Differential Equations for JEE: All Nine Methods, Worked Examples

Differential equations is the chapter where marks come from recognition, not from technique. Every equation JEE sets is solved by one of nine standard moves, and all nine are short. What is actually being tested is whether you can look at an equation and name the move inside ten seconds.

So this note is a catalogue. Each type gets the shape that triggers it, the substitution or factor it wants, and a recent JEE question worked in full.

What a differential equation is

An equation relating an unknown function to one or more of its derivatives is a differential equation. If the unknown depends on a single independent variable the equation is ordinary; if partial derivatives appear it is partial. Only ordinary equations are in the JEE syllabus.

So $2x\,y' - 3y = 5$ and $4y'' + (y')^3 = 0$ are differential equations, while $x^2 + 2x + y^2 = 0$ is not: no derivative appears in it.

A differential equation almost never has one solution. Solving

$$\frac{dy}{dx} = 2x$$

gives $y = x^2 + C$, one curve for every value of $C$, and each of them has the prescribed slope at every point. One extra condition, a point the curve must pass through, picks out one member.

A one-parameter family of solution curves, with the member through (0,2) picked out

Order and degree

Worked example

Find the order and degree of

$$\sqrt[3]{\frac{d^2y}{dx^2}} = \sqrt[4]{\frac{dy}{dx} + 5}$$

▸ SHOW SOLUTION

Step 1. The highest derivative present is $\dfrac{d^2y}{dx^2}$, so the order is $2$.

Step 2. The degree cannot be read yet, because both sides carry fractional powers. Raise both sides to the twelfth power, the LCM of $3$ and $4$:

$$\left(\frac{d^2y}{dx^2}\right)^{4} = \left(\frac{dy}{dx} + 5\right)^{3}$$

Step 3. Now the equation is a polynomial in the derivatives, and $\dfrac{d^2y}{dx^2}$ appears to the power $4$.

Answer: order $2$, degree $4$.

Forming the equation: eliminating the constants

The reverse problem. You are given a family of curves and asked for the differential equation it satisfies.

Working rule. For a family containing $n$ independent arbitrary constants:

1. Count the constants; call the number $n$.
2. Differentiate the relation $n$ times with respect to $x$.
3. Eliminate the $n$ constants using the $n+1$ equations now available.

The result has order exactly $n$.

Why exactly $n$. Each differentiation produces one new equation, so after $n$ of them you hold $n+1$ equations in $n$ unknown constants. Eliminating $n$ unknowns from $n+1$ equations leaves one relation, and the highest derivative it can contain is the $n$th. It cannot contain fewer, because $n$ equations are genuinely needed to remove $n$ independent constants.

Worked example

Find the differential equation of the family $y = C_1 e^{C_2 x}$, where $C_1$ and $C_2$ are arbitrary.

▸ SHOW SOLUTION

Step 1. Two constants, so expect order $2$.

Step 2. Take logarithms first, which is what removes the product rule:

$$\log_e y = \log_e C_1 + C_2 x$$

Step 3. Differentiate:

$$\frac{y'}{y} = C_2$$

so $C_1$ has already gone.

Step 4. Differentiate $y' = C_2 y$ again to get $y'' = C_2 y'$, and substitute $C_2 = y'/y$:

$$y\,y'' = \left(y'\right)^{2}$$

Answer: $y\,y'' = (y')^2$.

Two families worth knowing, because they recur:

Type 1 — Variable separable

The shape. Every $x$ can be pushed to one side and every $y$ to the other:

$$f(x)\,dx + g(y)\,dy = 0$$

Then integrate each term on its own.

Why it works. Write the equation as $g(y)\,y' = -f(x)$ and integrate both sides with respect to $x$. On the left, the substitution rule turns $\int g(y)\,y'\,dx$ into $\int g(y)\,dy$, because $y$ is a function of $x$. Moving everything to one side introduces the single constant $C$.

The slope field of dy/dx = −x/y, with three circular solution curves

Worked example 1 · JEE Main 2026, 2 April Shift 1

The curve $y = f(x)$ passes through $(1,e)$ and satisfies $dy = y\,(2 + \log_e x)\,dx$ for $x > 0$. Find $f(e)$.

▸ SHOW SOLUTION

Step 1. The $y$ stands alone against a function of $x$, so divide by $y$ and separate:

$$\frac{dy}{y} = \left(2 + \log_e x\right) dx$$

Step 2. On the right, $\displaystyle\int \log_e x\,dx = x\log_e x - x$ by parts, so

$$\log_e y = 2x + x\log_e x - x + C = x + x\log_e x + C$$

Step 3. At $(1,e)$: $\log_e e = 1$ and $\log_e 1 = 0$, so $1 = 1 + 0 + C$ and $C = 0$.

Step 4. Hence $\log_e y = x + x\log_e x$, and at $x = e$,

$$f(e) = e^{\,e + e} = e^{2e}$$

Answer: $e^{2e}$.

Worked example 2 · JEE Main 2026, 4 April Shift 1

Let $y = y(x)$ solve $\dfrac{dy}{dx} = (1 + x + x^2)(1 - y + y^2)$ with $y(0) = \dfrac12$. Find $2y(1) - 1$.

