Finding some Solutions

Excerpts

• Finding some Solutions

  He who would attain that facility must work out examples, and more examples, and yet more examples, such as are found abundantly in all the regular treatises on the Calculus.

• Finding some Solutions

  Now the test of the matter is this. If the expression is an exact differential, it must be true that

• Finding some Solutions

  There is no one rule for discovering such an integrating factor; but experience will usually suggest one.

• Finding some Solutions

  The only function we know that has this property is the exponential function (see unchanged), and we may be certain therefore that the solution of the equation will be of that form.

• Finding some Solutions

  The beginner, who now knows how easy most of those processes are in themselves, will here begin to realize that integration is *an art*. As in all arts, so in this, facility can be acquired only by diligent and regular practice.

• Finding some Solutions

  Now, as to the $C$, its meaning depends on the initial value of $y$.

• Finding some Solutions

  This is indeed none other than the equation of an alternating electric current, where $g$ represents the amplitude of the electromotive force, $n$ the frequency, $a$ the resistance, $b$ the coefficient of self-induction of the circuit, and $\phi$ is an angle of lag.

• Finding some Solutions

  In this case the simplified equation represents the propagation of a wave (of any form) at a uniform speed along the $x$ direction.

• Finding some Solutions

  Now the mere inspection of this relation tells us that we have got to do with a case in which $\dfrac{dy}{dx}$ is proportional to $y$. If we think of the curve which will represent $y$ as a function of $x$, it will be such that its slope at any point will be proportional to the ordinate at that point, and will be a negative slope if $y$ is positive.

• Finding some Solutions

  As both $y$ and $dy$ occur in the equation and on opposite sides, we can do nothing until we get both $y$ and $dy$ to one side, and $dx$ to the other. To do this, we must split our usually inseparable companions $dy$ and $dx$ from one another.

• Finding some Solutions

  Now, as it stands, the left side is not integrable. But it can be made so by the artifice---and this is where skill and practice suggest a plan---of multiplying all the terms by $\epsilon^{\efrac{a}{b} t}$, giving us:

• Finding some Solutions

  It is possible in such cases to discover, however, *an integrating factor*, that is to say, a factor such that if both are multiplied by this factor, the expression will become an exact differential. There is no one rule for discovering such an integrating factor; but experience will usually suggest one.

• Finding some Solutions

  You have now been personally conducted over the frontiers into the enchanted land.

Equations

• Finding some Solutions

  ay + b \frac{dy}{dx} = 0

  The differential equation of Example 1, in which dy/dx is proportional to y.

• Finding some Solutions

  y = C \epsilon^{-\efrac{a}{b} x}

  The solution of Example 1: y equals C times the exponential of minus (a/b) x.

• Finding some Solutions

  ay+b\frac{dy}{dt} = g · \sin 2\pi nt

  The differential equation of Example 3, a forced equation with a sinusoidal right-hand side, describing an alternating electric circuit.

• Finding some Solutions

  y = g \left\{\frac{ a · \sin 2 \pi n t - 2 \pi n b · \cos 2 \pi nt}{ a^2 + 4 \pi^2 n^2 b^2}\right\}

  The steady-state solution of Example 3 before its amplitude and phase are simplified.

• Finding some Solutions

  \ds\int u dv = uv - \int v du

  The general formula for integrating a product by parts.

• Finding some Solutions

  M\, dx + N\, dy = 0

  The general first-order differential equation in differential form, to be tested for exactness.

• Finding some Solutions

  2 \frac{d^2y}{dt^2}\, \frac{dy}{dt} = \frac{d \left(\dfrac{dy}{dt}\right)^2}{dt}

  Multiplying by 2 dy/dt turns the left side into the exact derivative of (dy/dt) squared.

• Finding some Solutions

  y = y_0 \epsilon^{-\efrac{a}{b} x}

  Example 1's solution rewritten with the constant C identified as the starting value y_0 of y at x = 0.

• Finding some Solutions

  ay + b \frac{dy}{dx} = g

  The differential equation of Example 2, with a constant term g on the right.

• Finding some Solutions

  y = \frac{g}{a} + C\epsilon^{-\efrac{a}{b}x}

  The general solution of Example 2: y equals g/a plus C times the decaying exponential.

• Finding some Solutions

  y = \frac{g}{a} (1-\epsilon^{-\efrac{a}{b} x})

  The solution of Example 2 satisfying y = 0 when x = 0.

• Finding some Solutions

  y_{\text{max.}} = \dfrac{g}{a}

  The maximum value that y approaches as x grows indefinitely, in Example 2.

• Finding some Solutions

  y = y_{\text{max.}}(1-\epsilon^{-\efrac{a}{b} x})

  Example 2's solution written in terms of its maximum value y_max; y grows toward that maximum as x increases.

• Finding some Solutions

  \tan \phi = \dfrac{2 \pi n b}{ a}

  Definition of the angle of lag phi in the alternating-current solution of Example 3.

• Finding some Solutions

  \sin \phi = \frac{2 \pi nb}{\sqrt{a^2 + 4 \pi^2 n^2 b^2}}

  The sine of the lag angle phi, expressed in the circuit constants.

