JEE Mains Maths · Differential Equations

Separating the Variables

Solving an equation by putting every y on one side and every x on the other, then integrating both sides; including the cases that separate only after a substitution or after spotting an exact differential.

All Differential Equations notes JEE Mains Maths strategy

Why this matters

Thirty-one PYQs, and 2024 alone has twelve. Most separate at once, and the work is the integral and the constant from the given point. A few need a substitution for x + y, or a regrouping such as x dy + y dx = d(xy), before they separate. Three ideas cover the page.

Concept 1 of 3: Separate, integrate, fix the constant

If the right side factors as

f(x)\,g(y)

, divide by

g(y)

and integrate:

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

. One given point fixes

C

. Keep the constant as a multiplier when logs appear:

\ln|y+3|=\ln|x|+C

y+3=Ax

, which is easier to use.

Definition

$\frac{dy}{dx}=f(x)g(y)\Rightarrow\int\frac{dy}{g(y)}=\int f(x)\,dx+C$ .
Factor first: $xy-1+x-y=(x-1)(y+1)$ , $2^{x+y}-2^x=2^x(2^y-1)$ .
$\int\frac{dy}{1+y^2}=\tan^{-1}y$ ; $\int\frac{e^x\,dx}{1+e^{2x}}=\tan^{-1}e^x$ .
Two solutions with different constants of $y'=y+k$ never meet.

Separable form

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

Worked example

Solve

\frac{dy}{dx}=\frac{x(1+y^2)}{y}

with

y(0)=1

, and find

y^2

x=1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 23 Jan 2025 · Q55Moderate

Example 1 · Differential Equations · Separating the Variables

Let a curve

y = f(x)

pass through the points

(0,5)

and

\left( \log_{e}2,k \right)

. If the curve satisfies the differential equation

2(3 + y)e^{2x}dx -\left( 7 +e^{2x} \right)dy = 0

, then

k

is equal to

Constant before exponentiating

From

\ln y=x^2+C

, the solution is

y=Ae^{x^2}

, not

e^{x^2}+C

. Fix the constant in whichever form you keep, and never add it after exponentiating.

Drill 20 more on separate, integrate, fix the constant

Concept 2 of 3: When the slope depends on x alone

\frac{dy}{dx}

is a function of

x

only, the equation is just an integral:

y=\int f(x)\,dx+C

. The work is the integration — often by parts, by

t=e^x

, or by recognising a derivative. A second-order equation with no

y

in it integrates twice.

Definition

$y'=f(x)\Rightarrow y=\int f(x)\,dx+C$ .
$f''=g''\Rightarrow f-g=ax+b$ .
$1+\sin2x=(\sin x+\cos x)^2=2\sin^2\left(x+\frac\pi4\right)$ .

Direct integration

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

Worked example

y'=x\cos x

and

y(0)=1

. Find

y(\pi)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 2 · Q70Moderate

Example 2 · Differential Equations · Separating the Variables

Let

f(x)

and

g(x)

be twice differentiable functions satisfying

f^{''}(x) =g^{''}(x)

for all

x \in R

f^{'}(1) = 2{\text{ }g}^{'}(1) = 4

and

g(2) = 3f(2) = 9

. Then

f(25) - g(25)

is equal to :

Two conditions for a second-order equation

Integrating twice brings two constants, so it needs two conditions — often one on the derivative and one on the value.

Drill 3 more on when the slope depends on x alone

Concept 3 of 3: Substituting for x + y, and exact differentials

When the right side is a function of

ax+by+c

, put

t=ax+by+c

: then

\frac{dt}{dx}=a+b\frac{dy}{dx}

and the equation separates in

t

and

x

. When the equation mixes

x\,dy

and

y\,dx

, look for the exact pieces

x\,dy+y\,dx=d(xy)

and

\frac{x\,dy-y\,dx}{x^2}=d\left(\frac yx\right)

Definition

$y'=f(ax+by+c)$ : put $t=ax+by+c$ , $t'=a+bf(t)$ .
$x\,dy+y\,dx=d(xy)$ .
$\frac{x\,dy-y\,dx}{x^2}=d\left(\frac yx\right)$ , $\frac{x\,dy-y\,dx}{xy}=d\left(\ln\frac yx\right)$ .
$M\,dx+N\,dy=0$ with $M_y=N_x$ is exact: integrate to one function $F(x,y)=C$ .

Substitution for a linear combination

t=ax+by+c\ \Rightarrow\ \frac{dt}{dx}=a+b\,\frac{dy}{dx}

Worked example

Solve

\frac{dy}{dx}=(x+y)^2

with

y(0)=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q83Moderate

Example 3 · Differential Equations · Separating the Variables

If the solution of the differential equation

(2x+ 3y- 2)dx+ (4x+ 6y- 7)dy= 0,y(0) = 3

, is

\alpha x+\beta y+ 3\log_{e}|2x+ 3y-\gamma| = 6

, then

\alpha+ 2\beta+ 3\gamma

is equal to

Differentiate the substitution fully

With

t=2x+3y

\frac{dt}{dx}=2+3\frac{dy}{dx}

, not

3\frac{dy}{dx}

. Dropping the 2 gives a separable equation with the wrong answer.

Drill 5 more on substituting for x + y, and exact differentials

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Separate, integrate, fix the constant
Separable form
$\frac{dy}{dx}=f(x)\,g(y)\ \Rightarrow\ \int\frac{dy}{g(y)}=\int f(x)\,dx+C$
When the slope depends on x alone
Direct integration
$\frac{dy}{dx}=f(x)\ \Rightarrow\ y=\int f(x)\,dx+C$
Substituting for x + y, and exact differentials
Substitution for a linear combination
$t=ax+by+c\ \Rightarrow\ \frac{dt}{dx}=a+b\,\frac{dy}{dx}$

Watch out for (3)

Constant before exponentiating
→ Separate, integrate, fix the constant
Two conditions for a second-order equation
→ When the slope depends on x alone
Differentiate the substitution fully
→ Substituting for x + y, and exact differentials

Test yourself on Differential Equations

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Differential Equations chapter test Drill all 31 questions