▸ SHOW SOLUTION

Step 1. The right side is already a product of a function of $x$ and a function of $y$, so separate directly:

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

Step 2. Complete the square in the denominator:

$$y^2 - y + 1 = \left(y - \tfrac12\right)^{2} + \tfrac34$$

Step 3. With $u = y - \tfrac12$ and $a = \dfrac{\sqrt3}{2}$ the standard integral $\displaystyle\int \frac{du}{u^2 + a^2} = \frac1a \tan^{-1}\frac ua$ gives

$$\frac{2}{\sqrt3}\tan^{-1}\!\left(\frac{2y-1}{\sqrt3}\right) = \frac{x^3}{3} + \frac{x^2}{2} + x + C$$

Step 4. At $x = 0$, $y = \tfrac12$, so the left side is $\tan^{-1}0 = 0$ and $C = 0$.

Step 5. At $x = 1$ the right side is $\dfrac13 + \dfrac12 + 1 = \dfrac{11}{6}$, so

$$\tan^{-1}\!\left(\frac{2y-1}{\sqrt3}\right) = \frac{\sqrt3}{2}\cdot\frac{11}{6} = \frac{11\sqrt3}{12}$$

Step 6. Therefore

$$2y(1) - 1 = \sqrt3\,\tan\!\left(\frac{11\sqrt3}{12}\right)$$

Answer: $\sqrt3\tan\left(\dfrac{11\sqrt3}{12}\right)$.

Type 2 — Reducible to separable

The shape. The same linear combination of $x$ and $y$ appears throughout:

$$\frac{dy}{dx} = f(ax + by + c)$$

The move. Put $z = ax + by + c$. Then $\dfrac{dz}{dx} = a + b\dfrac{dy}{dx} = a + b\,f(z)$, which separates:

$$\frac{dz}{a + b\,f(z)} = dx$$

Every step is reversible as long as $a + b\,f(z) \neq 0$.

Worked example · JEE Main 2026, 22 January Shift 2

Let $y = y(x)$ satisfy

$$16\sqrt{x}\,\sqrt{9 + \sqrt{x}}\left(4 + \sqrt{9 + \sqrt{x}}\right)\cos y\,dy = \left(1 + 2\sin y\right) dx$$

for $x > 0$, with $y(256) = \dfrac{\pi}{2}$. If $y(49) = \alpha$, find $2\sin\alpha$.

▸ SHOW SOLUTION

Step 1. Separate:

$$\frac{\cos y\,dy}{1 + 2\sin y} = \frac{dx}{16\sqrt{x}\,\sqrt{9+\sqrt{x}}\left(4 + \sqrt{9+\sqrt{x}}\right)}$$

Step 2. The left side integrates at once, since $\dfrac{d}{dy}\left(1 + 2\sin y\right) = 2\cos y$:

$$\int \frac{\cos y\,dy}{1 + 2\sin y} = \frac12 \log_e\left|1 + 2\sin y\right|$$

Step 3. For the right side put $t = 4 + \sqrt{9 + \sqrt{x}}$, the whole outer bracket. Then

$$\frac{dt}{dx} = \frac{1}{2\sqrt{9+\sqrt x}}\cdot\frac{1}{2\sqrt x} = \frac{1}{4\sqrt x\,\sqrt{9+\sqrt x}}$$

so $\dfrac{dx}{\sqrt x\sqrt{9+\sqrt x}} = 4\,dt$ and the right side becomes $\displaystyle\int\frac{4\,dt}{16\,t} = \frac14\log_e t$.

Step 4. Therefore

$$\frac12 \log_e\left(1 + 2\sin y\right) = \frac14 \log_e t + c$$

Step 5. Apply $y(256) = \dfrac{\pi}{2}$. Working outwards: $\sqrt{256} = 16$, then $\sqrt{9+16} = 5$, then $t = 9$; and $1 + 2\sin\dfrac{\pi}{2} = 3$. So $\dfrac12\log_e 3 = \dfrac14\log_e 9 + c = \dfrac12\log_e 3 + c$, giving $c = 0$ and

$$1 + 2\sin y = \sqrt{t}$$

Step 6. At $x = 49$: $\sqrt{49} = 7$, then $\sqrt{9+7} = 4$, then $t = 8$. Hence $1 + 2\sin\alpha = \sqrt8 = 2\sqrt2$.

Answer: $2\sin\alpha = 2\sqrt2 - 1$.

Two more of this type, worth recognising by sight:

Type 3 — Exact differentials

The shape. A group of terms is already the differential of something. This is the fastest method in the chapter when it applies, and the slowest to find if you have not memorised the list.

The identities worth knowing by sight

Worked example 1 · JEE Main 2026, 22 January Shift 1

Let the solution curve of $x\,dy - y\,dx = \sqrt{x^2+y^2}\,dx$ for $x > 0$ with $y(1) = 0$ be $y = y(x)$. Find $y(3)$.

