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2.6 Differential equations

Syllabus
9231–2028–2029
Topic
2.6
Level
A2

Learning objectives

2.6.1Linear differential equations• find an integrating factor for a first order linear differential equation, and use an integrating factor to find the general solution e.g. x y yx2d d 2- =, x x y yxd d 4- =, coth coshx y yx xd d + =.2.6.2Linear differential equations• recall the meaning of the terms 'complementary function' and 'particular integral' in the context of linear differential equations, and recall that the general solution is the sum of the complementary function and a particular integral2.6.3Linear differential equations• find the complementary function for a first or second order linear differential equation with constant coefficients For second order equations, including the cases where the auxiliary equation has distinct real roots, a repeated real root or conjugate complex roots.2.6.4Linear differential equations• recall the form of, and find, a particular integral for a first or second order linear differential equation in the cases where a polynomial or a ebx or a cos px + b sin px is a suitable form, and in other simple cases find the appropriate coefficient(s) given a suitable form of particular integral e.g. evaluate k given that kx cos 2x is a particular integral of sinx y yx42d d 2 2 + =.2.6.5Differential substitutions• use a given substitution to reduce a differential equation to a first or second order linear equation with constant coefficients or to a first order equation with separable variables e.g. the substitution x = et to reduce to linear form a differential equation with terms of the form ax x y bx x y cyd d d d2 2 2 ++, or the substitution y = ux to reduce x y xy xy d d -= + to separable form.2.6.6Initial conditions• use initial conditions to find a particular solution to a differential equation, and interpret a solution in terms of a problem modelled by a differential equation.

The integrating factor turns three terms into one derivative

FirstwriteFirst write\frac{dy}{dx}+P(x)y=Q(x).ThenThen\mu(x)=e^{\int P(x),dx},\qquad \frac{d}{dx}(\mu y)=\mu Q,sosoy=\frac{1}{\mu}\left(\int\mu Q,dx+C\right).Divide by the original coefficient of $dy/dx$ before identifying $P$.

For $x>0$, solvey'+y\coth x=\cosh x.Since $\mu=e^{\int\coth x\,dx}=\sinh x$,\frac{d}{dx}(y\sinh x)=\sinh x\cosh x.HenceHencey\sinh x=\tfrac12\sinh^2x+C,\qquad y=\tfrac12\sinh x+C\operatorname{cosech}x.

Differentiate the final expression and substitute it into the standardised equation. The two C-terms cancel because the homogeneous part is exactly what the integrating factor method carries as the constant of integration.

The integrating factor multiplies every term. Its antiderivative constant is omitted because it only rescales the final arbitrary constant; domain restrictions such as x>0x>0 still matter when logarithms or coth appear.

The CF carries free behaviour; the PI carries the forcing

part equation it satisfies constants meaning
complementary function ycy_c L[yc]=0L[y_c]=0 contains the full arbitrary constants free/homogeneous behaviour
particular integral ypy_p L[yp]=f(x)L[y_p]=f(x) one chosen solution, no arbitrary constants response to forcing
general solution y=yc+ypy=y_c+y_p constants come from ycy_c every solution of L[y]=f(x)L[y]=f(x)

For a linear operator L, L[yc+yp]=L[yc]+L[yp]=0+f(x)L[y_c+y_p]=L[y_c]+L[y_p]=0+f(x). Any two particular solutions differ by a homogeneous solution, so adding the complete CF supplies every possible solution.

ForFory''-3y'+2y=e^{3x},the homogeneous roots are 1 and 2, so $y_c=C_1e^x+C_2e^{2x}$. Trying $y_p=Ae^{3x}$ gives $(9-9+2)Ae^{3x}=e^{3x}$, hence $A=1/2$. Thereforey=C_1e^x+C_2e^{2x}+\tfrac12e^{3x}.

A PI is not the general solution and must not contain a new arbitrary constant. Initial conditions act on the assembled CF + PI solution, not on the forcing trial alone.

