Chapter 9 · Differential Equations

Example 8: Curve through $(-2, 3)$ with slope $\frac{2x}{y^2}$

ExamplesVariables separable, curve from its slope · Sections 9.3 and 9.4.1

The question

Find the equation of a curve passing through the point $(-2, 3)$ ⁠, given that the slope of the tangent to the curve at any point $(x, y)$ is $\dfrac{2x}{y^2}$ ⁠.

Exam answer

The slope of the tangent is $\frac{dy}{dx}$ ⁠, so

\frac{dy}{dx} = \frac{2x}{y^2}

Separating the variables,

y^2\,dy = 2x\,dx

Integrating both sides,

\begin{aligned} \int y^2\,dy &= \int 2x\,dx \\ \frac{y^3}{3} &= x^2 + C\quad\ldots(\text{i}) \end{aligned}

Putting $x = -2$ ⁠, $y = 3$ in (i): $9 = 4 + C$ ⁠, so $C = 5$ ⁠.

$\therefore$ The required curve is $\dfrac{y^3}{3} = x^2 + 5$ ⁠, or $y = (3x^2 + 15)^{\dfrac13}$ ⁠.

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

The one idea this problem is built on

"The slope of the tangent at $(x, y)$ " is $\frac{dy}{dx}$ ⁠. So the sentence in the question is a differential equation, $\frac{dy}{dx} = \frac{2x}{y^2}$ ⁠, and "passing through $(-2, 3)$ " is the condition that fixes the constant. The equation is separable: multiply by $y^2$ ⁠.

Given, to find, and the plan

Given: At every point $(x, y)$ of the curve, the slope of the tangent is $\frac{2x}{y^2}$ (so $y \ne 0$ on the curve). The curve passes through $(-2, 3)$ ⁠.
To find: The equation of the curve.
Plan: (1) Write the slope condition as a differential equation. (2) Separate. (3) Integrate, with one constant. (4) Use $(-2, 3)$ to find $C$ ⁠. (5) Solve for $y$ ⁠. (6) Check by substitution and look at the curve.

Basics you need

1Slope of the tangent

The slope of the tangent to a curve $y = f(x)$ at a point is $\frac{dy}{dx}$ there; the book's solution of this example (Section 9.4.1, Example 8) starts from this fact. Basics: dy/dx = F(x, y) draws an arrow at every point

2Variables separable

$\frac{dy}{dx} = g(x)\,h(y)$ with $h(y) \ne 0$ gives $\int\frac{dy}{h(y)} = \int g(x)\,dx$ (Section 9.4.1). Lesson: separate, then integrate

3Power rule

$\int t^n\,dt = \frac{t^{n+1}}{n + 1} + C$ ⁠, $n \ne -1$ (Section 7.2, formula (i)). Basics: every method ends with an integral

4Cube roots

Every real number has exactly one real cube root, so $y^3 = a$ gives $y = a^{\frac13}$ with no $\pm$ choice.

Solution

Turn the words into a differential equation

The slope of the tangent at $(x, y)$ is $\frac{dy}{dx}$ ⁠, so
$\frac{dy}{dx} = \frac{2x}{y^2},\quad y \ne 0.$
This is the step that makes it a differential equation question. The slope formula has $y^2$ in the denominator, so it only makes sense where $y \ne 0$ ⁠; the curve cannot touch the $x$ -axis.

Slope of the tangent = $\frac{dy}{dx}$ ⁠, as the book's solution states (Section 9.4.1, Example 8). Basics: a rule for slopes
Separate the variables

The right side is $2x\cdot\frac{1}{y^2}$ ⁠, separable. Multiply both sides by $y^2\,dx$ ⁠:
$y^2\,dy = 2x\,dx.$
The left side now has only $y$ ⁠, the right side only $x$ ⁠. Here $h(y) = \frac{1}{y^2}$ is never 0, so multiplying by $y^2$ loses nothing.

Section 9.4.1: $\frac{1}{h(y)}\,dy = g(x)\,dx$ ⁠, here with $\frac{1}{h(y)} = y^2$ ⁠. Lesson: separate, then integrate
Integrate both sides, with one constant
$\begin{aligned} \int y^2\,dy &= \int 2x\,dx \\ \frac{y^3}{3} &= x^2 + C\quad\ldots(1) \end{aligned}$
Check each integral by differentiating: $\frac{d}{dy}\frac{y^3}{3} = y^2$ and $\frac{d}{dx}x^2 = 2x$ ⁠.

Power rule on each side; the two constants combine into one $C$ ⁠. Equation (1) is the general solution: the family of all curves with the given slope.

Section 7.2, formula (i). Basics: every method ends with an integral
Use the point $(-2, 3)$ to find $C$

Put $x = -2$ ⁠, $y = 3$ in (1):
$\begin{aligned} \frac{3^3}{3} &= (-2)^2 + C \\ \Rightarrow\ 9 &= 4 + C \;\Rightarrow\; C = 5. \end{aligned}$
So the curve is
$\frac{y^3}{3} = x^2 + 5.$
"Passing through $(-2, 3)$ " means $y = 3$ when $x = -2$ ⁠. Note $(-2)^2 = 4$ ⁠: the minus sign disappears when squared.

