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Chapter 9 · Differential Equations

Exercise 9.3 · Q9: General solution of dydx=sin⁡−1x\frac{dy}{dx} = \sin^{-1}x

Exercise 9.3Variables separable · Section 9.4.1
The question

For each of the differential equations in Exercises 1 to 10, find the general solution:

9. dydx=sin⁡−1x\dfrac{dy}{dx} = \sin^{-1}x

Exam answer

Separating the variables,

dy=sin⁡−1x dxdy = \sin^{-1}x\,dx

Integrating both sides, by parts with sin⁡−1x\sin^{-1}x as the first function,

y=∫sin⁡−1x⋅1 dx=xsin⁡−1x−∫x dx1−x2\begin{aligned} y &= \int\sin^{-1}x\cdot 1\,dx \\ &= x\sin^{-1}x - \int\frac{x\,dx}{\sqrt{1 - x^2}} \end{aligned}

Put 1−x2=t1 - x^2 = t⁠. Then −2x dx=dt-2x\,dx = dt⁠.

∫x dx1−x2=−12∫dtt=−t=−1−x2\begin{aligned} \int\frac{x\,dx}{\sqrt{1 - x^2}} &= -\frac{1}{2}\int\frac{dt}{\sqrt{t}} \\ &= -\sqrt{t} = -\sqrt{1 - x^2} \end{aligned}

∴\therefore y=xsin⁡−1x+1−x2+Cy = x\sin^{-1}x + \sqrt{1 - x^2} + C is the general solution.

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

The one idea this problem is built on

The slope depends on xx alone, so yy is just ∫sin⁡−1x dx\int\sin^{-1}x\,dx⁠. There is no standard formula for that integral, but writing it as sin⁡−1x⋅1\sin^{-1}x\cdot 1 and integrating by parts swaps sin⁡−1x\sin^{-1}x for its derivative 11−x2\frac{1}{\sqrt{1 - x^2}}⁠, and what is left is a simple substitution.

Given, to find, and the plan
Given
dydx=sin⁡−1x\frac{dy}{dx} = \sin^{-1}x⁠, defined for −1≤x≤1-1 \le x \le 1⁠.
To find
The general solution, with one arbitrary constant.
Plan
(1) Separate: dy=sin⁡−1x dxdy = \sin^{-1}x\,dx⁠. (2) Integrate by parts, sin⁡−1x\sin^{-1}x first, 1 second. (3) Finish the remaining integral by t=1−x2t = 1 - x^2⁠. (4) Add one constant. (5) Check by differentiating back.

Basics you need

1Variables separable

If dydx=g(x) h(y)\frac{dy}{dx} = g(x)\,h(y) with h(y)≠0h(y) \ne 0⁠, then ∫dyh(y)=∫g(x) dx\int\frac{dy}{h(y)} = \int g(x)\,dx (Section 9.4.1). Here h(y)=1h(y) = 1⁠. Lesson: separate, then integrate

2Integration by parts

For a first function ff and a second function gg (Section 7.6),

∫f(x)g(x) dx=f(x)∫g(x) dx−∫[f′(x)∫g(x) dx]dx.\begin{aligned} &\int f(x)g(x)\,dx \\ &= f(x)\int g(x)\,dx \\ &\quad - \int\left[f'(x)\int g(x)\,dx\right]dx. \end{aligned}

Remark (iii) of Section 7.6 takes an inverse trigonometric function as the first function (the memory aid ILATE says the same: inverse trigonometric before algebraic). Basics: every method ends with an integral

3Derivative of sin⁡−1x\sin^{-1}x

ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}} for −1<x<1-1 < x < 1 (Chapter 5). Basics: y′ is the slope

4One arbitrary constant

A first order equation has one arbitrary constant in its general solution. Basics: order n gives n constants

Solution

  1. Separate: dydy in terms of dxdx

    Multiply both sides by dxdx⁠:

    dy=sin⁡−1x dx.dy = \sin^{-1}x\,dx.
    In the book's form dydx=g(x)⋅1\frac{dy}{dx} = g(x)\cdot 1⁠, with h(y)=1h(y) = 1⁠, never 0: nothing to divide by, no solution lost. sin⁡−1x\sin^{-1}x needs −1≤x≤1-1 \le x \le 1⁠; we work on −1<x<1-1 < x < 1⁠, where the derivatives below exist.
    Section 9.4.1: separate, then integrate. Lesson: separate, then integrate

  2. Integrate by parts: sin⁡−1x\sin^{-1}x first, 1 second

    Write the integrand as sin⁡−1x⋅1\sin^{-1}x\cdot 1⁠. With f=sin⁡−1xf = \sin^{-1}x⁠, g=1g = 1⁠, ∫g dx=x\int g\,dx = x⁠, f′=11−x2f' = \frac{1}{\sqrt{1 - x^2}}⁠:

    y=∫sin⁡−1x⋅1 dx=xsin⁡−1x−∫x1−x2 dx.\begin{aligned} y &= \int\sin^{-1}x\cdot 1\,dx \\ &= x\sin^{-1}x - \int\frac{x}{\sqrt{1 - x^2}}\,dx. \end{aligned}
    We cannot integrate sin⁡−1x\sin^{-1}x directly, but we can differentiate it. By parts moves the derivative onto sin⁡−1x\sin^{-1}x⁠, turning it into the algebraic 11−x2\frac{1}{\sqrt{1 - x^2}}⁠. ILATE agrees: the Inverse trigonometric function comes before the Algebraic one (the 1).
    Section 7.6, integration by parts. Basics: every method ends with an integral

