JEE Corner — Differential Equations for JEE Main and Advanced

Where Differential Equations Sit in JEE

JEE Main reliably asks 1-2 questions from this chapter every session: ordinary differential equations, their order and degree, the formation of differential equations, solution by the method of separation of variables, homogeneous and linear differential equations. That syllabus line is exactly the NCERT chapter — but JEE dresses the same five skills in heavier disguises and adds three tools the NCERT text only hints at:

1. The shift substitution. When the equation involves $x + y$ (or any $ax + by + c$) as a block — say $\dfrac{dy}{dx} = f(x + y)$ — substitute $t = x + y$, so $\dfrac{dt}{dx} = 1 + \dfrac{dy}{dx}$. The equation in $t$ and $x$ is separable. This converts equations that look hopeless, like $\dfrac{dy}{dx} = (x + y + 1)^2$ or $\dfrac{dy}{dx} = \sin(x+y) + \cos(x+y)$, into two-line solves.

2. Bernoulli's equation. $\dfrac{dy}{dx} + Py = Qy^n$ is linear except for the $y^n$. Divide by $y^n$ and substitute $v = y^{1-n}$, so $\dfrac{dv}{dx} = (1-n)y^{-n}\dfrac{dy}{dx}$ — the equation becomes genuinely linear in $v$. JEE Main loves $n = 2$ and $n = 6$.

3. Exact-differential recognition. Certain combinations are derivatives of famous quantities. Memorise this table:

Spotting one of these rows collapses a five-minute integrating-factor grind into a single integration.

Orthogonal trajectories (JEE Main pattern staple): to find curves cutting a family at right angles, form the family's differential equation, replace $\dfrac{dy}{dx}$ by $-\dfrac{dx}{dy}$, and solve the new equation.

The Speed Toolkit

1. Classify in ten seconds. Right side factors as $g(x)h(y)$ → separable. Every term the same total degree → homogeneous, $y = vx$. Shaped $\dfrac{dy}{dx} + Py = Q$ → linear I.F. A block $ax + by + c$ repeating → shift substitution $t = ax + by$. A stray $y^n$ on the right of a linear shape → Bernoulli.

2. I.F. speed table (compute $e^{\int P dx}$ from memory): $P = \dfrac{a}{x} \Rightarrow x^a$; $P = \tan x \Rightarrow \sec x$; $P = \cot x \Rightarrow \sin x$; $P = \sec x \Rightarrow \sec x + \tan x$; $P = -x \Rightarrow e^{-x^2/2}$; $P = \dfrac{1}{1+x^2}$… only if you must — check first whether the left side is already $d(\text{something})$.

3. The mirror reflex. If $\dfrac{dy}{dx}$ gives a nonlinear mess, immediately try $\dfrac{dx}{dy}$ — equations linear in $x$ (with $y$ as the variable) are a favourite JEE trap. Marker: terms like $x\,f(y)$ and $g(y)$ with a clean $\dfrac{dy}{dx}$ denominator.

4. Formation drill. To form the DE of an $n$-parameter family, differentiate $n$ times and eliminate all parameters. The order of the result always equals $n$ — an answer-check in itself.

5. Definite-condition shortcut. When a particular value like $y(\pi/4)$ or $x_0$ with $y(x_0) = e$ is asked, solve symbolically first, apply the condition once at the end. Substituting early breeds sign errors.

6. Blow-up awareness. Solutions of nonlinear equations can fail to exist beyond a point: $y' = y^2$, $y(0) = 1$ gives $y = \dfrac{1}{1 - x}$, which explodes at $x = 1$. JEE Advanced probes exactly this.

JEE-Pattern Worked Examples

Example 1: The squared block

Solve $\dfrac{dy}{dx} = (x + y + 1)^2$.

Solution:

1. Substitute the block: $t = x + y + 1$, so $\dfrac{dt}{dx} = 1 + \dfrac{dy}{dx} = 1 + t^2$.
2. Separate: $\dfrac{dt}{1 + t^2} = dx$.
3. Integrate: $\tan^{-1} t = x + C$.

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

Example 2: A first Bernoulli

Solve $\dfrac{dy}{dx} + \dfrac{y}{x} = y^2$.

Solution:

1. Divide by $y^2$: $\dfrac{1}{y^2}\dfrac{dy}{dx} + \dfrac{1}{xy} = 1$.
2. Substitute $v = \dfrac{1}{y}$: then $\dfrac{dv}{dx} = -\dfrac{1}{y^2}\dfrac{dy}{dx}$, and the equation becomes $-\dfrac{dv}{dx} + \dfrac{v}{x} = 1$, i.e. $\dfrac{dv}{dx} - \dfrac{v}{x} = -1$ — linear.
3. I.F.: $e^{-\int dx/x} = \dfrac{1}{x}$, so $\dfrac{d}{dx}\left(\dfrac{v}{x}\right) = -\dfrac{1}{x}$.
4. Integrate: $\dfrac{v}{x} = -\log|x| + C$.

