Jee Main Math Class 12 Differential Equations JEE Main Physics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: (b)

Explanation: $2 x^2 \;\frac{dy}{dx} - 2 x y + 3 y^2 = 0$ $\Rightarrow 2 x (x dy - y dx) + 3 y^3 dx = 0$ $\Rightarrow 2\left(\frac{x dy - y dx}{y^2}\right) + 3 \;\frac{dx}{x} = 0$ $\Rightarrow -\frac{2x}{y} + 3 \ln x = C$ $\because y(e) = \frac{e}{3} \Rightarrow -6+3 = C \Rightarrow C = -3$ Now, at $x=1, -\frac{2}{y} + 0 = -3$ $y = \frac{2}{3}$

Correct Answer: 6

Explanation: $\int\limits_{3}^{x} f(x) dx = \left(\frac{f(x)}{x}\right)^3$ $x^3 \cdot \int\limits_{3}^{x} f(x) dx = f^3(x)$ Differentiate w.r.t. $x$ $x^3 f(x) + 3 x^2 \cdot \frac{f^3(x)}{x^3} = 3 f^2(x) f'(x)$ $\Rightarrow 3 y^2 \frac{dy}{dx} = x^3 y + \frac{3 y^3}{x}$ $3 x y \frac{dy}{dx} = x^4 + 3 y^2$ $\text{ Let } y^2 = t$ $\frac{3}{2} \frac{dt}{dx} = x^3 + \frac{3t}{x}$ $\frac{dt}{dx} - \frac{2t}{x} = \frac{2 x^3}{3}$ $\text{ I. } F = e^{\int -\frac{2}{x} dx} = \frac{1}{x^2}$ Solution of differential equation t. $\frac{1}{x^2} = \int \frac{2}{3} x dx$ $\frac{y^2}{x^2} = \frac{x^2}{3} + C$ $y^2 = \frac{x^4}{3} + C x^2$ Curve passes through $(3,3) \Rightarrow C = -2$ $y^2 = \frac{x^4}{3} - 2 x^2$ Which passes through $(\alpha, 6\sqrt{10})$ $\frac{\alpha^4 - 6\alpha^2}{3} = 360$ $\alpha^4 - 6\alpha^2 - 1080 = 0$ $\alpha = 6$

Correct Answer: (a)

Explanation: Differential equation :- $x \cos \frac{y}{x} \frac{dy}{dx} = y \cos \frac{y}{x} + x$ $\cos \frac{y}{x} \left[ x \frac{dy}{dx} - y \right] = x$ Divide both sides by $x^2$ $\cos \frac{y}{x} \left( \frac{x \frac{dy}{dx} - y}{x^2} \right) = \frac{1}{x}$ Let $\frac{y}{x} = t$ $\cos t \left( \frac{dt}{dx} \right) = \frac{1}{x}$ $\cos t \, t = \frac{1}{x} dx$ Integrating both sides $\sin t = \ln |x| + c$ $\sin \frac{y}{x} = \ln |x| + c$ $\text{Using } y(1) = \frac{\pi}{3}$ , we get $c = \frac{\sqrt{3}}{2}$ So, $\alpha = \sqrt{3} \Rightarrow \alpha^2 = 3$

Correct Answer: (d)

Explanation: $x^2 f(x)-x=4 \int\limits_{0}^{x} t f(t) dt$ Differentiate w.r.t. $x$ $x^2 f'(x)+2 x f(x)-1=4 x f(x)$ $\;\text{ Let }\; y=f(x)$ $\Rightarrow x^2 \;\frac{d y}{d x}-2 x y-1=0$ $\;\frac{d y}{d x}-\;\frac{2}{x} y=\;\frac{1}{x^2}$ $\;\text{ I.F. }\;=e^{\int\;\frac{-2}{x} dx}=\;\frac{1}{x^2}$ Its solution is $\;\frac{y}{x^2}=\int\;\frac{1}{x^4} dx+C$ $\;\frac{y}{x^2}=\;\frac{-1}{3 x^3}+C$ $\because f(1)=\;\frac{2}{3} \Rightarrow y(1)=\;\frac{2}{3}$ $\Rightarrow \;\frac{2}{3}=-\;\frac{1}{3}+C$ $\Rightarrow C=1$ $\because y=-\;\frac{1}{3 x}+x^2$ $f(x)=-\;\frac{1}{3 x}+x^2$ $f(3)=-\;\frac{1}{9}+9=\;\frac{80}{9} \Rightarrow 18 f(3)=160$

Correct Answer: 3

Explanation: $(y-2 \ln x) dx + (2 x \ln x) dy = 0$ $dy (2 x \ln x) = [(2 \ln x) - y] dx$ $\frac{dy}{dx} = \frac{1}{x} - \frac{y}{2 x \ln x}$ $\frac{dy}{dx} + \frac{y}{2 x \ln x} = \frac{1}{x}$ $\text{I.F.} = e^{\int \frac{1}{2 x \ln x} dx}$ $= e^{\frac{1}{2} \int \frac{dt}{t}} = e^{\frac{1}{2} \ln(\ln x)}$ $\Rightarrow \text{I.F.} = (\ln x)^{1/2}$ $\therefore y \sqrt{\ln x} = \int \frac{\sqrt{\ln x}}{x} dx \qquad \text{(Let, } \ln x = u^2)$ $= 2 \int u^2 du$ $\frac{1}{x} dx = 2u du$ $y \sqrt{\ln x} = \frac{2}{3} (\ln x)^{3/2} + c \leftarrow \left(e, \frac{4}{3}\right)$ $\frac{4}{3} = \frac{2}{3} + c \Rightarrow c = \frac{2}{3}$ $y \sqrt{\ln x} = \frac{2}{3} (\ln x)^{3/2} + \frac{2}{3} \leftarrow (e^4, \alpha)$ $\alpha \cdot 2 = \frac{2}{3} \times 8 + \frac{2}{3}$ $\alpha = 3$