▸ SHOW SOLUTION

Step 1. Divide by $x^2$ so the left side becomes an exact differential. Using $\sqrt{x^2+y^2} = x\sqrt{1 + (y/x)^2}$ for $x > 0$,

$$d\!\left(\frac yx\right) = \sqrt{1 + \left(\frac yx\right)^{2}}\;\frac{dx}{x}$$

Step 2. With $v = \dfrac yx$ this is $\dfrac{dv}{\sqrt{1+v^2}} = \dfrac{dx}{x}$, now separable.

Step 3. Integrate:

$$\log_e\!\left(v + \sqrt{1+v^2}\right) = \log_e x + c$$

so $v + \sqrt{1+v^2} = Kx$.

Step 4. At $x = 1$, $y = 0$ gives $v = 0$, so $0 + 1 = K$ and $v + \sqrt{1+v^2} = x$.

Step 5. Rationalise: $\sqrt{1+v^2} = x - v$, and squaring gives $1 = x^2 - 2xv$, so $v = \dfrac{x^2-1}{2x}$ and

$$y = vx = \frac{x^2 - 1}{2}$$

Step 6. Hence $y(3) = \dfrac{9-1}{2} = 4$.

Answer: $4$.

Worked example 2 · JEE Main 2023, 11 April Shift I

Let $y = y(x)$ satisfy $\left(1 - x^2y^2\right)dx = y\,dx + x\,dy$. The line $x = 1$ meets the curve at $y = 2$ and the line $x = 2$ meets it at $y = \alpha$. Find $\alpha$.

▸ SHOW SOLUTION

Step 1. The right side is exactly $d(xy)$, so with $u = xy$ the equation is $\left(1 - u^2\right)dx = du$.

Step 2. Separate and integrate:

$$\frac12 \log_e\left|\frac{1+u}{1-u}\right| = x + c$$

Step 3. At $x = 1$, $y = 2$, so $u = 2$ and $\dfrac12\log_e 3 = 1 + c$, giving $c = \dfrac12\log_e 3 - 1$.

Step 4. At $x = 2$ write $u = 2\alpha$. Then

$$\frac12 \log_e\left|\frac{1+2\alpha}{1-2\alpha}\right| = 1 + \frac12\log_e 3$$

Step 5. So the modulus equals $3e^{2}$, and since $u > 1$ the quantity inside is negative:

$$\frac{1+2\alpha}{1-2\alpha} = -3e^{2}$$

Step 6. Solving, $\alpha\left(2 - 6e^2\right) = -\left(3e^2 + 1\right)$, so

$$\alpha = \frac{1 + 3e^{2}}{2\left(3e^{2} - 1\right)}$$

Answer: $\dfrac{1+3e^2}{2(3e^2-1)}$.

Type 4 — Homogeneous

The shape. The whole right side can be written in terms of $\dfrac yx$ alone:

$$\frac{dy}{dx} = F\!\left(\frac yx\right)$$

Equivalently, the equation is a ratio of two expressions homogeneous of the same degree.

The move. Put $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$, and the equation becomes

$$\frac{dv}{F(v) - v} = \frac{dx}{x}$$

which is separable.

Along each ray through the origin the prescribed slope is the same

Worked example · JEE Main 2026, 5 April Shift 1

Let $y = y(x)$ solve

$$x\sin\!\left(\frac yx\right)dy = \left(y\sin\!\left(\frac yx\right) - x\right)dx$$

with $y(1) = \dfrac{\pi}{2}$, and let $\alpha = \cos\!\left(\dfrac{y(e^{12})}{e^{12}}\right)$. Find $\alpha$.

▸ SHOW SOLUTION

Step 1. Every occurrence of $y$ sits inside $\dfrac yx$ or is multiplied by it, so put $y = vx$ and $dy = v\,dx + x\,dv$.

Step 2. Substituting,

$$x\sin v\left(v\,dx + x\,dv\right) = \left(vx\sin v - x\right) dx$$

Step 3. The term $vx\sin v\,dx$ appears on both sides and cancels, leaving $x^2\sin v\,dv = -x\,dx$.

Step 4. Separate: $\sin v\,dv = -\dfrac{dx}{x}$, so $-\cos v = -\log_e x + c$, that is

$$\cos\!\left(\frac yx\right) = \log_e x - c$$

Step 5. At $x = 1$, $v = \dfrac{\pi}{2}$: $\cos\dfrac{\pi}{2} = 0 = 0 - c$, so $c = 0$ and $\cos\left(\dfrac yx\right) = \log_e x$.

Step 6. Putting $x = e^{12}$ gives $\alpha = \log_e e^{12} = 12$.

Answer: $\alpha = 12$.

Two standard homogeneous results worth carrying:

Type 5 — Reducible to homogeneous

The shape. A ratio of two linear expressions, each carrying a constant term:

$$\frac{dy}{dx} = \frac{ax + by + c}{dx + ey + f}$$

The constants $c$ and $f$ are what stop it being homogeneous, so remove them by moving the origin to the point where the two lines meet.

The move, when $ae \neq bd$. Put $x = X + h$ and $y = Y + k$, choosing $h$ and $k$ so that

$$ah + bk + c = 0 \qquad\text{and}\qquad dh + ek + f = 0$$

Because the shift is by constants, $\dfrac{dY}{dX} = \dfrac{dy}{dx}$, and the equation becomes $\dfrac{dY}{dX} = \dfrac{aX + bY}{dX + eY}$, homogeneous. The two linear conditions have a unique solution exactly when $ae - bd \neq 0$.

