JEE Mains Maths · Differential Equations

Equations Reducible to Linear

Equations that become linear after a change of viewpoint: treating x as the unknown function of y, dividing a Bernoulli equation by a power of y, or substituting for a function of y such as tan y or e to the power sin y.

All Differential Equations notes JEE Mains Maths strategy

Why this matters

Nineteen PYQs. None is linear in y as printed, and each becomes the previous pages' routine after one move. Seven are linear in x once dx/dy is taken as the unknown; five are Bernoulli; seven hide a function of y whose derivative sits next to dy/dx. Three ideas cover the page.

Concept 1 of 3: Linear in x: dx/dy + P(y)x = Q(y)

y

appears in awkward places but

x

appears only to the first power, flip the derivative. Write the equation as

\frac{dx}{dy}+P(y)\,x=Q(y)

; the integrating factor is

e^{\int P\,dy}

and everything is integrated in

y

. The given point is still

(x,y)

, so read it the right way round.

Definition

$\frac{dx}{dy}+P(y)x=Q(y)\Rightarrow x\,e^{\int P\,dy}=\int Q\,e^{\int P\,dy}\,dy+C$ .
Signal: terms like $\tan^{-1}y$ , $y^3$ , $e^y$ multiplying $dx$ or standing alone, with $x$ linear.
$y^2\,dx+\left(x-\frac1y\right)dy=0\Rightarrow\frac{dx}{dy}+\frac{x}{y^2}=\frac{1}{y^3}$ .

Linear in x

\frac{dx}{dy}+P(y)\,x=Q(y)\ \Rightarrow\ x\,e^{\int P\,dy}=\int Q\,e^{\int P\,dy}\,dy+C

Worked example

Solve

(x+y^2)\,dy=y\,dx

for

x=x(y)

with

x(1)=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 June 2022 · Q160Moderate

Example 1 · Differential Equations · Equations Reducible to Linear

If the solution curve of the differential equation

\left( \left( \tan^{- 1}y \right)- x \right)dy =\left( 1 +y^{2} \right)dx

passes through the point

(1,0)

then the abscissa of the point on the curve whose ordinate is

\tan(1)

is:

Integrate in y throughout

Once

x

is the unknown, every integral is with respect to

y

, including the integrating factor. Writing

e^{\int P\,dx}

out of habit mixes the variables.

Drill 6 more on linear in x: dx/dy + p(y)x = q(y)

Concept 2 of 3: Bernoulli equations

An equation

\frac{dy}{dx}+Py=Qy^n

is linear except for the

y^n

. Divide by

y^n

and put

z=y^{1-n}

: then

\frac{dz}{dx}=(1-n)y^{-n}\frac{dy}{dx}

and the equation becomes linear in

z

. The commonest case is

n=2

, with

z=\frac1y

Definition

$y'+Py=Qy^n$ : put $z=y^{1-n}$ , giving $z'+(1-n)Pz=(1-n)Q$ .
$n=2$ : $z=\frac1y$ , $z'-Pz=-Q$ .
$n=3$ : $z=\frac1{y^2}$ , $z'-2Pz=-2Q$ .
Shift first if needed: $y'=(y+1)(\dots)$ suggests $Y=y+1$ .

Bernoulli substitution

y'+Py=Qy^n,\ z=y^{1-n}\ \Rightarrow\ z'+(1-n)P\,z=(1-n)Q

Worked example

Solve

\frac{dy}{dx}+\frac yx=y^2

with

y(1)=1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 4 · Q90Moderate

Example 2 · Differential Equations · Equations Reducible to Linear

If the curve

y=y(x)

is the solution of the differential equation

(2xy^2-y)\,dx + x\,dy = 0

which passes through the point of intersection of the lines

2x-3y=1

and

3x+2y=8

, then

|y(1)|

is equal to

The factor 1 - n

With

z=y^{-1}

z'=-y^{-2}y'

: the minus sign flips both

P

and

Q

. Forgetting it gives an integrating factor of the wrong sign.

Drill 4 more on bernoulli equations

Concept 3 of 3: Substituting for a function of y

Look for a function of

y

whose derivative already multiplies

\frac{dy}{dx}

. In

\sec^2y\,y'+P\tan y=Q

, put

u=\tan y

: the equation is linear in

u

. The same works for

u=\sin y

\cos y

e^{y}

e^{-y}

e^{\sin y}

x\ln x

— divide by whatever makes the derivative appear.

Definition

$u=\tan y$ : $u'=\sec^2y\,y'$ ; divide by $\cos^2y$ first.
$u=\cos y$ : $u'=-\sin y\,y'$ .
$u=e^{-y}$ : $u'=-e^{-y}y'$ .
$u=e^{\sin y}$ : $u'=e^{\sin y}\cos y\,y'$ .

Substitution

f'(y)\,\frac{dy}{dx}+P(x)\,f(y)=Q(x),\ u=f(y)\ \Rightarrow\ \frac{du}{dx}+Pu=Q

Worked example

Solve

\cos y\,\frac{dy}{dx}+\sin y=1

with

y(0)=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 8 April 2024 · Q160Moderate

Example 3 · Differential Equations · Equations Reducible to Linear

Let

y = y(x)

be the solution curve of the differential equation secy

\frac{dy}{dx}+ 2xsiny =x^{3}\cos y

y(1) = 0

. Then

y(\sqrt{3})

is equal to:

sin 2y becomes 2 tan y

After dividing by

\cos^2y

\sin2y

becomes

2\tan y

, not

\tan y

. The factor 2 goes straight into the integrating factor.

Drill 6 more on substituting for a function of y

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)

Linear in x: dx/dy + P(y)x = Q(y)
Linear in x
$\frac{dx}{dy}+P(y)\,x=Q(y)\ \Rightarrow\ x\,e^{\int P\,dy}=\int Q\,e^{\int P\,dy}\,dy+C$
Bernoulli equations
Bernoulli substitution
$y'+Py=Qy^n,\ z=y^{1-n}\ \Rightarrow\ z'+(1-n)P\,z=(1-n)Q$
Substituting for a function of y
Substitution
$f'(y)\,\frac{dy}{dx}+P(x)\,f(y)=Q(x),\ u=f(y)\ \Rightarrow\ \frac{du}{dx}+Pu=Q$

Watch out for (3)

Integrate in y throughout
→ Linear in x: dx/dy + P(y)x = Q(y)
The factor 1 - n
→ Bernoulli equations
sin 2y becomes 2 tan y
→ Substituting for a function of y

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 19 questions