A good motivation for the study of differential equations can be found in physics. For example, Newton's second law states that:
\[m \mathbf a = \sum \mathbf F\]
where \(m\) is the mass of a body, \(\mathbf a\) is the acceleration of said body, and \(\sum \mathbf F\) is the vector sum of all forces acting on the body. Say we have a raindrop falling vertically downwards close to the ground. (we are therefore only considering one-dimensional motion) Say it is subject to a drag force proportional to its velocity, (acting upwards on the drop) and a gravitational force \(g\) taken to be constant. (acting downwards on the drop) Let the raindrop's acceleration be \(a\), its velocity be \(v\) and its mass be \(m\). Then:
\[ma = mg - kv\]
for some constant \(v\). We know the acceleration to be the rate of change of velocity with time, so we can write this as:
\[m \frac {\mathrm dv} {\mathrm dt} = mg - kv\]
which you might prefer to write as:
\[m \frac {\mathrm dv} {\mathrm dt} + kv = mg\]
This is an example of a separable linear first-order differential equation, which we can solve for \(v\) as a function of \(t\). This would enable us to find the terminal velocity of the raindrop, and also its acceleration and displacement from its starting position as functions of \(t\).
Second order differential equations pop up in simple harmonic motion, situations where a body is in periodic motion about a fixed point, subject to a force proportional to its displacement from said point. From Newton's second law, we then have that the acceleration of this body will be proportional to its displacement from its starting point. (as at non-relativistic speeds, the increase in mass of the body will be negligible) We can write this as: \[a = kx\] We know that acceleration is the rate of change of velocity with respect to time, and that velocity is the rate of change of displacement with respect to time. So we can write \(a = \frac {\mathrm d^2 x}{\mathrm dt^2}\). (or as physicists may prefer, \(\ddot x\)) The constant of proportional is usually written \(-\omega^2\). (the utility of this will be seen when we cover SHM) So the above is most commonly seen as: \[\frac {\mathrm d^2 x}{\mathrm dt^2} = -\omega^2 x\] This is an example of a linear second-order differential equation, which we will look into solving in a later section.
Solving separable equations
Separable differential equations are those of the form: \[\frac {\mathrm dy} {\mathrm dx} = f(x) g(y)\] We can divide through \(g(y)\) to get: \[\frac 1 {g(y)} \frac {\mathrm dy} {\mathrm dx} = f(x)\] We can now integrate both sides with respect to \(x\): \[\int \frac 1 {g(y)} \frac {\mathrm dy} {\mathrm dx} \mathrm dx = \int f(x) \mathrm dx\] From integration by substitution, we know that: \[\int \frac 1 {g(y)} \frac {\mathrm dy} {\mathrm dx} \mathrm dx = \int \frac 1 {g(y)} \mathrm dy\] So effectively we are integrating the LHS with respect to \(y\) and the RHS with respect to \(x\). If we're lucky, we can now easily integrate \(\frac 1 {g(y)}\) and \(f(x)\), we can obtain a relationship between \(y\) and \(x\). If we're luckier yet, we may even be able to nicely write a solution in the form \(y = f\) where \(f\) depends only on \(x\). To take an example, let's look at: \[m \frac {\mathrm dv} {\mathrm dt} = mg - kv\] From earlier, we can divide through \(mg - kv\): \[\frac m {mg - kv} \frac {\mathrm dv} {\mathrm dt} = 1\] Which is in a form we can now solve as: \[\int \frac m {mg - kv} \mathrm dv = \frac m k \int \frac 1 {\frac {mg} k - v} \mathrm dv = - \frac m k \log \left({\frac {mg} k - v}\right)\] The RHS is simply \(t + C\) for some (as yet unknown) constant \(C\). Then: \[t + C = - \frac m k \log \left({\frac {mg} k - v}\right)\] This can be relatively straightforwardly rearranged for \(v\): \[-\frac k m t + C' = \log \left({\frac {mg} k - v}\right)\] (where \(C' = -\frac k m C\)) Giving: \[Ae^{-\frac k m t} = \frac {mg} k - v\] (where \(A = e^{C'}\)) So: \[v = \frac {mg} k - Ae^{-\frac k m t}\] We would need some more information about the body to find this constant \(A\), such as the body's velocity at a point in time. (this is called a boundary condition) We can see however, without that information, that as \(t \to \infty\), \(v \to \frac {mg} k\), giving us the terminal velocity of the body. We can also differentiate/integrate this function to obtain its acceleration or the displacement from its start point, in terms of \(t\), though this would introduce further constants which we would need even more information to determine. For example, if we knew that the body started from rest we would have: \[v(0) = 0 = \frac {mg} k - Ae^0 = \frac {mg} k - A\] So: \[A = \frac {mg} k\] Giving: \[v = \frac {mg} k - \frac {mg} k e^{-\frac k m t}\] Sometimes it won't be possible to separate the variables, for example if we had: \[\sin x \frac {\mathrm dy} {\mathrm dx} + y \cos x = e^x \sin x\] We would not be able to apply this technique, we must instead look at alternatives. We will look at these in the next section.