Worked example · JEE Main 2024, 27 January Shift I

The solution of $(2x + 3y - 2)\,dx + (4x + 6y - 7)\,dy = 0$ with $y(0) = 3$ is

$$\alpha x + \beta y + 3\log_e\left|2x + 3y - \gamma\right| = 6$$

Find $\alpha + 2\beta + 3\gamma$.

▸ SHOW SOLUTION

Step 1. Here $4x + 6y = 2(2x+3y)$, so the coefficients are proportional and the shift method does not apply. Put $z = 2x + 3y$ instead.

Step 2. Then $\dfrac{dz}{dx} = 2 + 3\dfrac{dy}{dx}$, and the equation gives $\dfrac{dy}{dx} = -\dfrac{z-2}{2z-7}$, so

$$\frac{dz}{dx} = \frac{2(2z-7) - 3(z-2)}{2z-7} = \frac{z-8}{2z-7}$$

Step 3. Separate and divide out, using $\dfrac{2z-7}{z-8} = 2 + \dfrac{9}{z-8}$:

$$\left(2 + \frac{9}{z-8}\right) dz = dx$$

Step 4. Integrate:

$$2z + 9\log_e\left|z - 8\right| = x + c$$

Step 5. Restore $z = 2x + 3y$ and collect:

$$3x + 6y + 9\log_e\left|2x + 3y - 8\right| = c$$

Step 6. At $x = 0$, $y = 3$: $z - 8 = 1$, so the logarithm vanishes and $c = 18$. Dividing throughout by $3$,

$$x + 2y + 3\log_e\left|2x + 3y - 8\right| = 6$$

Step 7. Comparing, $\alpha = 1$, $\beta = 2$, $\gamma = 8$, so $\alpha + 2\beta + 3\gamma = 1 + 4 + 24 = 29$.

Answer: $29$.

The decision in one line:

Type 6 — Linear

More JEE questions come from this one type than from any other in the chapter. The method never changes; the only work is putting the equation into the standard shape so that $P$ and $Q$ can be read off.

The shape.

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

The move. Multiply by the integrating factor

$$\text{I.F.} = e^{\int P\,dx}$$

which makes the left side an exact derivative, and then

$$y \cdot \text{I.F.} = \int Q \cdot \text{I.F.}\;dx + C$$

Why it works. Write $\mu = e^{\int P\,dx}$, so $\mu' = P\mu$. Then $\dfrac{d}{dx}(\mu y) = \mu y' + \mu' y = \mu\left(y' + Py\right) = \mu Q$, and one integration finishes it. Any antiderivative of $P$ will do: replacing $\int P\,dx$ by $\int P\,dx + k$ multiplies $\mu$ by the constant $e^{k}$, which cancels from both sides.

Solution curves of a linear equation never cross; the initial condition picks one

Integrating factors you should never have to compute

Worked example 1 · JEE Main 2026, 23 January Shift 1

Let $y = y(x)$ solve $x^4\,dy + \left(4x^3y + 2\sin x\right)dx = 0$ for $x > 0$, with $y\!\left(\dfrac{\pi}{2}\right) = 0$. Find $\pi^4 y\!\left(\dfrac{\pi}{3}\right)$.

▸ SHOW SOLUTION

Step 1. Group the two terms carrying $y$:

$$x^4\,dy + 4x^3y\,dx = -2\sin x\,dx$$

Step 2. The left side is already exact, because $d\left(x^4 y\right) = x^4\,dy + 4x^3y\,dx$.

Step 3. Integrate: $x^4 y = 2\cos x + c$.

Step 4. At $x = \dfrac{\pi}{2}$, $y = 0$, so $c = 0$ and $x^4y = 2\cos x$.

Step 5. At $x = \dfrac{\pi}{3}$: $\left(\dfrac{\pi}{3}\right)^{4} y = 2\cos\dfrac{\pi}{3} = 1$, and $\left(\dfrac{\pi}{3}\right)^{4} = \dfrac{\pi^4}{81}$.

Answer: $\pi^4 y\!\left(\dfrac{\pi}{3}\right) = 81$.

Worked example 2 · JEE Main 2026, 21 January Shift 1

Let $y = y(x)$ be the solution curve of $\left(1+x^2\right)dy + \left(y - \tan^{-1}x\right)dx = 0$ with $y(0) = 1$. Find $y(1)$.

▸ SHOW SOLUTION

Step 1. Divide by $\left(1+x^2\right)dx$:

$$\frac{dy}{dx} + \frac{1}{1+x^2}\,y = \frac{\tan^{-1}x}{1+x^2}$$

Step 2. Here $P = \dfrac{1}{1+x^2}$, so I.F. $= e^{\tan^{-1}x}$ and

$$\frac{d}{dx}\!\left(y\,e^{\tan^{-1}x}\right) = e^{\tan^{-1}x}\cdot\frac{\tan^{-1}x}{1+x^2}$$

Step 3. Put $t = \tan^{-1}x$, so $dt = \dfrac{dx}{1+x^2}$ and the right side integrates by parts:

$$\int t\,e^{t}\,dt = e^{t}\left(t-1\right)$$

Step 4. Hence $y = \tan^{-1}x - 1 + c\,e^{-\tan^{-1}x}$.

