial conditions). This is simple harmonic motion.
2
In resistor-capacitor circuit(RC) we are dealing with a system where we have the charge($q(t)$), the voltage($V(t)$) and the current($I(t)$) depends on time.
We know that the charge is nothing but $CV(t)$($C$ is a constant depends on the capacitor in the system).
Also the rate of change in the charge is just the current, so $$fracdqdt=CfracdVdt=I(t)$$Ohm's law gives us that $$I(t)=fracV(t)R$$In the resistor. Where $R$ is the resistance(another constant)
Now we can't compare the $2$ yet, because the first equation is about the capacitor and the second is about the resistor.
Now we look for help from kirchoff's law, which says that the sum of the currents must be equal $0$, so we get the ODE$$CfracdVdt+fracV(t)R=0$$Which solving it will result with $$V(t)=V(0)e^-frac tRC$$
This is just 2 examples of non-trivial results of ODEs in physics
answered Jul 29 at 17:56
Holo
4,1072528
answered Jul 29 at 17:56
Holo
4,1072528
answered Jul 29 at 17:56
Holo
4,1072528
answered Jul 29 at 17:56
Holo
4,1072528
4,1072528
add a comment |Â
add a comment |Â
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OP, @Simon gave great answer to the main question. Are you interest in an example where Newton law(using other laws) gave very non-trivial result by solving an ODE(ordinal differential equation)?
– Holo
Jul 29 at 4:29
@Holo yes, his answer is very useful. If you can show me how that's done, it'd be great.
– PartialDifferentials
Jul 29 at 15:13