Section 9.3: a particular solution from a given point. Lesson: general and particular solutions
Solve for $y$

Multiply by 3: $y^3 = 3x^2 + 15$ ⁠. Taking the real cube root,
$y = (3x^2 + 15)^{\frac13}.$
Since $3x^2 + 15 \ge 15$ ⁠, we get $y \ge 15^{\frac13} \approx 2.466$ for every real $x$ ⁠. So the curve is defined for all $x$ ⁠, lies above the $x$ -axis, and never meets $y = 0$ ⁠, where the slope would be undefined.

A cube root has no sign choice, so this formula is the whole curve. It agrees with the given point: $(12 + 15)^{\frac13} = 27^{\frac13} = 3$ ⁠.

Every real number has one real cube root; the curve must keep $y \ne 0$ (Step 1).
Check: substitute back, and see the curve

Differentiate $y^3 = 3x^2 + 15$ implicitly:
$3y^2\frac{dy}{dx} = 6x \;\Rightarrow\; \frac{dy}{dx} = \frac{2x}{y^2},$
the given slope; and $y = 3$ at $x = -2$ ⁠.

Slope field of $\frac{dy}{dx} = \frac{2x}{y^2}$ with equal scales (no marks on $y = 0$ ⁠, where the slope is undefined). The curve $y^3 = 3x^2 + 15$ through $(-2, 3)$ follows the marks; its lowest point is $(0, 15^{1/3})$ ⁠, where the slope $2x/y^2$ is 0.

The derivative of the answer reproduces the slope rule, and the point fits, so the curve is right. The picture shows why the curve is lowest at $x = 0$ ⁠: the slope $\frac{2x}{y^2}$ is negative for $x < 0$ and positive for $x > 0$ ⁠.

Implicit differentiation (Section 5.3.2). Basics: y′ is the slope

Result

The general solution is $\frac{y^3}{3} = x^2 + C$ ⁠; the point $(-2, 3)$ gives $C = 5$ ⁠. The required curve is $\frac{y^3}{3} = x^2 + 5$ ⁠, that is $y = (3x^2 + 15)^{\frac13}$ ⁠. (Textbook answer: $\frac{y^3}{3} = x^2 + 5$ or $y = (3x^2 + 15)^{\frac13}$ ⁠.)

Check it with numbers

At $x = 1$ ⁠: $y = 18^{\frac13} \approx 2.6207414$ ⁠, and the slope rule gives $\frac{2(1)}{2.6207414^2} \approx 0.29119$ ⁠.

The curve itself: $y(1.001) \approx 2.6210327009$ and $y(0.999) \approx 2.6204503140$ ⁠, so

\begin{aligned} &\frac{2.6210327009 - 2.6204503140}{0.002} \\ &\approx 0.29119. \end{aligned}

Common mistakes

Using $\frac{dx}{dy}$ for the slope of the tangent. The slope is $\frac{dy}{dx}$ ⁠.
$(-2)^2 = -4$ ⁠. It is $+4$ ⁠, so $C = 9 - 4 = 5$ ⁠.
Forgetting the 3: $\int y^2\,dy = \frac{y^3}{3}$ ⁠, not $y^3$ ⁠.
Writing $y = \pm(3x^2 + 15)^{\frac13}$ ⁠. A cube root has no $\pm$ ⁠; the minus version would not pass through $(-2, 3)$ ⁠.

What to take away

"Slope of the tangent at $(x, y)$ is …" means $\frac{dy}{dx} = \ldots$ ⁠.
Particular solution: integrate with $C$ ⁠, then substitute the point.
Check the domain of the answer: here $y \ge 15^{\frac13} > 0$ ⁠, so the curve never reaches $y = 0$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Slope of the tangent $= \frac{dy}{dx}$	Step 1	Section 9.4.1, Example 8 (the book's solution) (Basics)
Variables separable	Steps 2, 3	Section 9.4.1 (Lesson)
$\int t^n\,dt = \frac{t^{n+1}}{n + 1} + C$	Step 3	Section 7.2, formula (i) (Basics)
A particular solution from a given point	Step 4	Section 9.3 (Lesson)
Every real number has exactly one real cube root	Step 5	Class XI
Implicit differentiation	Step 6	Chapter 5, Section 5.3.2 (Basics)

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Example 8: Curve through (−2,3)(-2, 3)(−2,3) with slope 2xy2\frac{2x}{y^2}y22x​

Basics you need

Solution

Turn the words into a differential equation

Separate the variables

Integrate both sides, with one constant

Use the point (−2,3)(-2, 3)(−2,3) to find CCC

Solve for yyy

Check: substitute back, and see the curve

Every fact used, and where it comes from

Example 8: Curve through $(-2, 3)$ with slope $\frac{2x}{y^2}$

Use the point $(-2, 3)$ to find $C$

Solve for $y$