Class 12 Maths, Worked in Full Past papersGet updates
Chapter 9 · Differential Equations

Exercise 9.5 · Q2: Linear: dydx+3y=e−2x\frac{dy}{dx} + 3y = e^{-2x}

Exercise 9.5Linear differential equations · Section 9.4.3
The question

For each of the differential equations given in Exercises 1 to 12, find the general solution:

2. dydx+3y=e−2x\dfrac{dy}{dx} + 3y = e^{-2x}

Exam answer

This is of the form dydx+Py=Q\frac{dy}{dx} + Py = Q⁠, where P=3P = 3 and Q=e−2xQ = e^{-2x}⁠.

I.F.=e∫3 dx=e3x\text{I.F.} = e^{\int 3\,dx} = e^{3x}

So the solution is given by

ye3x=∫e−2x⋅e3x dx+C=∫ex dx+C=ex+C\begin{aligned} ye^{3x} &= \int e^{-2x}\cdot e^{3x}\,dx + C \\ &= \int e^{x}\,dx + C \\ &= e^{x} + C \end{aligned}

Dividing by e3xe^{3x}⁠,

∴\therefore y=e−2x+Ce−3xy = e^{-2x} + Ce^{-3x} is the general solution.

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

The one idea this problem is built on

The equation is linear with the constant P=3P = 3⁠, so the integrating factor is e3xe^{3x}⁠. Multiplying by it makes the left side ddx(ye3x)\frac{d}{dx}\left(ye^{3x}\right)⁠, and on the right the exponentials combine: e−2x⋅e3x=exe^{-2x}\cdot e^{3x} = e^{x}⁠, which integrates at once.

Given, to find, and the plan
Given
dydx+3y=e−2x\frac{dy}{dx} + 3y = e^{-2x}⁠, for all real xx⁠.
To find
The general solution (one arbitrary constant).
Plan
(1) Read off PP and QQ⁠. (2) I.F. =e∫P dx= e^{\int P\,dx}⁠. (3) Multiply and see the product rule. (4) Integrate, with one constant. (5) Solve for yy⁠. (6) Check by substitution.

Basics you need

1First order linear equations

dydx+Py=Q\frac{dy}{dx} + Py = Q⁠, where PP and QQ are constants or functions of xx only. Its solution is y (I.F.)=∫(Q×I.F.) dx+Cy\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C⁠, where I.F.=e∫P dx\text{I.F.} = e^{\int P\,dx} (Section 9.4.3). Lesson: linear equations, the integrating factor

2Why the integrating factor works

If g′=Pgg' = Pg⁠, then g y′+Pg y=(yg)′g\,y' + Pg\,y = (yg)' by the product rule; g=e3xg = e^{3x} has g′=3gg' = 3g⁠. Basics: (y · g)′ = g(y′ + Py)

3Laws of exponents

ea⋅eb=ea+be^a\cdot e^b = e^{a + b}⁠, so e−2x⋅e3x=exe^{-2x}\cdot e^{3x} = e^{x} and ex⋅e−3x=e−2xe^{x}\cdot e^{-3x} = e^{-2x}⁠. Basics: exponentials

4∫ex dx=ex+C\int e^x\,dx = e^x + C

Section 7.2, formula (xi). Basics: every method ends with an integral

Solution

  1. Recognise the linear form: P=3P = 3⁠, Q=e−2xQ = e^{-2x}

    dydx+(3) y=e−2x.\frac{dy}{dx} + (3)\,y = e^{-2x}.
    The coefficient of dydx\frac{dy}{dx} is 1, yy appears to the first power, P=3P = 3 is a constant and Q=e−2xQ = e^{-2x} depends on xx only: this is the standard linear form.
    Section 9.4.3, step (i). Lesson: linear equations

  2. The integrating factor: e3xe^{3x}

    I.F.=e∫P dx=e∫3 dx=e3x.\text{I.F.} = e^{\int P\,dx} = e^{\int 3\,dx} = e^{3x}.
    One antiderivative of P=3P = 3 is 3x3x⁠. No constant is needed: it would only multiply the I.F. by a nonzero constant.
    Section 9.4.3, step (ii). Basics: exponentials

  3. Multiply by e3xe^{3x}⁠: the left side is one derivative

    e3xdydx+3e3xy=e−2x⋅e3xddx(ye3x)=ex.\begin{aligned} e^{3x}\frac{dy}{dx} + 3e^{3x}y &= e^{-2x}\cdot e^{3x} \\ \frac{d}{dx}\left(ye^{3x}\right) &= e^{x}. \end{aligned}
    By the product rule, ddx(ye3x)=e3xdydx+y⋅3e3x\frac{d}{dx}\left(ye^{3x}\right) = e^{3x}\frac{dy}{dx} + y\cdot 3e^{3x}⁠, the left side. On the right, add the exponents: −2x+3x=x-2x + 3x = x⁠.
    Product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