Auxiliary roots determine the complete homogeneous basis

For a constant-coefficient homogeneous equation, try $y=e^{mx}$. A first-order equation $ay'+by=0$ gives $am+b=0$. A second-order equationay''+by'+cy=0givestheauxiliaryequationgives the auxiliary equationam^2+bm+c=0.

auxiliary roots complementary function
first-order root mm CemxCe^{mx}
distinct real m1,m2m_1,m_2 C1em1x+C2em2xC_1e^{m_1x}+C_2e^{m_2x}
repeated real mm (C1+C2x)emx(C_1+C_2x)e^{mx}
conjugate α±iβ\alpha\pm i\beta eαx(C1cos⁡βx+C2sin⁡βx)e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)

ForFory''-4y'+13y=0,m^2-4m+13=0\quad\Rightarrow\quad m=2\pm3i.HenceHencey_c=e^{2x}(C_1\cos3x+C_2\sin3x).

Two identical copies of emxe^{mx} are not independent when a root repeats; the second solution needs the factor x. Conjugate roots combine into a real sine-cosine form when the equation has real coefficients.

Match the forcing, then remove any overlap with the CF

forcing term initial PI trial
polynomial of degree nn general polynomial of degree nn
aebxae^{bx} AebxAe^{bx}
acos⁡px+bsin⁡pxa\cos px+b\sin px Acos⁡px+Bsin⁡pxA\cos px+B\sin px

If the trial overlaps the CF, multiply the whole trial by x; repeat with another x only if the overlap has higher multiplicity.

Differentiate the trial, substitute it into the differential equation, and equate coefficients of independent powers, exponentials, sine and cosine terms. The solved trial is one PI; the general solution is still CF + PI.

ForthesyllabuscaseFor the syllabus casey''+4y=\sin2x,sine and cosine of 2x already belong to the CF, so use the supplied form $y_p=kx\cos2x$. Theny_p''+4y_p=-4k\sin2x.Matching $\sin2x$ givesk=-\tfrac14,so one PI is $y_p=-\tfrac14x\cos2x$.

For sinusoidal forcing, include both sine and cosine in an unrestricted trial because differentiation mixes them. A resonant unmodified trial collapses into the homogeneous equation and cannot determine its coefficients.

A useful substitution must transform every derivative

For the given substitution $x=e^t$ (so $t=\ln x$, $x>0$),x\frac{dy}{dx}=\frac{dy}{dt},\qquad x^2\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}-\frac{dy}{dt}.ThereforeThereforeax^2y''+bxy'+cy=F(x)becomesbecomesa\ddot y+(b-a)\dot y+cy=F(e^t),aconstant−coefficientequationwhenthetransformedrightsidepermitsit.a constant-coefficient equation when the transformed right side permits it.

For a given homogeneous substitution $y=ux$,\frac{dy}{dx}=u+x\frac{du}{dx}.InIn\frac{dy}{dx}=\frac{x+y}{x-y},thisgivesthis givesu+xu'=\frac{1+u}{1-u},\qquad x\frac{du}{dx}=\frac{1+u^2}{1-u}.

Nowseparateandintegrate:Now separate and integrate:\frac{1-u}{1+u^2},du=\frac{dx}{x},\tan^{-1}u-\tfrac12\ln(1+u^2)=\ln|x|+C.Finally substitute $u=y/x$ and retain any domain restrictions or separately lost constant solutions.

A substitution is not a relabelling. Derive the first and second derivative transformations before replacing terms, and inspect any division by x, u or another expression for excluded solutions.

Initial conditions select one trajectory; the model decides what it means

First obtain the full general solution CF + PI. Differentiate it as many times as the conditions require, substitute all conditions at their stated input value, and solve the simultaneous equations for every arbitrary constant. A second-order equation normally needs two independent conditions.

SupposeSupposey=3+C_1e^{-t}+C_2e^{-2t},\qquad y(0)=5,\quad y'(0)=-6.ThenThenC_1+C_2=2,\qquad -C_1-2C_2=-6,so $C_1=-2$ and $C_2=4$. The particular solution isy=3-2e^{-t}+4e^{-2t}.

Interpret the variables and units; restrict the independent variable to the model's stated domain; check whether quantities that represent size, mass or population remain admissible; and compare dominant terms for long-term behaviour. Here both exponentials decay, so the model approaches the equilibrium value 3 as t→∞t\to\infty.

An exact algebraic solution is not automatically a valid real-world prediction outside the model's interval or assumptions. Do not discard transient terms before applying the initial conditions: they are what allow the trajectory to match the starting state.