Page:EB1911 - Volume 08.djvu/243

Another simple form is

dy/dx + y P＝Q,

where P, Q are functions of x only; this is known as the linear equation, since it contains y and dy/dx only to the first degree. If ƒPdx = u, we clearly have

${\frac {d}{dx}}(ye^{u})=e^{u}\left({\frac {dy}{dx}}+\mathrm {P} y\right)=e^{u}\mathrm {Q} ,$

so that y = e^−u(ƒe^uQdx + A) solves the equation, and is the only possible solution, A being an arbitrary constant. The rule for the solution of the linear equation is thus to multiply the equation by e^u, where u = ƒPdx.

A third simple and important form is that denoted by

y＝px + ƒ(p),

where p is an abbreviation for dy/dx; this is known as Clairaut’s form. By differentiation in regard to x it gives

$p=p+x{\frac {dp}{dx}}+f'(p){\frac {dp}{dx}},$

where

$f'(p)={\frac {d}{dp}}f(p);$

thus, either (i.) dp/dx＝ 0, that is, p is constant on the curve satisfying the differential equation, which curve is thus any one of the straight lines y = cx = ƒ(c), where c is an arbitrary constant, or else, (ii.) x + ƒ′(p) = 0; if this latter hypothesis be taken, and p be eliminated between x + ƒ′(p) = 0 and y = px + ƒ(p), a relation connecting x and y, not containing an arbitrary constant, will be found, which obviously represents the envelope of the straight lines y = cx + ƒ(c).

In general if a differential equation φ(x, y, dy/dx) = 0 be satisfied by any one of the curves F(x, y, c) = 0, where c is an arbitrary constant, it is clear that the envelope of these curves, when existent, must also satisfy the differential equation; for this equation prescribes a relation connecting only the co-ordinates x, y and the differential coefficient dy/dx, and these three quantities are the same at any point of the envelope for the envelope and for the particular curve of the family which there touches the envelope. The relation expressing the equation of the envelope is called a singular solution of the differential equation, meaning an isolated solution, as not being one of a family of curves depending upon an arbitrary parameter.

An extended form of Clairaut’s equation expressed by

y＝xF(p) + ƒ(p)

may be similarly solved by first differentiating in regard to p, when it reduces to a linear equation of which x is the dependent and p the independent variable; from the integral of this linear equation, and the original differential equation, the quantity p is then to be eliminated.

Other types of solvable differential equations of the first order are (1)

M dy/dx＝N,

where M, N are homogeneous polynomials in x and y, of the same order; by putting v = y/x and eliminating y, the equation becomes of the first type considered above, in v and x. An equation (aB≷bA)

(ax+by+c)dy/dx＝Ax+By+C

may be reduced to this rule by first putting x+h, y+k for x and y, and determining h, k so that ah+bk+c = 0, Ah+Bk+C = 0.

(2) An equation in which y does not explicitly occur,

ƒ(x, dy/dx)＝0,

may, theoretically, be reduced to the type dy/dx = F(x); similarly an equation F(y, dy/dx) = 0.

(3) An equation

ƒ(dy/dx, x, y)＝0,

which is an integral polynomial in dy/dx, may, theoretically, be solved for dy/dx, as an algebraic equation; to any root dy/dx = F₁(x, y) corresponds, suppose, a solution φ₁(x, y, c) = 0, where c is an arbitrary constant; the product equation φ₁(x, y, c)φ₂(x, y, c) . . . = 0, consisting of as many factors as there were values of dy/dx, is effectively as general as if we wrote φ₁(x, y, c₁)φ₂(x, y, c₂) . . . = 0; for, to evaluate the first form, we must necessarily consider the factors separately, and nothing is then gained by the multiple notation for the various arbitrary constants. The equation φ₁(x, y, c)φ₂(x, y, c) . . . = 0 is thus the solution of the given differential equation.

In all these cases there is, except for cases of singular solutions, one and only one arbitrary constant in the most general solution of the differential equation; that this must necessarily be so we may take as obvious, the differential equation being supposed to arise by elimination of this constant from the equation expressing its solution and the equation obtainable from this by differentiation in regard to x.

A further type of differential equation of the first order, of the form

dy/dx＝A + By + Cy ²

in which A, B, C are functions of x, will be briefly considered below under differential equations of the second order.

When we pass to ordinary differential equations of the second order, that is, those expressing a relation between x, y, dy/dx and d ²y/dx², the number of types for which the solution can be found by a known procedure is very considerably reduced. Consider the general linear equation

${\frac {d^{2}y}{dx^{2}}}+\mathrm {P} {\frac {dy}{dx}}+\mathrm {Q} =\mathrm {R} ,$

where P, Q, R are functions of x only. There is no method always effective; the main general result for such a linear equation is that if any particular function of x, say y ₁, can be discovered, for which

${\frac {d^{2}y_{1}}{dx^{2}}}+\mathrm {P} {\frac {dy_{1}}{dx}}+\mathrm {Q} y_{1}=0,$

then the substitution y = y ₁η in the original equation, with R on the right side, reduces this to a linear equation of the first order with the dependent variable dη/dx. In fact, if y = y ₁η we have