  3. The remaining integral: substitute t=1−x2t = 1 - x^2

    Put t=1−x2t = 1 - x^2⁠, so dt=−2x dxdt = -2x\,dx and x dx=−12 dtx\,dx = -\frac{1}{2}\,dt⁠:

    ∫x dx1−x2=−12∫t−12 dt=−12⋅t1212=−t=−1−x2.\begin{aligned} \int\frac{x\,dx}{\sqrt{1 - x^2}} &= -\frac{1}{2}\int t^{-\frac{1}{2}}\,dt = -\frac{1}{2}\cdot\frac{t^{\frac{1}{2}}}{\frac{1}{2}} \\ &= -\sqrt{t} = -\sqrt{1 - x^2}. \end{aligned}

    Check by differentiating: ddx(−1−x2)=−−2x21−x2=x1−x2\frac{d}{dx}\left(-\sqrt{1 - x^2}\right) = -\frac{-2x}{2\sqrt{1 - x^2}} = \frac{x}{\sqrt{1 - x^2}}⁠.

    The top x dxx\,dx is, up to the factor −12-\frac{1}{2}⁠, the differential of 1−x21 - x^2 under the root, so the substitution leaves a plain power of tt⁠, done by the power rule with n=−12n = -\frac{1}{2}⁠. Here t=1−x2>0t = 1 - x^2 > 0 for −1<x<1-1 < x < 1⁠.
    Section 7.3.1 (substitution) and Section 7.2, formula (i). Basics: every method ends with an integral

  4. Put it together, with one constant

    y=xsin⁡−1x−(−1−x2)+C=xsin⁡−1x+1−x2+C.\begin{aligned} y &= x\sin^{-1}x - \left(-\sqrt{1 - x^2}\right) + C \\ &= x\sin^{-1}x + \sqrt{1 - x^2} + C. \end{aligned}
    Subtracting −1−x2-\sqrt{1 - x^2} gives +1−x2+\sqrt{1 - x^2}⁠. All constants of integration (from ∫dy\int dy⁠, from the by-parts step and from the substitution) combine into one arbitrary CC⁠.
    Section 9.3: one arbitrary constant. Basics: order n gives n constants

  5. Check: differentiate the answer back

    By the product rule and Step 3,

    dydx=sin⁡−1x+x1−x2−x1−x2=sin⁡−1x.\begin{aligned} \frac{dy}{dx} &= \sin^{-1}x + \frac{x}{\sqrt{1 - x^2}} - \frac{x}{\sqrt{1 - x^2}} \\ &= \sin^{-1}x. \end{aligned}
    The two fractions cancel and the given equation comes back, for every CC and every xx in −1<x<1-1 < x < 1⁠. So every member of the family is a solution (Section 9.3).
    Product rule; derivative of sin⁡−1x\sin^{-1}x (Chapter 5). Basics: y′ is the slope
Result

The general solution is y=xsin⁡−1x+1−x2+Cy = x\sin^{-1}x + \sqrt{1 - x^2} + C⁠, where CC is an arbitrary constant, for −1<x<1-1 < x < 1⁠. (Textbook answer: y=xsin⁡−1x+1−x2+Cy = x\sin^{-1}x + \sqrt{1 - x^2} + C⁠.)

Check it with numbers

Take C=0C = 0⁠. At x=0.5x = 0.5⁠, y≈1.1278247916y \approx 1.1278247916⁠, and the equation gives the slope sin⁡−10.5=π6≈0.52359878\sin^{-1}0.5 = \frac{\pi}{6} \approx 0.52359878⁠.

The curve itself: y(0.501)≈1.1283489678y(0.501) \approx 1.1283489678 and y(0.499)≈1.1273017700y(0.499) \approx 1.1273017700⁠, so

1.1283489678−1.12730177000.002≈0.52360.\begin{aligned} &\frac{1.1283489678 - 1.1273017700}{0.002} \\ &\approx 0.52360. \end{aligned}
Common mistakes
  1. Taking 1 as the first function. Then you would need ∫sin⁡−1x dx\int\sin^{-1}x\,dx⁠, the very integral you are looking for. sin⁡−1x\sin^{-1}x goes first.
  2. Sign of the root term: −∫x dx1−x2=+1−x2-\int\frac{x\,dx}{\sqrt{1 - x^2}} = +\sqrt{1 - x^2}⁠.
  3. Forgetting the factor −12-\frac{1}{2} from dt=−2x dxdt = -2x\,dx⁠.
What to take away
  1. When dydx\frac{dy}{dx} depends on xx only, the work is all in one integral.
  2. Integrate an inverse trigonometric function alone by parts, with 1 as the second function.
  3. Check by differentiating: the extra terms must cancel.

Every fact used, and where it comes from

FactUsed inSource
Variables separable: separate, then integrateStep 1Section 9.4.1 (Lesson)
Integration by parts; an inverse trigonometric function as the first function (ILATE)Step 2Section 7.6 and its Remark (iii) (Basics)
ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}Steps 2, 5Chapter 5; Section 7.2, formula (viii) (Basics)
Substitution; ∫tn dt=tn+1n+1+C\int t^n\,dt = \frac{t^{n + 1}}{n + 1} + CStep 3Section 7.3.1; Section 7.2, formula (i)
One arbitrary constant for a first order equationStep 4Section 9.3 (Basics)
Product ruleStep 5Chapter 5
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