Variables Separable Method

The First Solving Engine

A first order, first degree differential equation has the shape $\frac{dy}{dx} = F(x, y)$. When $F$ factors as a product of a pure-$x$ function and a pure-$y$ function,

$$\frac{dy}{dx} = g(x)\cdot h(y)$$

the equation is of variables separable type — and it solves in four moves.

The method

1. Recognise the product form $g(x)h(y)$.
2. Separate (for $h(y) \neq 0$): $\dfrac{dy}{h(y)} = g(x)\,dx$ — every $y$ travels with $dy$, every $x$ with $dx$.
3. Integrate both sides: $\displaystyle\int\frac{dy}{h(y)} = \int g(x)\,dx$, giving $H(y) = G(x) + C$ — the general solution, usually left implicit.
4. Apply the condition (if given) to find $C$ — the particular solution.

Worked template

$\dfrac{dy}{dx} = \dfrac{x + 1}{2 - y}$ (with $y \neq 2$): separate to $(2 - y)\,dy = (x + 1)\,dx$, integrate:

$$2y - \frac{y^2}{2} = \frac{x^2}{2} + x + C_1 \;\Rightarrow\; x^2 + y^2 + 2x - 4y + C = 0$$

(where $C = 2C_1$ absorbed the doubling — constants merge freely).

Housekeeping rules

1. One constant is enough: write $+C$ on the $x$-side only; two constants merge into one.
2. Constants can be reshaped: $2C_1 \to C$, $e^{C_1} \to C$, $\pm C \to C$ — any invertible renaming is legal and keeps answers tidy.
3. Implicit answers are complete answers: $\tan^{-1}y = \tan^{-1}x + C$ need not be solved for $y$ unless asked.
4. Note excluded values: the separation divided by $h(y)$, so cases like $y = 2$ above are excluded by hypothesis (the problem states them).

Key Point: Separability is a structural property — spot the product $g(x)h(y)$. Sums like $\frac{dy}{dx} = x + y$ do NOT separate (that one is linear, Section 5); products, quotients and exponentials like $e^{x+y} = e^x e^y$ do.

[JEE Tip] $e^{x+y}$ splitting into $e^x\cdot e^y$ is the most-recycled separable disguise in MCQs: $\frac{dy}{dx} = e^{x+y}$ separates to $e^{-y}dy = e^x dx$, giving $-e^{-y} = e^x - C$, i.e. $e^x + e^{-y} = C$. Recognise the family.

Curves and Growth — Separable Equations at Work

Curve problems

"The slope of the tangent at $(x, y)$ is …" translates directly to $\frac{dy}{dx} = \ldots$ — then the routine is: solve generally, substitute the given point, extract $C$.

Template: the curve through $(-2, 3)$ with slope $\frac{2x}{y^2}$: separate $y^2dy = 2x\,dx$, integrate $\frac{y^3}{3} = x^2 + C$; the point gives $9 = 4 + C$, so $C = 5$ and the curve is $\frac{y^3}{3} = x^2 + 5$.

The exponential growth model

"Principal increases continuously at the rate of 5% per year" means the rate of change is proportional to the current amount:

$$\frac{dP}{dt} = \frac{5}{100}P = \frac{P}{20}$$

Separate and integrate: $\log P = \frac{t}{20} + C_1$, so $P = Ce^{t/20}$ (with $C = e^{C_1}$). An initial value pins down $C$: starting from ₹1000, $P = 1000e^{t/20}$, and doubling ($P = 2000$) gives $e^{t/20} = 2$, i.e.

$$t = 20\log_e 2 \text{ years}$$

The same skeleton handles every "proportional growth" model — bacteria counts, populations, radioactive decay (with a minus sign): $\frac{dN}{dt} = kN \Rightarrow N = N_0e^{kt}$.

Key Point: In growth problems the answer usually stays in exact log/exponential form ($20\log 2$, not a decimal). Two data points determine the two unknowns $N_0$ and $k$; a question for "when does $N$ reach $\ldots$" then solves an exponential equation by taking logs.

[JEE Tip] Boards love the bank/bacteria stories; JEE prefers the same mathematics stripped bare ("if $y' = ky$ and $y(0) = A$…"). Either way, jump straight to $y = (\text{initial value})\,e^{kt}$ — deriving it costs a minute, quoting it costs nothing, and the mark scheme accepts both.

Solved Examples

Example 1: The model separation

Find the general solution of $\dfrac{dy}{dx} = \dfrac{x + 1}{2 - y}$ (where $y \neq 2$).

