(a) Solve the following differential equation : $\frac{dy}{dx} = \frac{2 - y}{x + 1}$

OR

(b) If $\int_{a}^{b} x^3 dx = 0$ and $\int_{a}^{b} x^2 dx = \frac{2}{3}$ , then find the values of 'a' and 'b'.

Q: (a) Solve the following differential equation : $\frac{dy}{dx} = \frac{2 - y}{x + 1}$ OR (b) If $\int_{a}^{b} x^3 dx = 0$ and $\int_{a}^{b} x^2 dx = \frac{2}{3}$, then find the values of 'a' and 'b'.

**Part (a):** The solution is **$(2 - y)(x + 1) = C$**. - Separate: $\frac{dy}{2 - y} = \frac{dx}{x + 1}$. - Integrate: $-\ln|2 - y| = \ln|x + 1| + c$ ⇒ $\ln|(2 - y)(x + 1)| = \text{constant}$ ⇒ $(2 - y)(x + 1) = C$. **Part (b):** **$a = -1$ and $b = 1$.** - $\int_a^b x^3 dx = \frac{b^4 - a^4}{4} = 0$ ⇒ $b^4 = a^4$ ⇒ $b = -a$ (since $a \ne b$). - $\int_a^b x^2 dx = \frac{b^3 - a^3}{3} = \frac{2}{3}$ ⇒ with $b = -a$: $\frac{-2a^3}{3} = \frac{2}{3}$ ⇒ $a^3 = -1$ ⇒ $a = -1, b = 1$.

Differential Equations

Differential Equations / Definite Integrals

Official Answer

Part (a): The solution is $(2 - y)(x + 1) = C$ .

Separate: $\frac{dy}{2 - y} = \frac{dx}{x + 1}$ .
Integrate: $-\ln|2 - y| = \ln|x + 1| + c$ ⇒ $\ln|(2 - y)(x + 1)| = \text{constant}$ ⇒ $(2 - y)(x + 1) = C$ .

Part (b): $a = -1$ and $b = 1$ .

$\int_a^b x^3 dx = \frac{b^4 - a^4}{4} = 0$ ⇒ $b^4 = a^4$ ⇒ $b = -a$ (since $a \ne b$ ).
$\int_a^b x^2 dx = \frac{b^3 - a^3}{3} = \frac{2}{3}$ ⇒ with $b = -a$ : $\frac{-2a^3}{3} = \frac{2}{3}$ ⇒ $a^3 = -1$ ⇒ $a = -1, b = 1$ .

variable separabledifferential equationdefinite integralodd functionsymmetric limitsintegrationgeneral solution

Marking Scheme

1Part (a) — 1 mark: correct separation $\frac{dy}{2-y}=\frac{dx}{x+1}$ ; 1 mark: correct integration to $(2-y)(x+1)=C$ (or equivalent log form).
2Part (b) — 1 mark: using $\int x^3=0$ to get $b=-a$ ; 1 mark: using $\int x^2=\frac{2}{3}$ to obtain $a=-1, b=1$ .
3Accept equivalent constant forms in (a) and either order of (a,b) if consistent.

Hint

For (a) separate variables and integrate both sides; for (b) note $x^3$ is odd so the zero condition forces $b = -a$ .

Quick Oral Answer

For part a, separating variables and integrating gives 2 minus y times x plus 1 equals a constant; for part b, since x cubed is odd the limits must be symmetric, giving a equal to minus 1 and b equal to 1.

Analysis & Explanation

This question offers a choice between a first-order differential equation and a definite-integral condition problem.

Part (a) concept — variable separable: Since $\frac{dy}{dx}$ factors into a function of y times a function of x, we move all y-terms to one side and x-terms to the other, then integrate both sides. The integral of $\frac{1}{2-y}$ introduces a minus sign, which students often miss.

Part (b) concept — even/odd symmetry: The condition $\int_a^b x^3 dx = 0$ with $a \ne b$ forces the limits to be symmetric ( $b = -a$ ), because $x^3$ is an odd function whose integral over a symmetric interval vanishes. Then the second condition fixes the magnitude.

Exam trap: In (a), forgetting the negative sign gives the wrong sign of C. In (b), taking $b = a$ is invalid because that collapses the interval; the meaningful solution is $b = -a$ .

Application: Separable equations model growth and decay; symmetric-limit reasoning is central to Fourier analysis and mean-value computations.

Common Mistakes

1In (a), dropping the negative sign when integrating $1/(2-y)$ , giving an incorrectly signed solution.
2In (b), accepting $b = a$ , which makes the interval zero and is not a valid distinct-limit solution.
3In (b), forgetting to check the second condition and leaving the answer only as $b = -a$ without finding numerical values.

Interesting Facts

Any odd function integrated over a symmetric interval $[-a, a]$ gives zero — this is why $\int_a^b x^3 dx = 0$ immediately signals symmetric limits.

Variable-separable equations were among the first differential equations solved historically, used by Leibniz in the 1690s to model curves.

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Frequently Asked Questions

How do I know part (b) requires symmetric limits?

Because x³ is an odd function, $\int_a^b x^3 dx$ equals zero only when the interval is symmetric about the origin, i.e. $b = -a$ (assuming $a \ne b$ ). This symmetry condition is the key that unlocks the problem before the second integral fixes the exact values.

What does 'variable separable' mean in part (a)?

It means the equation can be rearranged so that all terms in y (with dy) are on one side and all terms in x (with dx) are on the other. Each side is then integrated independently. This works whenever $\frac{dy}{dx}$ can be written as a product or quotient of a function of x and a function of y.