Differential Equations: JEE Main 2026 PYQs

Answer: Option (3)

Key idea: First-order linear differential equation with variable separable method and integration by parts

1. Separate variables: $\frac{dy}{y} = (2 + \ln x)dx$
2. Integrate both sides: $\ln y = \int(2 + \ln x)dx$
3. Use integration by parts for $\int \ln x\, dx = x\ln x - x$, giving $\ln y = 2x + x\ln x - x + C = x(1 + \ln x) + C$
4. Apply initial condition $(1, e)$: $\ln e = 1(1 + \ln 1) + C \Rightarrow 1 = 1 + C \Rightarrow C = 0$
5. Particular solution: $\ln y = x(1 + \ln x)$
6. Find $f(e)$: $\ln y = e(1 + \ln e) = e(1 + 1) = 2e$, so $y = e^{2e}$

Answer: Option (4)

Key idea: First-order linear differential equation solved by variable separation

1. Rearrange and separate variables: $\frac{dy}{y+1} = \frac{-\cos x}{1+\sin x}dx$
2. Integrate both sides: $\ln|y+1| = -\ln|1+\sin x| + C$
3. Simplify to get: $(y+1)(1+\sin x) = K$
4. Apply initial condition $y(0)=0$ to find $K=1$, giving $(y+1)(1+\sin x) = 1$
5. Solve for $y$: $y = \frac{-\sin x}{1+\sin x}$
6. Substitute $y = -\frac{1}{2}$: $\frac{-\sin \alpha}{1+\sin \alpha} = -\frac{1}{2}$
7. Solve to get $\sin \alpha = 1$, therefore $\alpha = \frac{\pi}{2}$

Answer: Option (2)

Key idea: Bernoulli Differential Equation with Substitution

1. Step 1: Use substitution $x = vy$ where $v$ is a function of $y$, so $\frac{dx}{dy} = v + y\frac{dv}{dy}$
2. Step 2: Substitute into the differential equation: $2y^2(v + y\frac{dv}{dy}) - 2vy^2 + v^2y^2 = 0$
3. Step 3: Simplify to get $2y^3\frac{dv}{dy} + v^2y^2 = 0$, which gives $\frac{dv}{v^2} = -\frac{dy}{2y}$
4. Step 4: Integrate both sides: $-\frac{1}{v} = -\frac{1}{2}\ln y + C$, so $\frac{1}{v} = \frac{1}{2}\ln y + C_1$
5. Step 5: Apply initial condition $x(e) = e$ to find $C_1 = \frac{1}{2}$, giving $v = \frac{2}{\ln y + 1}$
6. Step 6: Since $x = vy$, we have $x = \frac{2y}{\ln y + 1}$
7. Step 7: Calculate $x(e^2) = \frac{2e^2}{\ln(e^2) + 1} = \frac{2e^2}{3}$

Answer: Option (3)

Key idea: Differential Equations with Polynomial Functions

1. Step 1: Compare degrees: Since $f(x) = f'(x)f''(x)$, we have $n = (n-1) + (n-2)$, giving $n = 3$. So $f(x) = ax^3 + bx^2 + cx$ (using $f(0) = 0$).
2. Step 2: Compute derivatives: $f'(x) = 3ax^2 + 2bx + c$ and $f''(x) = 6ax + 2b$.
3. Step 3: Expand and compare coefficients in $ax^3 + bx^2 + cx = (3ax^2 + 2bx + c)(6ax + 2b)$. From $x^3$ coefficient: $a = 18a^2$, so $a = \frac{1}{18}$. From constant term: $bc = 0$. Solving all equations gives $b = c = 0$.
4. Step 4: Therefore $f(x) = \frac{1}{18}x^3$, so $f'(x) = \frac{1}{6}x^2$ and $f''(x) = \frac{1}{3}x$.
5. Step 5: Calculate: $f'(2) = \frac{2}{3}$, $f''(2) = \frac{2}{3}$, and $\int_0^2 f(x)dx = \frac{1}{72} \cdot 16 = \frac{2}{9}$.
6. Step 6: Sum: $\frac{2}{3} + \frac{2}{3} + \frac{2}{9} = \frac{14}{9}$.
7. Step 7: Final answer: $36 \cdot \frac{14}{9} = 56$.

Answer: Option (3)

Key idea: Separable Differential Equations with Trigonometric Integration

1. Separate variables: $\frac{dy}{1-y+y^2} = (1+x+x^2)dx$
2. Integrate RHS: $\int(1+x+x^2)dx = x + \frac{x^2}{2} + \frac{x^3}{3} + C$
3. For LHS, complete the square: $1-y+y^2 = (y-\frac{1}{2})^2 + \frac{3}{4}$, then integrate to get $\frac{2}{\sqrt{3}}\arctan\left(\frac{2y-1}{\sqrt{3}}\right)$
4. Apply initial condition $y(0) = \frac{1}{2}$ to find $C = 0$
5. At $x=1$: $\frac{2}{\sqrt{3}}\arctan\left(\frac{2y(1)-1}{\sqrt{3}}\right) = \frac{11}{6}$
6. Solve: $\arctan\left(\frac{2y(1)-1}{\sqrt{3}}\right) = \frac{11\sqrt{3}}{12}$
7. Therefore: $2y(1)-1 = \sqrt{3}\tan\left(\frac{11\sqrt{3}}{12}\right)$