• Finding some Solutions

  \cos \phi = \frac{a}{\sqrt{a^2 + 4 \pi^2 n^2 b^2}}

  The cosine of the lag angle phi, expressed in the circuit constants.

• Finding some Solutions

  y = g \frac{\sin(2 \pi nt - \phi)}{\sqrt{a^2 + 4 \pi^2 n^2 b^2}}

  The steady solution of Example 3: the alternating current y is a sine wave lagging the electromotive force by the angle phi.

• Finding some Solutions

  \int u dv = uv - \int v du

  The rule of integration by parts, turning the integral of u dv into uv minus the integral of v du.

• Finding some Solutions

  \frac{dM}{dy} = \frac{dN}{dx}

  The test for an exact differential: M dx + N dy is exact only if dM/dy equals dN/dx.

• Finding some Solutions

  \frac{\partial U}{\partial x} = M

  If M dx + N dy is exact, M is the partial derivative of a common function U with respect to x.

• Finding some Solutions

  \frac{\partial U}{\partial y} = N

  If M dx + N dy is exact, N is the partial derivative of the common function U with respect to y.

• Finding some Solutions

  w = 2x^3y

  The function w whose differential is the exact expression 6x^2y dx + 2x^3 dy in Example 4.

• Finding some Solutions

  U = x^2 + 2x^3y + C

  The integrated common function U of the exact differential in Example 4, with constant C.

• Finding some Solutions

  \left(\frac{dy}{dt}\right)^2 + n^2 (y^2-C^2) = 0

  The first integral of Example 5's equation, obtained by multiplying by 2 dy/dt and integrating. Erratum flagged, not corrected: the chapter then writes dy/dt = -n sqrt(y^2 - C^2), but this relation gives dy/dt = -n sqrt(C^2 - y^2), so the radicand sign in the chapter's next line does not match.

• Finding some Solutions

  \frac{1}{\sqrt{C^2 - y^2}} = \frac{d (\arcsin \dfrac{y}{C})}{dy}

  The derivative of arcsin(y/C) with respect to y equals 1 over the square root of C squared minus y squared.

• Finding some Solutions

  y = A \sin nt + B \cos nt

  The solution of Example 5 written as a sum of sine and cosine terms with constants A and B.

• Finding some Solutions

  y = C \sin (nt + C_1)

  Equivalent form of Example 5's solution with amplitude C and constant angle C_1.

• Finding some Solutions

  \frac{dw}{dy} = \frac{1}{\sqrt{ y^2 + c^2}}

  The derivative of w = log(y + sqrt(y^2 + c^2)) with respect to y, used in Example 6.

• Finding some Solutions

  y + \sqrt{y^2 + c^2} = C \epsilon^{nx}

  Result (1) of Example 6: integrating gives y + sqrt(y^2 + c^2) as C times the growing exponential.

• Finding some Solutions

  -y + \sqrt{y^2 + c^2} = \dfrac{c^2}{C} \epsilon^{-nx}

  Result (2) of Example 6, obtained from (1) by multiplying through by the conjugate expression.

• Finding some Solutions

  y = \frac{1}{2} C \epsilon^{nx} - \frac{1}{2}\, \frac{c^2}{C} \epsilon^{-nx}

  Half the difference of results (1) and (2), giving y as a combination of growing and decaying exponentials.

• Finding some Solutions

  y = A \epsilon^{nx} + B \epsilon^{-nx}

  The solution of Example 6 as a sum of a growing and a dying exponential with constants A and B.

• Finding some Solutions

  b \frac{d^2y}{dt^2} + a \frac{dy}{dt} + gy = 0

  The damped second-order differential equation of Example 7, combining the forms of Examples 1 and 6.

• Finding some Solutions

  m = \frac{a}{2b}

  Definition of m used in the solution of Example 7.

• Finding some Solutions

  n = \sqrt{\frac{a^2}{4b^2}} - \frac{g}{b}

  Definition of n in Example 7 as printed. Erratum flagged, not corrected: the square root covers only a^2/(4b^2), so the printed expression differs from the usual sqrt(a^2/(4b^2) - g/b).

• Finding some Solutions

  y = (\epsilon^{-mt})(A \epsilon^{nt} + B \epsilon^{-nt})

  The solution of Example 7 for the damped equation, with m and n as defined there.

• Finding some Solutions

  \frac{d^2y}{dt^2} = a^2 \frac{d^2y}{dx^2}

  The wave equation of Example 8: the second derivative in time equals a^2 times the second derivative in space.

• Finding some Solutions

  y = F(x+at) + f(x-at)

  The general solution of the wave equation in Example 8, with F and f arbitrary functions.

• Finding some Solutions

  a = \sqrt{\frac{k}{m}}

  The velocity of propagation of a wave for the more general equation m d^2y/dt^2 = k d^2y/dx^2.

• Finding some Solutions

  m \frac{d^2y}{dt^2} = k\, \frac{d^2y}{dx^2}

  The more general wave equation of Example 8, whose propagation velocity is sqrt(k/m).

Problems

No exercises in this chapter.