Variables separable

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2013, 2014, 2017)
• Priority tier: T3
• Marks (count): 10 (2), 8 (1)
• Average solve time: ~7 min
• Difficulty mix: medium 2, easy 1
• Section: B | Dominant type: computation

Why This Chapter Matters

Variables-separable ODEs underlie every other first-order method: growth/decay models, geometric tangent problems, and substitution-reducible equations all end as separations. UPSC has tested three distinct flavours — geometric (tangent bisected by axes), applied (exponential growth), and substitution (trig ODE reducible by $v=x+y$) — at 8–10 marks each. Mastering the core separation + integration algorithm and the two standard substitutions ($v=x+y$ for $F(x+y)$ type; $v=y/x$ for homogeneous type) covers every question from 2013–2017 in under eight minutes.

Minimum Theory

Separation of variables. A first-order ODE $\dfrac{dy}{dx}=f(x)\,g(y)$ is separable. Rewrite as $\dfrac{dy}{g(y)}=f(x)\,dx$ and integrate both sides:

$\int\frac{dy}{g(y)} = \int f(x)\,dx + C.$

The solution is a relation $G(y)=F(x)+C$ where $G'=1/g$ and $F'=f$. Add the initial condition if given to determine $C$.

Substitution for $F(x+y)$ type. If the ODE has the form $dy/dx=F(x+y)$, set $v=x+y$. Then $dv/dx=1+dy/dx$, so $dy/dx=dv/dx-1$, and the ODE becomes $dv/dx = 1+F(v)$ — always separable.

Geometric interpretation. The tangent at $(x_0,y_0)$ to $y=f(x)$ has $x$-intercept $(x_0-y_0/f',0)$ and $y$-intercept $(0,y_0-x_0f')$. Any property of these intercepts (bisection, equal intercepts, etc.) translates into an ODE that is typically separable.

Question Archetypes

geometric-ode (1 question(s); 2014)

Recognition Cues

• “Find the curve such that the tangent cut-off by axes is bisected at the point of tangency.”
• “The portion of the tangent between axes has some length/ratio property.”
• “Tangent at every point passes through the origin / makes equal intercepts.”
• You must set up the ODE from the geometry of the tangent line, then solve.

Solution Template

1. Write the tangent line at $(x_0,y_0)$: $y-y_0=y'(x-x_0)$.
2. Find the intercepts (set $y=0$ for $x$-intercept, $x=0$ for $y$-intercept).
3. Translate the geometric condition (midpoint, ratio, equal intercepts, etc.) into an equation in $x_0,y_0,y'$.
4. Drop the subscripts to get the ODE. Recognise it as separable.
5. Separate and integrate. Determine the constant if initial data are given.
6. Verify that a member of the solution family satisfies the original geometric condition.

Worked Example

2014 Paper 1, 2014-P1-Q5b (10 marks)

Find the curve for which the part of the tangent cut-off by the axes is bisected at the point of tangency.

Step 1 — Tangent line. At $(x_0,y_0)$ with slope $y'=f'(x_0)$:

$y - y_0 = y'(x-x_0).$

Step 2 — Intercepts.

• $x$-intercept ($y=0$): $x = x_0 - y_0/y'$.
• $y$-intercept ($x=0$): $y = y_0 - x_0 y'$.

Step 3 — Bisection condition. The point of tangency $(x_0,y_0)$ is the midpoint of $\bigl(x_0-y_0/y',\;0\bigr)$ and $\bigl(0,\;y_0-x_0 y'\bigr)$:

$x_0 = \frac{(x_0 - y_0/y') + 0}{2} \qquad\Longrightarrow\qquad 2x_0 = x_0 - \frac{y_0}{y'} \qquad\Longrightarrow\qquad y' = -\frac{y_0}{x_0}.$

(The $y$-coordinate condition gives the same result.)

Step 4 — Separate and integrate. The ODE is $\dfrac{dy}{dx}=-\dfrac{y}{x}$, i.e.\ $\dfrac{dy}{y}=-\dfrac{dx}{x}$. Integrate:

$\ln|y| = -\ln|x| + \ln|k| \qquad\Longrightarrow\qquad xy = k.$

$\boxed{\;xy = k\quad\text{(rectangular hyperbolas with coordinate axes as asymptotes).}\;}$

Verify: On $xy=k$, implicit differentiation gives $y+xy'=0$, so $y'=-y/x$ ✓. The tangent at $(x_0,y_0)$ has $x$-intercept $(2x_0,0)$ and $y$-intercept $(0,2y_0)$; their midpoint is $(x_0,y_0)$ ✓.

Common Traps

• Do not forget to divide by 2 in the midpoint formula: the midpoint of $(X,0)$ and $(0,Y)$ is $(X/2, Y/2)$, not $(X,Y)$.
• Both coordinate equations of the midpoint condition yield the same ODE; you only need one, but verifying the other provides a free check.
• The family $xy=k$ consists of rectangular hyperbolas — do not confuse them with general hyperbolas.

growth-decay (1 question(s); 2017)

Recognition Cues

• “Growth rate proportional to amount present at time $t$.”
• “Population doubles in [one period].”
• “Radioactive decay; half-life is [given].”
• Two pieces of data: initial amount and one time-point measurement; asked for amount at another time.

Solution Template

1. Write the model. $dN/dt = kN$ (growth) or $dN/dt = -kN$ (decay).
2. Solve. Separate: $dN/N = k\,dt$; integrate: $N(t) = N_0 e^{kt}$.
3. Use the first data point to find $k$ (e.g., doubling: $N_0 e^{kT}=2N_0 \Rightarrow k=\ln 2/T$).
4. Evaluate at the required time. If $T$ is the doubling period and the question asks for $nT$, $N(nT)=2^n N_0$.
5. State units and any proportionality constant absorbed.