Answer: Option (4)

Key idea: First-order linear differential equation with integrating factor

1. Identify the differential equation as first-order linear: $\frac{dy}{dx} + P(x)y = Q(x)$ where $Q(x) = 2 + e^{-2x}$
2. Find the integrating factor by comparing coefficients: $\mu(x) = \frac{x^3 + 2}{2 + e^{-2x}}$
3. Multiply both sides by $\mu(x)$ and integrate: $y \cdot \frac{x^3 + 2}{2 + e^{-2x}} = \int (x^3 + 2) dx = \frac{x^4}{4} + 2x + C$
4. Apply initial condition $y(0) = \frac{3}{2}$ to find $C = 1$
5. Write the solution: $y = \left(\frac{x^4}{4} + 2x + 1\right) \cdot \frac{2 + e^{-2x}}{x^3 + 2}$
6. Evaluate at $x=1$: $y(1) = \frac{13}{4} \cdot \frac{2 + e^{-2}}{3} = \frac{13}{12}(2 + e^{-2})$
7. Therefore $a = \frac{13}{12}$

Answer: 6

Key idea: Homogeneous Differential Equations and Circle Properties

1. Step 1: Recognize the homogeneous differential equation. Substitute $v = \frac{y}{x}$, so $y = vx$ and $dy = vdx + xdv$.
2. Step 2: Simplify to get $\sin(v)dv = -\frac{dx}{x}$. Integrate: $-\cos(v) = -\ln|x| + C$, giving $\cos\left(\frac{y}{x}\right) = \ln|x| + C$.
3. Step 3: Apply initial condition $y(1) = \frac{\pi}{2}$: $\cos\left(\frac{\pi}{2}\right) = 0 = C$. Solution: $\cos\left(\frac{y}{x}\right) = \ln|x|$.
4. Step 4: Find $a = \cos\left(\frac{y(e)}{e}\right) = \ln(e) = 1$.
5. Step 5: Circle equation becomes $x^2 + y^2 - 2px + 2py + 3 = 0$. Standard form: $(x-p)^2 + (y+p)^2 = 2p^2 - 3$ with radius $r = \sqrt{2p^2 - 3}$.
6. Step 6: For a valid circle: $2p^2 - 3 > 0 \Rightarrow p^2 > 1.5$. For $r \leq 6$: $2p^2 - 3 \leq 36 \Rightarrow p^2 \leq 19.5$.
7. Step 7: Integral values satisfying $1.5 < p^2 \leq 19.5$ are $p \in \{-4, -3, -2, 2, 3, 4\}$, giving 6 values.

Answer: 48

Key idea: First-order linear differential equation with integrating factor method

1. Rewrite the equation as $\frac{dy}{dx} + (\tan x) y = \frac{\sec^3 x}{(\tan x)^{1/2}}$
2. Find integrating factor: $I.F. = e^{\int \tan x dx} = \sec x$
3. Multiply by I.F. and integrate: $y \sec x = \int \frac{\sec^4 x}{(\tan x)^{1/2}} dx$
4. Use substitution $u = \tan x$ to get $\int (u^{-1/2} + u^{3/2}) du = 2u^{1/2} + \frac{2u^{5/2}}{5} + C$
5. Apply initial condition $y(\pi/4) = \frac{6\sqrt{2}}{5}$ to find $C = 0$
6. Evaluate $y(\pi/3) = 3^{1/4} \cdot \frac{8}{5} = \frac{4\alpha}{5}$ to get $\alpha = 2 \cdot 3^{1/4}$
7. Calculate $\alpha^4 = 16 \times 3 = 48$

Answer: 3

Key idea: First-order linear differential equation with integrating factor

1. Step 1: Rewrite the differential equation by factoring: $x(x - \sqrt{x^2-1})dy = (x - y(x - \sqrt{x^2-1}))dx$
2. Step 2: Simplify to get the standard form: $\frac{dy}{dx} + \frac{y}{x} = \frac{1}{x - \sqrt{x^2-1}}$
3. Step 3: Use integrating factor $\mu(x) = x$ and rationalize the RHS: $\frac{d}{dx}(xy) = x^2 + x\sqrt{x^2-1}$
4. Step 4: Integrate both sides: $xy = \frac{x^3}{3} + \frac{(x^2-1)^{3/2}}{3} + C$
5. Step 5: Apply initial condition $y(1) = 1$ to find $C = \frac{2}{3}$
6. Step 6: Substitute $x = \sqrt{5}$ to get $y(\sqrt{5}) = \frac{5 + 2\sqrt{5}}{3} \approx 3.157$
7. Step 7: The greatest integer less than $y(\sqrt{5})$ is $3$

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Key idea: First-order linear differential equation with integrating factor method