  4. Integrate both sides, with one constant

    ye3x=∫ex dx+C=ex+C.ye^{3x} = \int e^{x}\,dx + C = e^{x} + C.
    This is y (I.F.)=∫(Q×I.F.) dx+Cy\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C⁠, with Q×I.F.=exQ\times\text{I.F.} = e^x⁠. The CC is the one arbitrary constant of the general solution.
    Section 9.4.3, step (iii); Section 7.2, formula (xi). Basics: one constant for order 1

  5. Solve for yy

    y=ex⋅e−3x+Ce−3x=e−2x+Ce−3x.y = e^{x}\cdot e^{-3x} + Ce^{-3x} = e^{-2x} + Ce^{-3x}.
    Multiply both sides by e−3xe^{-3x} (never zero), and add exponents: x−3x=−2xx - 3x = -2x⁠. Every term gets the factor, so CC becomes Ce−3xCe^{-3x}⁠. The solution holds for all real xx⁠.
    Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

  6. Check by substituting back

    From y=e−2x+Ce−3xy = e^{-2x} + Ce^{-3x}⁠: dydx=−2e−2x−3Ce−3x\frac{dy}{dx} = -2e^{-2x} - 3Ce^{-3x}⁠, so

    dydx+3y=−2e−2x−3Ce−3x+3e−2x+3Ce−3x=e−2x,\begin{aligned} \frac{dy}{dx} + 3y &= -2e^{-2x} - 3Ce^{-3x} \\ &\quad + 3e^{-2x} + 3Ce^{-3x} \\ &= e^{-2x}, \end{aligned}

    for every CC⁠.

    The Ce−3xCe^{-3x} part solves dydx+3y=0\frac{dy}{dx} + 3y = 0 on its own, and e−2xe^{-2x} supplies the right side: −2e−2x+3e−2x=e−2x-2e^{-2x} + 3e^{-2x} = e^{-2x}⁠.
    A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions
Result

The general solution is y=e−2x+Ce−3xy = e^{-2x} + Ce^{-3x}⁠, where CC is an arbitrary constant, for all real xx⁠. (Textbook answer: y=e−2x+Ce−3xy = e^{-2x} + Ce^{-3x}⁠.)

Check it with numbers

Take C=1C = 1 and x=1x = 1⁠: e−2≈0.1353353e^{-2} \approx 0.1353353 and e−3≈0.0497871e^{-3} \approx 0.0497871⁠, so y≈0.1851224y \approx 0.1851224⁠. The equation says dydx=e−2x−3y≈−0.4200318\frac{dy}{dx} = e^{-2x} - 3y \approx -0.4200318⁠.

The curve itself: y(1.001)≈0.1847028141y(1.001) \approx 0.1847028141 and y(0.999)≈0.1855428785y(0.999) \approx 0.1855428785⁠, so

0.1847028141−0.18554287850.002≈−0.42003.\begin{aligned} &\frac{0.1847028141 - 0.1855428785}{0.002} \\ &\approx -0.42003. \end{aligned}
Common mistakes
  1. Multiplying exponents instead of adding them: e−2x⋅e3x=exe^{-2x}\cdot e^{3x} = e^{x}⁠, not e−6x2e^{-6x^2}⁠.
  2. Taking the I.F. as e−2xe^{-2x} (from QQ⁠); the I.F. comes from PP only.
  3. Writing y=e−2x+Cy = e^{-2x} + C: when dividing by e3xe^{3x}⁠, the constant is divided too.
What to take away
  1. The I.F. depends on PP only; QQ only enters the integral.
  2. With exponential PP and QQ⁠, the product Q×I.F.Q\times\text{I.F.} is often a single exponential.
  3. The answer is "one particular solution + CC times the I.F.'s reciprocal": here e−2x+Ce−3xe^{-2x} + Ce^{-3x}⁠.

Every fact used, and where it comes from

FactUsed inSource
Linear equation; I.F. =e∫P dx= e^{\int P\,dx}⁠; y (I.F.)=∫(Q×I.F.) dx+Cy\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + CSteps 1, 2, 4Section 9.4.3, steps (i) to (iii) (Lesson)
Product rule; ddxekx=kekx\frac{d}{dx}e^{kx} = ke^{kx}Steps 3, 6Chapter 5, Sections 5.3 and 5.4 (Basics)
ea⋅eb=ea+be^a\cdot e^b = e^{a + b}Steps 3, 5Laws of exponents (Basics)
∫ex dx=ex+C\int e^x\,dx = e^x + CStep 4Section 7.2, formula (xi) (Basics)
Substituting back into the equationStep 6Section 9.3 (Lesson)
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →