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

Exercise 9.3 · Q17: Curve through (0,−2)(0, -2) with ydydx=xy\frac{dy}{dx} = x

Exercise 9.3Variables separable, curve from a slope condition · Sections 9.3 and 9.4.1
The question

17. Find the equation of a curve passing through the point (0,−2)(0, -2) given that at any point (x,y)(x, y) on the curve, the product of the slope of its tangent and yy coordinate of the point is equal to the xx coordinate of the point.

Exam answer

Let (x,y)(x, y) be any point on the curve. The slope of the tangent there is dydx\frac{dy}{dx}⁠. Given

ydydx=xy\frac{dy}{dx} = x

Separating the variables, y dy=x dxy\,dy = x\,dx⁠.

Integrating both sides,

y22=x22+Cy2−x2=2C…(i)\begin{aligned} \frac{y^2}{2} &= \frac{x^2}{2} + C \\ y^2 - x^2 &= 2C\quad\ldots(\text{i}) \end{aligned}

The curve passes through (0,−2)(0, -2)⁠. Putting x=0x = 0⁠, y=−2y = -2 in (i): 4−0=2C4 - 0 = 2C⁠, so 2C=42C = 4⁠.

∴\therefore The equation of the curve is y2−x2=4y^2 - x^2 = 4⁠.

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

The one idea this problem is built on

A condition on the tangent at every point is a differential equation: the slope of the tangent is dydx\frac{dy}{dx}⁠, so the sentence says ydydx=xy\frac{dy}{dx} = x⁠. This separates at once into y dy=x dxy\,dy = x\,dx⁠, integrates to a hyperbola y2−x2=constanty^2 - x^2 = \text{constant}⁠, and the point (0,−2)(0, -2) fixes the constant.

Given, to find, and the plan
Given
At every point (x,y)(x, y) of the curve, (slope of the tangent) × y=x\times\, y = x⁠; the curve passes through (0,−2)(0, -2)⁠.
To find
The equation of the curve.
Plan
(1) Turn the sentence into a differential equation. (2) Separate. (3) Integrate, one constant. (4) Use the point. (5) Identify the branch through the point, check, and draw it.

Basics you need

1Slope of the tangent

At a point (x,y)(x, y) of a curve, the slope of the tangent is dydx\frac{dy}{dx} (Chapter 6). Basics: an equation draws an arrow at every point

2Variables separable

Collect yy with dydy and xx with dxdx⁠, then integrate both sides (Section 9.4.1). Lesson: separate, then integrate

3Power rule for integrals

∫y dy=y22+C\int y\,dy = \frac{y^2}{2} + C and ∫x dx=x22+C\int x\,dx = \frac{x^2}{2} + C (Section 7.2, formula (i)). Basics: every method ends with an integral

4A curve through a point

"Passes through (0,−2)(0, -2)" means y=−2y = -2 when x=0x = 0⁠; it fixes the constant (Section 9.3). Basics: one constant, one condition

Solution

  1. Turn the sentence into an equation

    At (x,y)(x, y) the slope of the tangent is dydx\frac{dy}{dx}⁠. "The product of the slope and the yy coordinate equals the xx coordinate" reads

    ydydx=x.y\frac{dy}{dx} = x.
    Each phrase becomes one symbol: slope of the tangent →dydx\to \frac{dy}{dx}⁠, yy coordinate →y\to y⁠, xx coordinate →x\to x⁠, "product" →\to multiply, "is equal to" →\to =. The equation has order 1 and degree 1.
    Slope of the tangent is dydx\frac{dy}{dx} (Chapter 6). Basics: an arrow at every point

  2. Separate the variables

    Multiply both sides by dxdx⁠:

    y dy=x dx.y\,dy = x\,dx.
    The yy is already beside dydy⁠, so no division is needed and no solution is lost. (In the form dydx=xy\frac{dy}{dx} = \frac{x}{y} we would need y≠0y \ne 0⁠; Step 5 shows yy is never 0 on our curve anyway.)
    Section 9.4.1: separate, then integrate. Lesson: separate, then integrate

  3. Integrate both sides, with one constant

    ∫y dy=∫x dxy22=x22+Cy2−x2=2C.\begin{aligned} \int y\,dy &= \int x\,dx \\ \frac{y^2}{2} &= \frac{x^2}{2} + C \\ y^2 - x^2 &= 2C. \end{aligned}
    The power rule gives y22\frac{y^2}{2} and x22\frac{x^2}{2}⁠. The constants of both sides combine into one CC⁠; multiplying by 2 gives the family of curves y2−x2=2Cy^2 - x^2 = 2C⁠, one constant for a first order equation.
    Section 7.2, formula (i). Basics: every method ends with an integral