Worked Example

2017 Paper 1, 2017-P1-Q6a-ii (8 marks)

If the growth rate of the population of bacteria at any time $t$ is proportional to the amount present at that time and the population doubles in one week, then how many bacteria can be expected after 4 weeks?

Step 1 — Model. Let $N(t)$ = population at time $t$ (weeks). “Growth rate proportional to amount”:

$\frac{dN}{dt} = kN.$

Step 2 — Solve. Separate: $\displaystyle\int\frac{dN}{N}=\int k\,dt$, giving $N(t)=N_0 e^{kt}$.

Step 3 — Find $k$. Population doubles in one week: $N(1)=2N_0$:

$N_0 e^k = 2N_0 \qquad\Longrightarrow\qquad e^k = 2 \qquad\Longrightarrow\qquad k = \ln 2.$

Step 4 — Evaluate at $t=4$.

$N(4) = N_0 e^{4\ln 2} = N_0\cdot 2^4 = 16\,N_0.$

$\boxed{\;N(4) = 16\,N_0.\;}$

After 4 weeks the population is 16 times the initial amount.

Common Traps

• Doubling in 1 week for 4 weeks compounds to $2^4=16$, not $2\times4=8$. Exponential growth is multiplicative, not additive.
• No numerical value of $N_0$ is given; the answer must be expressed as a multiple of the initial population.
• The constant $k=\ln 2$, not $k=2$; confusing the two gives $N(4)=e^8 N_0$ (incorrect).

reducible-substitution (1 question(s); 2013)

Recognition Cues

• The RHS depends only on $x+y$ (not on $x$ and $y$ separately): $dy/dx = F(x+y)$.
• After substitution $v=x+y$, the equation becomes separable.
• Trigonometric RHS involving $\cos(x+y)$, $\sin(x+y)$, or $\tan(x+y)$ almost always signals this type.

Solution Template

1. Identify that RHS is a function of $v=x+y$.
2. Substitute $v=x+y$: then $dv/dx = 1 + dy/dx$, so $dy/dx = dv/dx - 1$.
3. New ODE: $dv/dx = 1 + F(v)$ — now separable.
4. Separate: $\dfrac{dv}{1+F(v)}=dx$.
5. Simplify the LHS (often via trig identities) before integrating.
6. Integrate both sides. Back-substitute $v=x+y$.
7. State the solution free from derivatives if asked.

Worked Example

2013 Paper 1, 2013-P1-Q5a (10 marks)

$y$ is a function of $x$ such that $\dfrac{dy}{dx}=\cos(x+y)+\sin(x+y)$. Find a relation between $x$ and $y$ free from any derivative/differential.

Step 1 — Substitution. RHS depends only on $x+y$. Let $v=x+y$, so $\dfrac{dv}{dx}=1+\dfrac{dy}{dx}$. The ODE becomes:

$\frac{dv}{dx} = 1 + \cos v + \sin v.$

Step 2 — Half-angle simplification.

$1 + \cos v = 2\cos^2\!\frac{v}{2}, \qquad \sin v = 2\sin\frac{v}{2}\cos\frac{v}{2}.$

$\frac{dv}{dx} = 2\cos\frac{v}{2}\!\left(\cos\frac{v}{2}+\sin\frac{v}{2}\right).$

Step 3 — Separate.

$\frac{dv}{2\cos\tfrac{v}{2}\!\left(\cos\tfrac{v}{2}+\sin\tfrac{v}{2}\right)} = dx.$

Set $u = v/2$, $du = dv/2$:

$\frac{du}{\cos u\,(\cos u+\sin u)} = dx.$

Step 4 — Integrate. Divide numerator and denominator by $\cos^2 u$:

$\frac{\sec^2 u\,du}{1+\tan u} = dx.$

Let $w = 1+\tan u$, $dw = \sec^2 u\,du$:

$\int\frac{dw}{w} = \int dx \qquad\Longrightarrow\qquad \ln|1+\tan u| = x + C.$

Step 5 — Back-substitute. $u=(x+y)/2$:

$\ln\!\left|1+\tan\frac{x+y}{2}\right| = x + C.$

Exponentiate:

$\boxed{\;1 + \tan\frac{x+y}{2} = A\,e^{x},\quad A \text{ a non-zero constant.}\;}$

Common Traps

• The substitution $v=x+y$ is specifically for ODEs where the RHS is a function of $x+y$. Do not attempt it on a general ODE.
• Without the half-angle identities $1+\cos v=2\cos^2(v/2)$ and $\sin v=2\sin(v/2)\cos(v/2)$, the trig integral has no clean form. Recognising the need for them is the key step.
• After dividing by $\cos^2 u$, the denominator becomes $1+\tan u$ — the derivative of $\tan u$. The integrand is then $d(\tan u)/(1+\tan u)=d(1+\tan u)/(1+\tan u)$, which integrates to $\ln|1+\tan u|$.
• The final answer must be “free from derivatives,” as the question requests; back-substituting $u=(x+y)/2$ and exponentiating completes this.

Marks-Aware Writing

8-mark question (growth/decay, 2017): State the model equation $dN/dt=kN$ (1 mark), solve to get $N=N_0 e^{kt}$ (2 marks), compute $k=\ln2$ from the doubling condition (2 marks), evaluate $N(4)=16N_0$ (2 marks), state the answer clearly (1 mark). Any missing step costs the marks attached to everything that follows from it.

10-mark questions (geometric and substitution): Two marks for setting up correctly (tangent intercepts / substitution $v=x+y$), two marks for translating into the ODE, three marks for the integration (include the trig simplification step for the 2013 question), two marks for the final answer, one mark for a brief verification or check. A clean boxed answer without derivation earns at most 3 marks.

Practice Set