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Chapter 5 · Continuity and Differentiability

Exercise 5.7 · Q14: y=Aemx+Benxy=Ae^{mx}+Be^{nx} satisfies y′′−(m+n)y′+mny=0y''-(m+n)y'+mny=0

Exercise 5.7Second order derivative · Section 5.7
The question

14. If y=Aemx+Benxy = Ae^{mx} + Be^{nx}⁠, show that d2ydx2−(m+n)dydx+mny=0\dfrac{d^2y}{dx^2} - (m+n)\dfrac{dy}{dx} + mny = 0⁠.

Exam answer

Here y=Aemx+Benxy = Ae^{mx} + Be^{nx}⁠.

dydx=mAemx+nBenxd2ydx2=m2Aemx+n2Benx\begin{aligned} \frac{dy}{dx} &= mAe^{mx} + nBe^{nx} \\ \frac{d^2y}{dx^2} &= m^2Ae^{mx} + n^2Be^{nx} \end{aligned}

Substituting, the LHS is

m2Aemx+n2Benx−(m+n)(mAemx+nBenx)+mn(Aemx+Benx)=Aemx(m2−m2−mn+mn)+Benx(n2−mn−n2+mn)\begin{aligned} &m^2Ae^{mx} + n^2Be^{nx} \\ &\quad - (m + n)(mAe^{mx} \\ &\qquad + nBe^{nx}) \\ &\quad + mn(Ae^{mx} + Be^{nx}) \\ &= Ae^{mx}(m^2 - m^2 - mn + mn) \\ &\quad + Be^{nx}(n^2 - mn - n^2 + mn) \end{aligned}

∴ LHS=0=RHS\therefore\ \text{LHS} = 0 = \text{RHS}⁠. Hence proved.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Each differentiation of ekxe^{kx} just multiplies it by kk⁠. So in the left side the AemxAe^{mx} part is AemxAe^{mx} times

m2−(m+n)m+mn=0,m^2 - (m+n)m + mn = 0,

and the BenxBe^{nx} part is BenxBe^{nx} times n2−(m+n)n+mn=0n^2 - (m+n)n + mn = 0⁠. Both numbers are zero, because mm and nn are exactly the two roots of k2−(m+n)k+mn=(k−m)(k−n)k^2 - (m+n)k + mn = (k-m)(k-n)⁠.

Given, to show, and the plan
Given
y=Aemx+Benxy = Ae^{mx} + Be^{nx}⁠, where AA⁠, BB⁠, mm⁠, nn are constants. It is defined and differentiable for every real xx⁠.
To show
d2ydx2−(m+n)dydx+mny=0\dfrac{d^2y}{dx^2} - (m+n)\dfrac{dy}{dx} + mny = 0 for every real xx⁠.
Plan
(1) Find dydx\frac{dy}{dx} and d2ydx2\frac{d^2y}{dx^2} by the chain rule. (2) Substitute into the left side, group the emxe^{mx} terms and the enxe^{nx} terms, and simplify each coefficient to 00⁠.

Basics you need

1Derivative of ekxe^{kx}

ddxex=ex\frac{d}{dx}e^x = e^x (Theorem 5, Section 5.4). By the chain rule with inner function u=kxu = kx (kk a constant, dudx=k\frac{du}{dx} = k⁠):

ddxekx=ekx⋅k=k ekx.\frac{d}{dx}e^{kx} = e^{kx}\cdot k = k\,e^{kx}.

Basics: exponentials and logarithms

2Sum rule and constant multiples

(u+v)′=u′+v′(u + v)' = u' + v' (Section 5.3, rule (1)) and (cu)′=cu′(cu)' = cu' for a constant cc (Class XI, Limits and Derivatives). Basics: derivative rules

3Second order derivative (Section 5.7)

d2ydx2=ddx(dydx)\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)⁠. Lesson: second order derivative

Solution

  1. First derivative

    Term by term, with Basic 1 (k=mk = m for the first term, k=nk = n for the second):

    dydx=mAemx+nBenx.\frac{dy}{dx} = mAe^{mx} + nBe^{nx}.
    The constants AA and BB stay in front. Each exponential is differentiated by the chain rule, which brings down the constant in its exponent.
    Basic 1; Section 5.3, rule (1). Lesson: chain rule

  2. Second derivative

    Differentiate again in the same way:

    d2ydx2=m2Aemx+n2Benx.\frac{d^2y}{dx^2} = m^2Ae^{mx} + n^2Be^{nx}.
    mAemxmAe^{mx} is a constant mAmA times emxe^{mx}⁠, so its derivative is mA⋅m emx=m2AemxmA\cdot m\,e^{mx} = m^2Ae^{mx}⁠. Likewise for the other term.
    Section 5.7. Lesson: second order derivative

  3. Substitute into the left side

    d2ydx2−(m+n)dydx+mny=(m2Aemx+n2Benx)−(m+n)(mAemx+nBenx)+mn(Aemx+Benx).\begin{aligned} &\frac{d^2y}{dx^2} - (m+n)\frac{dy}{dx} + mny \\ &= \big(m^2Ae^{mx} + n^2Be^{nx}\big) \\ &\quad - (m+n)\big(mAe^{mx} + nBe^{nx}\big) \\ &\quad + mn\big(Ae^{mx} + Be^{nx}\big). \end{aligned}
    We now compute the left side completely. Brackets keep each derivative together so that the factors −(m+n)-(m+n) and mnmn multiply every term inside.

