FP2.5 - Second order differential equations

Learning objectives

—FP2.5.1 - Linear second order differentialThe linear second order differential The auxiliary equation may have real distinct, equal or d2y dy complex roots. f(x) will have one of the forms equation a + b + cy = f(x) dx2 dx k epx, A + Bx, p + qx + cx2 or m cos ωx + n sin ωx. where a, b and c are real constants Students should be familiar with the terms ‘complementary and the particular integral can be function’ and ‘particular integral’. found by inspection or trial.; Students should be able to solve equations of the form d2y + 4y = sin 2x. dx2.—FP2.5.2 - Reducible second order differential equationsDifferential equations reducible to the above types by means of a given substitution.

For a constant-coefficient equation $a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=f(x),\qquad a\ne0,$ the general solution is the sum of the complementary function (CF), which solves the homogeneous equation, and one particular integral (PI), which accounts for $f(x)$ : $y=\mathrm{CF}+\mathrm{PI}$ .

Roots of $am^2+bm+c=0$	Complementary function
Distinct real $m_1,m_2$	$Ae^{m_1x}+Be^{m_2x}$
Repeated real $m$	$(A+Bx)e^{mx}$
Complex $\alpha\pm i\beta$	$e^{\alpha x}(A\cos\beta x+B\sin\beta x)$

The auxiliary equation comes from trying $y=e^{mx}$ in the homogeneous equation. Each independent root solution contributes an arbitrary constant, so a second-order equation has two constants before conditions are imposed.

Form of $f(x)$	Trial form for the PI
$ke^{px}$	$\lambda e^{px}$
$A+Bx$	$\lambda x+\mu$
$p+qx+cx^2$	$\lambda x^2+\mu x+\nu$
$m\cos\omega x+n\sin\omega x$	$\lambda\cos\omega x+\mu\sin\omega x$

Differentiate the trial, substitute it into the full equation and compare coefficients to determine its unknowns. If any part of the trial is already in the CF, multiply the whole trial by $x$ ; use another factor of $x$ if the overlap is repeated.

For the official resonance model $y''+4y=\sin2x,$ the auxiliary roots are $\pm2i$ , so $\mathrm{CF}=A\cos2x+B\sin2x.$ A trial $C\cos2x+D\sin2x$ duplicates the CF and contributes zero. Use an $x$ factor instead: $y_p=Kx\cos2x$ . Since $(x\cos2x)''+4x\cos2x=-4\sin2x$ , $K=-\tfrac14$ . Hence $y=A\cos2x+B\sin2x-\frac14x\cos2x.$

Apply values of $y$ and $y'$ only after CF and PI have been combined, producing two simultaneous equations for $A$ and $B$ . A final substitution into the differential equation checks both the PI coefficients and any resonance factor.

Do not use the auxiliary equation on the non-homogeneous right side, and do not call the CF alone the general solution when $f(x)\ne0$ . A repeated root requires the factor $x$ in the CF; overlap between the PI trial and the CF also requires an extra factor $x$ . Keep the frequency $\omega$ , exponential rate $p$ and all derivative signs exact.

A given substitution can turn a variable-coefficient second-order equation into the constant-coefficient type just solved. The decisive step is to transform both derivatives with the chain rule before simplifying; every derivative must be taken with respect to the variable displayed in its denominator.

Given substitution	Required derivative relations
$t=g(x)$	$\dfrac{dy}{dx}=\dfrac{dy}{dt}g'(x)$ and $\dfrac{d^2y}{dx^2}=\dfrac{d^2y}{dt^2}[g'(x)]^2+\dfrac{dy}{dt}g''(x)$
$x=h(t)$	$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ and $\dfrac{d^2y}{dx^2}=\dfrac{y_{tt}x_t-y_tx_{tt}}{(x_t)^3}$
$y=q(x)v(x)$	$y'=q'v+qv'$ and $y''=q''v+2q'v'+qv''$

Differentiate the supplied relation, replace $y'$ , $y''$ and every explicit occurrence of the old variable, then cancel common non-zero factors using the stated domain. Check that the resulting equation is exactly the claimed constant-coefficient form. Solve it as CF + PI in the new variable, and finally back-substitute both the variable and any dependent-variable relation.

For $x^2y''+3xy'-3y=2\ln x,\qquad x>0,$ take the given substitution $t=\ln x$ . Then $y'=\frac1x y_t,\qquad y''=\frac1{x^2}(y_{tt}-y_t),$ so the equation becomes $y_{tt}+2y_t-3y=2t.$ Its auxiliary roots are $1$ and $-3$ . A PI $at+b$ gives $a=-\tfrac23$ and $b=-\tfrac49$ , hence $y=Ae^t+Be^{-3t}-\frac23t-\frac49.$ Returning to $x$ gives $y=Ax+Bx^{-3}-\frac23\ln x-\frac49,\qquad x>0.$

The transformed equation and the final answer can be checked independently: substitute the derivative relations into the original equation for the first check, then replace the final $y$ and its $x$ -derivatives in the original equation for the second.

Do not replace $x$ in the coefficients while leaving derivatives in the old variable. For $t=\ln x$ the condition $x>0$ is essential; for a square-root substitution, track the stated branch and domain. Constants $A$ and $B$ remain arbitrary through back-substitution, but the same letter must not be used ambiguously for both a coefficient and a new variable.

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—FP2.6 - Maclaurin and Taylor seriesRead

FP2.5 - Second order differential equations

Learning objectives

Solve a linear second-order differential equation

Reduce a second-order equation using a given substitution

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