Answer: $\dfrac{1}{xy} = C - \log|x|$, i.e. $xy\left(C - \log|x|\right) = 1$.

Example 3: Product recognition

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

Solution:

1. Spot the row: $x\,dy + y\,dx = d(xy)$. The equation is $d(xy) = (xy)^2\,dx$.
2. Treat $u = xy$ as the unknown: $\dfrac{du}{u^2} = dx$.
3. Integrate: $-\dfrac{1}{u} = x + C$.

Answer: $\dfrac{1}{xy} + x + C = 0$ — three written lines, thanks to the exact-differential table.

Example 4: The other quotient

Solve $y(1 + xy)\,dx = x\,dy$.

Solution:

1. Expand and regroup: $y\,dx + xy^2\,dx = x\,dy \Rightarrow y\,dx - x\,dy = -xy^2\,dx$.
2. Divide by $y^2$: $\dfrac{y\,dx - x\,dy}{y^2} = -x\,dx$ — the left side is exactly $d\left(\dfrac{x}{y}\right)$.
3. Integrate: $\dfrac{x}{y} = -\dfrac{x^2}{2} + C$.

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

Example 5: A logarithmic homogeneous equation

Solve $x\dfrac{dy}{dx} = y\left(\log y - \log x + 1\right)$.

Solution:

1. Rewrite: $\dfrac{dy}{dx} = \dfrac{y}{x}\left(\log\dfrac{y}{x} + 1\right)$ — a function of $\dfrac{y}{x}$ alone: homogeneous.
2. Substitute $y = vx$: $v + x\dfrac{dv}{dx} = v(\log v + 1)$, so $x\dfrac{dv}{dx} = v\log v$.
3. Separate: $\dfrac{dv}{v \log v} = \dfrac{dx}{x}$.
4. Integrate (put $w = \log v$): $\log|\log v| = \log|x| + \log C$, so $\log v = Cx$.

Answer: $\log\dfrac{y}{x} = Cx$, i.e. $y = x\,e^{Cx}$.

Example 6: Linear with a designed right side

If $\dfrac{dy}{dx} + y\tan x = 2x + x^2\tan x$ and $y(0) = 1$, find $y$.

Solution:

1. I.F.: $e^{\int \tan x\,dx} = \sec x$.
2. Solve: $y\sec x = \displaystyle\int \left(2x + x^2\tan x\right)\sec x\,dx = \int \left(2x\sec x + x^2 \sec x \tan x\right)dx$.
3. Spot the product rule: the integrand is exactly $\dfrac{d}{dx}\left(x^2\sec x\right)$, so $y \sec x = x^2\sec x + C$.
4. Apply $y(0) = 1$: $1 = 0 + C$, so $C = 1$.

Answer: $y = x^2 + \cos x$. The right side was built to be $\left(x^2\sec x\right)'$ — JEE setters design such gifts, and step 3 is where you either see it or lose four minutes.

Example 7: Homogeneous with a target value

If $\dfrac{dy}{dx} = \dfrac{xy}{x^2 + y^2}$ with $y(1) = 1$, find $x_0 > 0$ such that $y(x_0) = e$.

Solution:

1. Mirror: $\dfrac{dx}{dy} = \dfrac{x^2 + y^2}{xy} = \dfrac{x}{y} + \dfrac{y}{x}$ — homogeneous in the mirror direction.
2. Substitute $x = vy$: $v + y\dfrac{dv}{dy} = v + \dfrac{1}{v}$, so $v\,dv = \dfrac{dy}{y}$.
3. Integrate: $\dfrac{v^2}{2} = \log|y| + C$, i.e. $\dfrac{x^2}{2y^2} = \log|y| + C$.
4. Apply $y(1) = 1$: $\dfrac{1}{2} = 0 + C$, so $x^2 = y^2\left(2\log|y| + 1\right)$.
5. Set $y = e$: $x_0^2 = e^2(2 + 1) = 3e^2$.

Answer: $x_0 = \sqrt{3}\,e$.

Example 8: Sine plus cosine of the block

Solve $\dfrac{dy}{dx} = \sin(x + y) + \cos(x + y)$.

Solution:

1. Substitute $t = x + y$: $\dfrac{dt}{dx} = 1 + \sin t + \cos t$.
2. Half-angle everything: $1 + \cos t = 2\cos^2\dfrac{t}{2}$ and $\sin t = 2\sin\dfrac{t}{2}\cos\dfrac{t}{2}$, so $\dfrac{dt}{dx} = 2\cos^2\dfrac{t}{2}\left(1 + \tan\dfrac{t}{2}\right)$.
3. Separate: $\dfrac{\sec^2(t/2)\,dt}{2\left(1 + \tan(t/2)\right)} = dx$.
4. Integrate (put $u = 1 + \tan\dfrac{t}{2}$, $du = \dfrac{1}{2}\sec^2\dfrac{t}{2}\,dt$): $\log|u| = x + C$.

Answer: $\log\left|1 + \tan\dfrac{x + y}{2}\right| = x + C$.