Step 5. At $x = 0$: $1 = 0 - 1 + c$, so $c = 2$.

Step 6. At $x = 1$, with $\tan^{-1}1 = \dfrac{\pi}{4}$,

$$y(1) = \frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1$$

Answer: $\dfrac{2}{e^{\pi/4}} + \dfrac{\pi}{4} - 1$.

Worked example 3 · JEE Main 2026, 6 April Shift 1

Let $y = y(x)$ solve $\left(x^2 - x\sqrt{x^2-1}\right)dy + \left(y\left(x - \sqrt{x^2-1}\right) - x\right)dx = 0$ for $x \ge 1$, with $y(1) = 1$. Find $\left\lfloor y\!\left(\sqrt5\right)\right\rfloor$.

▸ SHOW SOLUTION

Step 1. Factor the coefficient of $dy$ as $x\left(x - \sqrt{x^2-1}\right)$ and divide throughout:

$$\frac{dy}{dx} + \frac yx = \frac{1}{x - \sqrt{x^2-1}}$$

Step 2. Rationalise. Multiplying above and below by $x + \sqrt{x^2-1}$ leaves the denominator $x^2 - (x^2-1) = 1$, so

$$\frac{dy}{dx} + \frac yx = x + \sqrt{x^2-1}$$

Step 3. I.F. $= x$, so $\dfrac{d}{dx}(xy) = x^2 + x\sqrt{x^2-1}$.

Step 4. Integrate, using $u = x^2 - 1$ in the second piece:

$$xy = \frac{x^{3}}{3} + \frac{\left(x^2-1\right)^{3/2}}{3} + C$$

Step 5. At $x = 1$, $y = 1$: $1 = \dfrac13 + 0 + C$, so $C = \dfrac23$.

Step 6. At $x = \sqrt5$: $x^3 = 5\sqrt5$ and $\left(x^2-1\right)^{3/2} = 8$, so

$$y\!\left(\sqrt5\right) = \frac{5}{3} + \frac{2\sqrt5}{3} \approx 3.16$$

Answer: $\left\lfloor y(\sqrt5)\right\rfloor = 3$.

Type 7 — Linear in $x$ instead of in $y$

The shape. The equation is not linear in $y$, but $x$ appears only to the first power once you treat $x$ as the unknown function of $y$:

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

Everything from Type 6 then applies with the roles swapped, and I.F. $= e^{\int P(y)\,dy}$.

Worked example · Spectrum Master Strokes

Solve $\left(1+y^2\right) + \left(x - e^{\tan^{-1}y}\right)\dfrac{dy}{dx} = 0$.

▸ SHOW SOLUTION

Step 1. The unknown $x$ appears to the first power while $y$ sits inside $\tan^{-1}y$ and $1+y^2$, so treat $x$ as the function. Multiply by $\dfrac{dx}{dy}$:

$$\left(1+y^2\right)\frac{dx}{dy} + x - e^{\tan^{-1}y} = 0$$

Step 2. Divide by $1 + y^2$ to reach standard form in $x$:

$$\frac{dx}{dy} + \frac{x}{1+y^{2}} = \frac{e^{\tan^{-1}y}}{1+y^{2}}$$

Step 3. $P(y) = \dfrac{1}{1+y^2}$, so I.F. $= e^{\tan^{-1}y}$ and

$$\frac{d}{dy}\!\left(x\,e^{\tan^{-1}y}\right) = \frac{e^{2\tan^{-1}y}}{1+y^{2}}$$

Step 4. With $t = \tan^{-1}y$ the right side is $e^{2t}$, whose integral is $\dfrac{e^{2t}}{2}$.

Answer: $2x\,e^{\tan^{-1}y} = e^{2\tan^{-1}y} + C$.

Three more that only work this way round:

Type 8 — Bernoulli, and other reductions to linear

The shape.

$$\frac{dy}{dx} + P(x)\,y = Q(x)\,y^{n}, \qquad n \neq 0,\,1$$

The move. Divide by $y^{n}$ and put $v = y^{1-n}$. Since $v' = (1-n)y^{-n}y'$, the equation becomes

$$\frac{dv}{dx} + (1-n)P(x)\,v = (1-n)Q(x)$$

which is linear in $v$. The excluded cases need no work at all: $n = 0$ is already linear, $n = 1$ is separable.

The wider family: anything whose derivative you can see

Bernoulli is one case of a general habit. Whenever the equation contains a block $t(y)$ together with $t'(y)\dfrac{dy}{dx}$, put $t$ equal to that block.

Worked example 1 · JEE Main 2026, 28 January Shift 1

Let $y = y(x)$ solve $x\dfrac{dy}{dx} - \sin 2y = x^3\left(2 - x^3\right)\cos^2 y$ for $x \neq 0$, with $y(2) = 0$. Find $\tan\left(y(1)\right)$.