Solution:

1. Separate: $(2 - y)\,dy = (x + 1)\,dx$.
2. Integrate: $2y - \frac{y^2}{2} = \frac{x^2}{2} + x + C_1$.
3. Tidy (multiply by 2, absorb constants): $x^2 + y^2 + 2x - 4y + C = 0$.

Final Answer: $x^2 + y^2 + 2x - 4y + C = 0$.

Example 2: Symmetric inverse-tangent pair

Find the general solution of $\dfrac{dy}{dx} = \dfrac{1 + y^2}{1 + x^2}$.

Solution:

1. Separate (safe: $1 + y^2 \neq 0$ always): $\frac{dy}{1 + y^2} = \frac{dx}{1 + x^2}$.
2. Integrate: $\tan^{-1}y = \tan^{-1}x + C$.

Final Answer: $\tan^{-1}y = \tan^{-1}x + C$ — a perfectly complete implicit answer.

Example 3: A particular solution

Find the particular solution of $\dfrac{dy}{dx} = -4xy^2$ given $y = 1$ when $x = 0$.

Solution:

1. Separate (for $y \neq 0$): $\frac{dy}{y^2} = -4x\,dx$.
2. Integrate: $-\frac1y = -2x^2 + C$, so $y = \frac{1}{2x^2 - C}$.
3. Apply the condition: $1 = \frac{1}{-C}$ gives $C = -1$.

Final Answer: $y = \dfrac{1}{2x^2 + 1}$.

Example 4: Curve through a point

Find the equation of the curve through $(1, 1)$ whose differential equation is $x\,dy = \left(2x^2 + 1\right)dx$ (with $x \neq 0$).

Solution:

1. Separate: $dy = \left(2x + \frac1x\right)dx$.
2. Integrate: $y = x^2 + \log|x| + C$.
3. Apply the point $(1,1)$: $1 = 1 + 0 + C$, so $C = 0$.

Final Answer: $y = x^2 + \log|x|$.

Example 5: Slope condition through a point

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

Solution:

1. Translate: $\frac{dy}{dx} = \frac{2x}{y^2}$; separate: $y^2dy = 2x\,dx$.
2. Integrate: $\frac{y^3}{3} = x^2 + C$.
3. Apply $(-2, 3)$: $9 = 4 + C$, so $C = 5$.

Final Answer: $\dfrac{y^3}{3} = x^2 + 5$, i.e. $y = \left(3x^2 + 15\right)^{1/3}$.

Example 6: Continuous compounding

In a bank, principal increases continuously at the rate of 5% per year. In how many years will ₹1000 double itself?

Solution:

1. Model: $\frac{dP}{dt} = \frac{5}{100}P = \frac{P}{20}$.
2. Separate and integrate: $\log P = \frac{t}{20} + C_1$, so $P = Ce^{t/20}$.
3. Initial value: $P = 1000$ at $t = 0$ gives $C = 1000$; so $P = 1000e^{t/20}$.
4. Doubling: $2000 = 1000e^{t/20}$ gives $e^{t/20} = 2$, so $t = 20\log_e 2$.

Final Answer: $t = 20\log_e 2$ years (about $13.86$ years).

Example 7: Trig products

Solve $\sec^2x\tan y\,dx + \sec^2y\tan x\,dy = 0$.

Solution:

1. Separate (divide by $\tan x\tan y$): $\frac{\sec^2x}{\tan x}dx = -\frac{\sec^2y}{\tan y}dy$.
2. Integrate (each side is $\frac{f'}{f}$): $\log|\tan x| = -\log|\tan y| + C_1$.
3. Combine the logs: $\log|\tan x\tan y| = C_1$.

Final Answer: $\tan x\tan y = C$ — the exponentiated constant $e^{C_1}$ renamed $C$.

Example 8: A trig-integrating condition

Find the particular solution of $\dfrac{dy}{dx} = y\tan x$, given $y = 1$ when $x = 0$.

Solution:

1. Separate: $\frac{dy}{y} = \tan x\,dx$.
2. Integrate: $\log|y| = \log|\sec x| + C_1$, so $y = C\sec x$.
3. Apply the condition: $1 = C\sec 0 = C$.

Final Answer: $y = \sec x$.

Takeaway: When both integrals produce logs, exponentiate and merge: $\log y = \log(\sec x) + C_1$ becomes $y = C\sec x$ in one step — cleaner than carrying the log form to the end.