  4. Group the AemxAe^{mx} terms and the BenxBe^{nx} terms

    =Aemx[m2−(m+n)m+mn]+Benx[n2−(m+n)n+mn].\begin{aligned} &= Ae^{mx}\big[m^2 - (m+n)m + mn\big] \\ &\quad + Be^{nx}\big[n^2 - (m+n)n + mn\big]. \end{aligned}

    Expand each bracket:

    m2−(m+n)m+mn=m2−m2−mn+mn=0,n2−(m+n)n+mn=n2−mn−n2+mn=0.\begin{aligned} m^2 - (m+n)m + mn &= m^2 - m^2 - mn + mn = 0, \\ n^2 - (m+n)n + mn &= n^2 - mn - n^2 + mn = 0. \end{aligned}

    So the left side is Aemx⋅0+Benx⋅0=0Ae^{mx}\cdot 0 + Be^{nx}\cdot 0 = 0⁠, which is the right side. The relation holds for every real xx and for all constants AA⁠, BB⁠, mm⁠, nn⁠.

    Grouping by the two exponentials turns a long expression into two small brackets of constants. Each bracket is zero identically, whatever mm and nn are, so no condition on xx or on the constants is needed. In factor form the brackets are (m−m)(m−n)(m-m)(m-n) and (n−m)(n−n)(n-m)(n-n)⁠.
    Expanding brackets: (m+n)m=m2+mn(m+n)m = m^2 + mn and (m+n)n=mn+n2(m+n)n = mn + n^2⁠.
Result

dydx=mAemx+nBenx\dfrac{dy}{dx} = mAe^{mx} + nBe^{nx}⁠, d2ydx2=m2Aemx+n2Benx\dfrac{d^2y}{dx^2} = m^2Ae^{mx} + n^2Be^{nx}⁠, and for every real xx

d2ydx2−(m+n)dydx+mny=0.\frac{d^2y}{dx^2} - (m+n)\frac{dy}{dx} + mny = 0.

(The answer key gives no entry for this "show that" question.)

Check it with numbers

Take A=2A = 2⁠, B=−1B = -1⁠, m=1m = 1⁠, n=3n = 3⁠, so m+n=4m + n = 4⁠, mn=3mn = 3 and the relation reads y′′−4y′+3y=0y'' - 4y' + 3y = 0⁠.

At x=0x = 0 (where e0=1e^0 = 1⁠): y=2−1=1y = 2 - 1 = 1⁠, y′=2−3=−1y' = 2 - 3 = -1⁠, y′′=2−9=−7y'' = 2 - 9 = -7⁠, and

−7−4(−1)+3(1)=−7+4+3=0.-7 - 4(-1) + 3(1) = -7 + 4 + 3 = 0.

At x=0.5x = 0.5: e0.5≈1.6487213e^{0.5} \approx 1.6487213⁠, e1.5≈4.4816891e^{1.5} \approx 4.4816891⁠, so y≈−1.1842465y \approx -1.1842465⁠, y′≈−10.1476247y' \approx -10.1476247⁠, y′′≈−37.0377591y'' \approx -37.0377591⁠, and

−37.0377591+40.5904987−3.5527396≈0.\begin{aligned} &-37.0377591 + 40.5904987 \\ &\quad - 3.5527396 \approx 0. \end{aligned}
Common mistakes
  1. Forgetting the chain-rule factor. ddxemx=memx\frac{d}{dx}e^{mx} = me^{mx}⁠, not emxe^{mx}⁠.
  2. Multiplying only the first term by −(m+n)-(m+n)⁠. The factor multiplies the whole of dydx\frac{dy}{dx}⁠; use brackets.
  3. Treating mm⁠, nn as variables. They are constants; only xx varies.
What to take away
  1. ekxe^{kx} satisfies y′′−(m+n)y′+mny=0y'' - (m+n)y' + mny = 0 exactly when k2−(m+n)k+mn=0k^2 - (m+n)k + mn = 0⁠, that is k=mk = m or k=nk = n⁠. So any combination Aemx+BenxAe^{mx} + Be^{nx} works.
  2. Example 37 is the case m=2m = 2⁠, n=3n = 3⁠, A=3A = 3⁠, B=2B = 2⁠: y′′−5y′+6y=0y'' - 5y' + 6y = 0⁠. Question 15 is the case m=7m = 7⁠, n=−7n = -7⁠, A=500A = 500⁠, B=600B = 600⁠: there m+n=0m + n = 0 and mn=−49mn = -49⁠, so the relation becomes y′′=49yy'' = 49y⁠.

Every fact used, and where it comes from

FactUsed inSource
ddxex=ex\frac{d}{dx}e^x = e^xSteps 1, 2Theorem 5, Section 5.4 (Lesson: exponential and logarithmic functions)
Chain rule: ddxekx=kekx\frac{d}{dx}e^{kx} = ke^{kx}Steps 1, 2Theorem 4, Section 5.3.1 (Lesson: chain rule)
Sum rule and constant multiplesSteps 1, 2Section 5.3, rule (1) (sum); Class XI, Limits and Derivatives (constant multiple) (Basics: derivative rules)
d2ydx2=ddx(dydx)\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)Step 2Section 5.7 (Lesson: second order derivative)
Similar relation for 3e2x+2e3x3e^{2x} + 2e^{3x}Take-awayExample 37, Section 5.7
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