▸ SHOW SOLUTION

Step 1. Use $\sin 2y = 2\sin y\cos y = 2\tan y\cos^2 y$ so that every term carries $\cos^2 y$:

$$x\frac{dy}{dx} - 2\tan y\cos^{2}y = x^{3}\left(2-x^{3}\right)\cos^{2}y$$

Step 2. Divide by $x\cos^2 y$:

$$\sec^{2}y\,\frac{dy}{dx} - \frac{2\tan y}{x} = x^{2}\left(2 - x^{3}\right)$$

Step 3. The combination $\sec^2 y\,y'$ is the derivative of $\tan y$, so put $t = \tan y$:

$$\frac{dt}{dx} - \frac{2}{x}\,t = x^{2}\left(2-x^{3}\right)$$

now linear in $t$.

Step 4. $P = -\dfrac2x$ gives I.F. $= \dfrac{1}{x^2}$, so $\dfrac{d}{dx}\!\left(\dfrac{t}{x^2}\right) = 2 - x^3$.

Step 5. Integrate:

$$\tan y = 2x^{3} - \frac{x^{6}}{4} + Cx^{2}$$

Step 6. At $x = 2$, $y = 0$ so $\tan y = 0$: $0 = 16 - 16 + 4C$, giving $C = 0$.

Step 7. At $x = 1$: $\tan\left(y(1)\right) = 2 - \dfrac14 = \dfrac74$.

Answer: $\dfrac74$.

Worked example 2 · JEE Main 2026, 2 April Shift 2

Let $x = x(y)$ solve $2y^2\dfrac{dx}{dy} - 2xy + x^2 = 0$ for $y > 1$, with $x(e) = e$. Find $x\!\left(e^2\right)$.

▸ SHOW SOLUTION

Step 1. The unknown here is $x$, and $x$ appears squared, so this is Bernoulli in $x$. Divide by $2y^2$:

$$\frac{dx}{dy} - \frac xy = -\frac{x^{2}}{2y^{2}}, \qquad n = 2$$

Step 2. Divide by $x^2$ and put $v = x^{-1}$, so $\dfrac{dv}{dy} = -x^{-2}\dfrac{dx}{dy}$:

$$\frac{dv}{dy} + \frac vy = \frac{1}{2y^{2}}$$

Step 3. I.F. $= y$, so $\dfrac{d}{dy}(vy) = \dfrac{1}{2y}$ and $vy = \dfrac12\log_e y + c$.

Step 4. Restore $v = \dfrac1x$: $\dfrac yx = \dfrac12\log_e y + c$.

Step 5. At $y = e$, $x = e$: $1 = \dfrac12 + c$, so $c = \dfrac12$ and $\dfrac yx = \dfrac{\log_e y + 1}{2}$.

Step 6. At $y = e^2$: $\dfrac{e^2}{x} = \dfrac32$, so $x = \dfrac23 e^{2}$.

Answer: $\dfrac23 e^{2}$.

Type 9 — Curves: tangents, normals and orthogonal trajectories

Every question here is a translation exercise. A sentence about a tangent, a normal, an intercept or an area becomes a differential equation, and from there the work is one of the eight methods above.

The dictionary

Let $P(x,y)$ lie on $y = f(x)$ and write $y'$ for the slope there. Use $X, Y$ for the running point on the tangent.

Tangent, normal, subtangent and subnormal at a point of a curve

Worked example · a tangent condition

A curve $y = f(x)$ passes through $(1,1)$, and at every point the $y$-intercept of the tangent equals the cube of the abscissa of the point of contact. Find $f(-3)$.

▸ SHOW SOLUTION

Step 1. From the dictionary the $y$-intercept is $y - xy'$, so the condition is $y - xy' = x^{3}$.

Step 2. Divide by $-x$ to reach standard linear form:

$$\frac{dy}{dx} - \frac yx = -x^{2}$$

Step 3. $P = -\dfrac1x$, so I.F. $= \dfrac1x$ and $\dfrac{d}{dx}\!\left(\dfrac yx\right) = -x$.

Step 4. Integrate: $\dfrac yx = -\dfrac{x^{2}}{2} + C$.

Step 5. At $(1,1)$: $C = \dfrac32$, so $y = \dfrac{3x}{2} - \dfrac{x^{3}}{2}$.

Step 6. Hence $f(-3) = -\dfrac92 + \dfrac{27}{2} = 9$.

Answer: $9$.

Orthogonal trajectories

Given a family with one arbitrary constant:

1. Form its differential equation, eliminating the constant, to get $\dfrac{dy}{dx} = F(x,y)$.
2. Replace $\dfrac{dy}{dx}$ by $-\dfrac{1}{\,dy/dx\,}$, giving $\dfrac{dy}{dx} = -\dfrac{1}{F(x,y)}$.
3. Solve. Its solutions cut every member of the original family at right angles.

Why. At a point of intersection the two curves meet at right angles exactly when $m_1m_2 = -1$. Step 1 produces $m_1 = F(x,y)$ as a function of position alone, because the constant has been eliminated, so the slope of whichever member passes through $(x,y)$ is known from the point itself. Substituting $m_2$ for $m_1$ therefore describes exactly the curves perpendicular at every point they meet.