  4. Use the point to find the constant

    Put x=0x = 0⁠, y=−2y = -2⁠: (−2)2−02=2C(-2)^2 - 0^2 = 2C⁠, so 2C=42C = 4 and

    y2−x2=4.y^2 - x^2 = 4.
    One condition fixes the one constant. It is enough to find 2C2C⁠, the number that appears in the equation; CC itself is 2.
    Section 9.3: a particular solution gives the constant a value. Basics: one constant, one condition

  5. The branch through the point, the check, and the picture

    y2=x2+4≥4y^2 = x^2 + 4 \ge 4⁠, so y≥2y \ge 2 or y≤−2y \le -2⁠: the hyperbola has two branches. The point (0,−2)(0, -2) is on the lower one, y=−x2+4y = -\sqrt{x^2 + 4}⁠.

    Check, differentiating y2−x2=4y^2 - x^2 = 4 implicitly: 2ydydx−2x=02y\frac{dy}{dx} - 2x = 0⁠, so ydydx=xy\frac{dy}{dx} = x⁠. At (0,−2)(0, -2)⁠: 4−0=44 - 0 = 4⁠.

    X Y X′ Y′ (0, −2) P y² − x² = 4 y = x y = −x O
    The curve y2−x2=4y^2 - x^2 = 4⁠, drawn with equal scales: the lower branch y=−x2+4y = -\sqrt{x^2 + 4} through (0,−2)(0, -2) is bold, the upper branch is dashed, and the asymptotes y=±xy = \pm x are dotted. At P(1.5,−2.5)P(1.5, -2.5) the tangent has slope xy=−0.6\frac{x}{y} = -0.6⁠, and −0.6×(−2.5)=1.5-0.6 \times (-2.5) = 1.5⁠, the xx coordinate.
    Implicit differentiation returns the equation, so the whole curve satisfies the condition, and the point lies on it. Since ∣y∣≥2|y| \ge 2⁠, yy is never 0 on the curve, so dividing by yy would also have been safe. The textbook's answer is the equation of the hyperbola; the curve through (0,−2)(0, -2) is its lower branch.
    Implicit differentiation (Chapter 5). Lesson: general and particular solutions
Result

The curve is y2−x2=4y^2 - x^2 = 4⁠; the part through (0,−2)(0, -2) is the lower branch y=−x2+4y = -\sqrt{x^2 + 4}⁠. (Textbook answer: y2−x2=4y^2 - x^2 = 4⁠.)

Check it with numbers

At x=1.5x = 1.5⁠: y=−2.25+4=−2.5y = -\sqrt{2.25 + 4} = -2.5⁠, and the condition asks for the slope xy=1.5−2.5=−0.6\frac{x}{y} = \frac{1.5}{-2.5} = -0.6⁠.

The curve itself: y(1.501)≈−2.5006001y(1.501) \approx -2.5006001 and y(1.499)≈−2.4994001y(1.499) \approx -2.4994001⁠, so

−2.5006001+2.49940010.002≈−0.6000.\begin{aligned} &\frac{-2.5006001 + 2.4994001}{0.002} \\ &\approx -0.6000. \end{aligned}

Then slope × y=(−0.6)(−2.5)=1.5=x\times\, y = (-0.6)(-2.5) = 1.5 = x⁠.

Common mistakes
  1. Writing dydx=xy\frac{dy}{dx} = xy or xdydx=yx\frac{dy}{dx} = y⁠. The product is slope times yy⁠, and it equals xx⁠.
  2. Forgetting the halves: y22=x22+C\frac{y^2}{2} = \frac{x^2}{2} + C gives y2−x2=2Cy^2 - x^2 = 2C⁠, and the point gives 2C=42C = 4⁠.
  3. Taking the upper branch. y=+x2+4y = +\sqrt{x^2 + 4} passes through (0,2)(0, 2)⁠, not (0,−2)(0, -2)⁠.
What to take away
  1. "Slope of the tangent" always becomes dydx\frac{dy}{dx}⁠.
  2. y dy=x dxy\,dy = x\,dx gives the hyperbolas y2−x2=constanty^2 - x^2 = \text{constant}⁠.
  3. When the equation has several branches, the given point says which one is the curve.

Every fact used, and where it comes from

FactUsed inSource
Slope of the tangent at (x,y)(x, y) is dydx\frac{dy}{dx}Step 1Chapter 6 (Basics)
Variables separable: separate, then integrateStep 2Section 9.4.1 (Lesson)
∫x dx=x22+C\int x\,dx = \frac{x^2}{2} + CStep 3Section 7.2, formula (i) (Basics)
A particular solution gives the constant a valueStep 4Section 9.3 (Basics)
Implicit differentiation; a solution makes L.H.S. = R.H.S.Step 5Chapter 5; Section 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 →