${\frac {dy}{dx}}=y_{1}{\frac {d\eta }{dx}}+\eta {\frac {dy_{1}}{dx}}$ and ${\frac {d^{2}y}{dx^{2}}}=y_{1}{\frac {d^{2}\eta }{dx^{2}}}+2{\frac {dy_{1}}{dx}}{\frac {d\eta }{dx}}+\eta {\frac {d^{2}y_{1}}{dx^{2}}},$

and thus

${\frac {d^{2}y}{dx^{2}}}+\mathrm {P} {\frac {dy}{dx}}+\mathrm {Q} =y_{1}{\frac {d^{2}\eta }{dx^{2}}}+\left(2{\frac {dy_{1}}{dx}}+\mathrm {P} y_{1}\right){\frac {d\eta }{dx}}+\left({\frac {d^{2}y_{1}}{dx^{2}}}+\mathrm {P} {\frac {dy_{1}}{dx}}+\mathrm {Q} y_{1}\right)\eta ;$

if then

${\frac {d^{2}y_{1}}{dx^{2}}}+\mathrm {P} {\frac {dy_{1}}{dx}}+\mathrm {Q} y_{1}=0,$

and z denote dη/dx, the original differential equation becomes

$y_{1}{\frac {dz}{dx}}+\left(2{\frac {dy_{1}}{dx}}\mathrm {P} y_{1}\right)=\mathrm {R} .$

From this equation z can be found by the rule given above for the linear equation of the first order, and will involve one arbitrary constant; thence y = y ₁ η = y ₁ ∫ zdx + Ay ₁, where A is another arbitrary constant, will be the general solution of the original equation, and, as was to be expected, involves two arbitrary constants.

The case of most frequent occurrence is that in which the coefficients P, Q are constants; we consider this case in some detail. If θ be a root of the quadratic equation θ² + θP + Q = 0, it can be at once seen that a particular integral of the differential equation with zero on the right side is y ₁ = e^θx. Supposing first the roots of the quadratic equation to be different, and φ to be the other root, so that φ + θ = −P, the auxiliary differential equation for z, referred to above, becomes dz/dx + (θ − φ)z = Re^−θx which leads to ze^(θ−φ) = B + ∫ Re^−θxdx, where B is an arbitrary constant, and hence to

$y=Ae^{\theta x}+e^{\theta x}\int \mathrm {B} e^{(\phi -\theta )x}dx+e^{\theta x}\int e^{(\phi -\theta )x}\int Re^{-\phi x}dxdx,$

or say to $y=\mathrm {A} e^{\theta x}+\mathrm {C} e^{\phi x}+\mathrm {U}$, where A, C are arbitrary constants and U is a function of x, not present at all when R = 0. If the quadratic equation θ² + Pθ + Q = 0 has equal roots, so that 2θ = −P, the auxiliary equation in z becomes ${\frac {dz}{dx}}=Re^{\theta x}$ giving $z=B+\int \mathrm {R} e^{\theta x}dx$, where B is an arbitrary constant, and hence

$y=(\mathrm {A} +\mathrm {B} x)e^{\theta x}+e^{\theta x}\iint \mathrm {R} e^{-\theta x}dxdx,$

or, say, $y=(\mathrm {A} +\mathrm {B} x)e^{\theta x}+U$, where A, B are arbitrary constants, and U is a function of x not present at all when R = 0. The portion $\mathrm {A} e^{\theta x}+\mathrm {B} e^{\phi x}$ or $(\mathrm {A} +\mathrm {B} x)e^{\theta x}$ of the solution, which is known as the complementary function, can clearly be written down at once by inspection of the given differential equation. The remaining portion U may, by taking the constants in the complementary function properly, be replaced by any particular solution whatever of the differential equation

${\frac {d^{2}y}{dx^{2}}}+\mathrm {P} {\frac {dy}{dx}}+\mathrm {Q} =\mathrm {R} ;$

for if u be any particular solution, this has a form

$\displaystyle u=A_{0}e^{\theta x}+B_{0}e^{\phi x}+U,$

or a form

$\displaystyle u=(A_{0}+B_{0}x)e^{\theta x}+U;$

thus the general solution can be written

$\displaystyle (A-A_{0})e^{\theta x}+(B-B_{0})e^{\theta x}+u,$ or $\displaystyle \{A-A_{0}+(B-B_{0})x\}e^{\theta x}+u,$

where A − A₀, B − B₀, like A, B, are arbitrary constants.

A similar result holds for a linear differential equation of any order, say

${\frac {d^{n}y}{dx^{n}}}+\mathrm {P} _{1}{\frac {d^{n-1}y}{dx^{n-1}}}+\ \ldots \ +\mathrm {P} _{n}=\mathrm {R} ,$

where P₁, P₂, . . . P_n are constants, and R is a function of x. If we form the algebraic equation θⁿ + P₁θⁿ⁻¹ + . . . + P_n = 0, and all the roots of this equation be different, say they are θ₁, θ₂, . . . θ_n, the general solution of the differential equation is

$\displaystyle y=A_{1}e^{\theta _{1}x}+A_{2}e^{\theta _{2}x}+\ ...\ +A_{n}e^{\theta _{n}x}+u,$

where A₁, A₂, . . . A_n are arbitrary constants, and u is any