Second order differential equations pop up in simple harmonic motion, situations where a body is in periodic motion about a fixed point, subject to a force proportional to its displacement from said point. From Newton's second law, we then have that the acceleration of this body will be proportional to its displacement from its starting point. (as at non-relativistic speeds, the increase in mass of the body will be negligible) We can write this as: \[a = kx\] We know that acceleration is the rate of change of velocity with respect to time, and that velocity is the rate of change of displacement with respect to time. So we can write \(a = \frac {\mathrm d^2 x}{\mathrm dt^2}\). (or as physicists may prefer, \(\ddot x\)) The constant of proportional is usually written \(-\omega^2\). (the utility of this will be seen when we cover SHM) So the above is most commonly seen as: \[\frac {\mathrm d^2 x}{\mathrm dt^2} = -\omega^2 x\] This is an example of a linear second-order differential equation, which we will look into solving in a later section.
Solving separable equations
Separable differential equations are those of the form: \[\frac {\mathrm dy} {\mathrm dx} = f(x) g(y)\] We can divide through \(g(y)\) to get: \[\frac 1 {g(y)} \frac {\mathrm dy} {\mathrm dx} = f(x)\] We can now integrate both sides with respect to \(x\): \[\int \frac 1 {g(y)} \frac {\mathrm dy} {\mathrm dx} \mathrm dx = \int f(x) \mathrm dx\] From integration by substitution, we know that: \[\int \frac 1 {g(y)} \frac {\mathrm dy} {\mathrm dx} \mathrm dx = \int \frac 1 {g(y)} \mathrm dy\] So effectively we are integrating the LHS with respect to \(y\) and the RHS with respect to \(x\). If we're lucky, we can now easily integrate \(\frac 1 {g(y)}\) and \(f(x)\), we can obtain a relationship between \(y\) and \(x\). If we're luckier yet, we may even be able to nicely write a solution in the form \(y = f\) where \(f\) depends only on \(x\). To take an example, let's look at: \[m \frac {\mathrm dv} {\mathrm dt} = mg - kv\] From earlier, we can divide through \(mg - kv\): \[\frac m {mg - kv} \frac {\mathrm dv} {\mathrm dt} = 1\] Which is in a form we can now solve as: \[\int \frac m {mg - kv} \mathrm dv = \frac m k \int \frac 1 {\frac {mg} k - v} \mathrm dv = - \frac m k \log \left({\frac {mg} k - v}\right)\] The RHS is simply \(t + C\) for some (as yet unknown) constant \(C\). Then: \[t + C = - \frac m k \log \left({\frac {mg} k - v}\right)\] This can be relatively straightforwardly rearranged for \(v\): \[-\frac k m t + C' = \log \left({\frac {mg} k - v}\right)\] (where \(C' = -\frac k m C\)) Giving: \[Ae^{-\frac k m t} = \frac {mg} k - v\] (where \(A = e^{C'}\)) So: \[v = \frac {mg} k - Ae^{-\frac k m t}\] We would need some more information about the body to find this constant \(A\), such as the body's velocity at a point in time. (this is called a boundary condition) We can see however, without that information, that as \(t \to \infty\), \(v \to \frac {mg} k\), giving us the terminal velocity of the body. We can also differentiate/integrate this function to obtain its acceleration or the displacement from its start point, in terms of \(t\), though this would introduce further constants which we would need even more information to determine. For example, if we knew that the body started from rest we would have: \[v(0) = 0 = \frac {mg} k - Ae^0 = \frac {mg} k - A\] So: \[A = \frac {mg} k\] Giving: \[v = \frac {mg} k - \frac {mg} k e^{-\frac k m t}\] Sometimes it won't be possible to separate the variables, for example if we had: \[\sin x \frac {\mathrm dy} {\mathrm dx} + y \cos x = e^x \sin x\] We would not be able to apply this technique, we must instead look at alternatives. We will look at these in the next section.
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