Parabolas y² = 4ax and their orthogonal family of ellipses 2x² + y² = C

Worked example · the classical pair

Find the orthogonal trajectories of the family of parabolas $y^2 = 4ax$.

▸ SHOW SOLUTION

Step 1. Differentiating, $2yy' = 4a$, so $4a = 2yy'$.

Step 2. Substitute back into $y^2 = 4ax$: $y^2 = 2xyy'$, and cancelling one $y$,

$$\frac{dy}{dx} = \frac{y}{2x}$$

Step 3. Replace $y'$ by $-\dfrac{1}{y'}$:

$$\frac{dy}{dx} = -\frac{2x}{y}$$

Step 4. Separate: $y\,dy = -2x\,dx$, so $\dfrac{y^{2}}{2} = -x^{2} + c$.

Answer: $2x^{2} + y^{2} = C$, a family of ellipses centred at the origin with major axis along $Oy$ and $a^2 : b^2 = 1 : 2$.

Three standard orthogonal pairs:

Growth, decay and modelling

These questions hand you the differential equation in words. Once written down it is always separable or linear, so the marks sit entirely in the translation and in the two conditions used to fix the constants.

The two standard laws

Exponential growth and decay. If a quantity changes at a rate proportional to its own size,

$$\frac{dN}{dt} = kN \qquad\Longrightarrow\qquad N(t) = N_0 e^{kt}$$

with $N_0 = N(0)$. Growth means $k > 0$, decay means $k < 0$.

Newton's law of cooling. A body at temperature $T(t)$ in surroundings held at $T_s$ satisfies

$$\frac{dT}{dt} = -k\left(T - T_s\right) \qquad\Longrightarrow\qquad T(t) = T_s + \left(T_0 - T_s\right)e^{-kt}$$

The proof is one line: put $u = T - T_s$, note $T_s$ is constant so $u' = T' = -ku$, and apply the first law to $u$.

Newton cooling: the excess over room temperature halves every fifteen minutes

Worked example 1 · cooling

A thermometer reading $160^\circ$F is placed in a room kept at $80^\circ$F. After $15$ minutes it reads $120^\circ$F. What does it read after $45$ minutes?

▸ SHOW SOLUTION

Step 1. Here $T_s = 80$ and $T_0 = 160$, so $T - 80 = 80\,e^{-kt}$.

Step 2. At $t = 15$, $T = 120$: $40 = 80e^{-15k}$, so $e^{-15k} = \dfrac12$.

Step 3. Do not solve for $k$. Since $45 = 3 \times 15$,

$$e^{-45k} = \left(e^{-15k}\right)^{3} = \frac18$$

Step 4. Therefore $T - 80 = 80 \cdot \dfrac18 = 10$.

Answer: $90^\circ$F.

Worked example 2 · a bounded model

In a closed community of $1000$ people, the number $P(t)$ who have heard a piece of news grows at a rate proportional to the product of those who know it and those who do not:

$$\frac{dP}{dt} = kP\left(1000 - P\right)$$

If $P(0) = 100$ and $P(1) = 200$, with $t$ in days, find $P(2)$.

▸ SHOW SOLUTION

Step 1. Separate and use partial fractions:

$$\frac{1}{P(1000-P)} = \frac{1}{1000}\left(\frac1P + \frac{1}{1000-P}\right)$$

Step 2. Integrating,

$$\frac{1}{1000}\log_e\!\left(\frac{P}{1000-P}\right) = kt + c$$

Step 3. Write $u = \dfrac{P}{1000-P}$. The last line says $u = u_0 e^{1000kt}$, so $u$ is multiplied by the same factor $r$ every day: $u(t) = u_0\,r^{\,t}$.

Step 4. Read $u_0$ and $r$ from the data. At $t = 0$, $u_0 = \dfrac{100}{900} = \dfrac19$; at $t = 1$, $u = \dfrac{200}{800} = \dfrac14$, so $r = \dfrac94$.

Step 5. Hence $u(2) = \dfrac19 \cdot \dfrac{81}{16} = \dfrac{9}{16}$.

Step 6. Invert $u$ to get $P = \dfrac{1000u}{1+u}$:

$$P(2) = 1000 \cdot \frac{9/16}{25/16} = 360$$

Answer: $360$ people.

Logistic growth flattens off, while the odds P/(1000−P) stay exponential

When the equation is hidden

At JEE Advanced level the differential equation is rarely handed over. It sits inside a limit, inside an integral identity, or behind a substitution you have to invent. Only after you have unearthed it do the nine methods apply.

Inside a limit

Worked example

Let $f$ be differentiable on $(0,\infty)$ with $f(1) = 2$, and suppose

$$\lim_{t\to x}\frac{t^{10}f(x) - x^{10}f(t)}{t^{9} - x^{9}} = 1 \quad\text{for every } x > 0$$

Find $f(x)$.

▸ SHOW SOLUTION

Step 1. At $t = x$ the numerator is $x^{10}f(x) - x^{10}f(x) = 0$ and the denominator is $0$, so the limit is $\dfrac00$ and L'Hôpital applies with $t$ as the variable.

Step 2. Differentiate top and bottom with respect to $t$, then set $t = x$:

$$\frac{10x^{9}f(x) - x^{10}f'(x)}{9x^{8}} = 1$$

Step 3. Clear the denominator and divide by $x^{10}$:

$$\frac{df}{dx} - \frac{10}{x}f = -\frac{9}{x^{2}}$$

Step 4. I.F. $= x^{-10}$, so $\dfrac{d}{dx}\!\left(\dfrac{f}{x^{10}}\right) = -9x^{-12}$.

Step 5. Integrate: $\dfrac{f}{x^{10}} = \dfrac{9}{11}x^{-11} + C$, that is $f(x) = \dfrac{9}{11x} + Cx^{10}$.

Step 6. Impose $f(1) = 2$: $\dfrac9{11} + C = 2$, so $C = \dfrac{13}{11}$.

Answer: $f(x) = \dfrac{9}{11x} + \dfrac{13x^{10}}{11}$.

Inside an integral

Worked example

Let $f$ be differentiable with $f(1) = \dfrac13$ and

$$3\int_{1}^{x} f(t)\,dt = x\,f(x) - \frac{x^{3}}{3} \quad (x > 0)$$

Find $f(e)$.

▸ SHOW SOLUTION

Step 1. Differentiate both sides. The left gives $3f(x)$; the right gives $f(x) + xf'(x) - x^2$.

Step 2. So $3f = f + xf' - x^2$, that is $xf' - 2f = x^2$.

Step 3. Divide by $x$: $\dfrac{df}{dx} - \dfrac2x f = x$, linear with I.F. $= x^{-2}$.

Step 4. Then $\dfrac{d}{dx}\!\left(\dfrac{f}{x^{2}}\right) = \dfrac1x$, so $\dfrac{f}{x^{2}} = \log_e x + C$.

Step 5. At $x = 1$: $\dfrac13 = 0 + C$, so $f(x) = x^{2}\left(\log_e x + \dfrac13\right)$.

Step 6. Check against the original identity at $x = 1$: the left side is $0$ and the right is $f(1) - \dfrac13 = 0$. Consistent.

Step 7. At $x = e$: $f(e) = e^{2}\left(1 + \dfrac13\right) = \dfrac{4e^{2}}{3}$.

Answer: $\dfrac{4e^{2}}{3}$.

Behind a substitution you invent

Nested radicals telescope. Each substitution is chosen to cancel the very factor it introduces, so start from the innermost and work outwards rather than trying to simplify the whole expression at once. The JEE Advanced equation

$$8\sqrt{x}\,\sqrt{9+\sqrt{x}}\;dy = \frac{dx}{\sqrt{4 + \sqrt{9+\sqrt{x}}}}$$

collapses in three moves: $u = \sqrt x$ kills the outer $\sqrt x$, then $w = \sqrt{9+u}$ kills the next layer, then $s = \sqrt{4+w}$ kills the last, leaving $dy = ds$. With $y(0) = \sqrt7$ the answer is $y = \sqrt{4 + \sqrt{9+\sqrt x}}$, so $y(256) = 3$.

A block substitution does the same job when a combination repeats. If

$$\frac{dy}{dx} = 2x(x+y)^{3} - x(x+y) - 1$$

the isolated $-1$ is the clue: put $u = x + y$, so $u' = 1 + y'$ and the equation becomes $u' = xu\left(2u^{2} - 1\right)$, which is Bernoulli with $n = 3$.

Which method? A decision chart

Run down this list in order and stop at the first "yes". The order matters: it is arranged so that the cheapest test comes first, and so that an equation which admits two methods is caught by the shorter one.

A nine-step decision chart for choosing the method

Where the marks are

Across the nineteen JEE Main 2026 shifts, differential equations produced 14 questions, roughly $0.7$ per shift, so essentially one question in every paper. The split by method:

Two patterns worth planning around:

1. The linear type dominates. Nearly half the 2026 questions were solved by writing the equation in standard form and reading off $P$. If you drill one thing, drill that.
2. The arithmetic is deliberately heavy at the end. Several 2026 questions ask not for $y$ but for $\pi^4 y(\pi/3)$, or $\alpha^4$, or $\left\lfloor y(\sqrt5)\right\rfloor$. Solve the equation, then read the question again before computing.

Quick recall

The five traps

1. Read $P$ only after making the coefficient of $y'$ equal to $1$. $x^2y' + 3xy = \sin x$ has $P = \dfrac3x$, not $3x$.
2. Clear radicals before naming the degree. And if a derivative sits inside $\sin$, $\log$ or an exponent, the degree is undefined, not fractional.
3. Eliminate the arbitrary constant before flipping the slope in an orthogonal-trajectory question.
4. When the two linear forms are parallel, do not try to shift. Substitute $z = ax + by$ instead.
5. A relation can outrun its own domain. Check that the value the question asks for is one the solution curve actually attains, and say so if it is not.

The one habit that pays

Before touching the algebra, write the equation as $\dfrac{dy}{dx} = \dots$ and spend ten seconds naming the type. Almost every mark lost in this chapter is lost by starting the wrong method and discovering it four lines in. Naming the type is free; unwinding a wrong substitution is not.

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