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Hard Differential Equations MCQs for JEE

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Hard Differential Equations MCQs train multi-concept problem solving — attempt after mastering easy and medium sets (14+ on Goodmarks).

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Q1MathsUnit 9: Differential Equations
The order, degree of the differential equation satisfying the relation 1+x2+1+y2=λ(x1+y2)\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=\lambda(x \sqrt{1+y^{2}}) y1+x2)\left.y \sqrt{1}+x^{2}\right) is
Q2MathsUnit 9: Differential Equations
The family of curves represented by dy1dx=x2+x+1y2+y+1\frac{d y_{1}}{d x}=\frac{x^{2}+x+1}{y^{2}+y+1} and the family represented by dy2dx+y2+y+1x2+x+1=0\frac{\boldsymbol{d} \boldsymbol{y}_{2}}{\boldsymbol{d} \boldsymbol{x}}+\frac{\boldsymbol{y}^{2}+\boldsymbol{y}+\mathbf{1}}{\boldsymbol{x}^{2}+\boldsymbol{x}+\mathbf{1}}=\mathbf{0}
Q3MathsUnit 9: Differential Equations
The order and degree of the differential equation, (d2ydx2)3=sin⁡y+3x\left(\frac{d^{2} y}{d x^{2}}\right)^{3}=\sin y+3 x \quad are
Q4MathsUnit 9: Differential Equations
Solution of the differential equation tan⁡y⋅sec⁡2xdx+tan⁡x⋅sec⁡2ydy=0\tan y \cdot \sec ^{2} x d x+\tan x \cdot \sec ^{2} y d y=0 is
Q5MathsUnit 9: Differential Equations
(1−xy+x2y2)dx=x2dy\left(1-x y+x^{2} y^{2}\right) d x=x^{2} d y
Q6MathsUnit 9: Differential Equations
The order and degree of the differential equation. (d2ydx2)3+(dydx)=∫ydx\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)=\int y d x are respectively.
Q7MathsUnit 9: Differential Equations
The normal at any point P(x,y)\boldsymbol{P}(\boldsymbol{x}, \boldsymbol{y}) of a\mathbf{a} curve meets the xx -axis at QQ and NN is the foot of the ordinate at PP If NQ=x(1+y2)1+x2,N Q=\frac{x\left(1+y^{2}\right)}{1+x^{2}}, then equation of such curve, given that it passes through the point (3,1) is:
Q8MathsUnit 9: Differential Equations
Assertion A normal is drawn at a point P(x,y)\boldsymbol{P}(\boldsymbol{x}, \boldsymbol{y}) of a\mathbf{a} curve. It meets the xx -axis and the yy -axis in point AA and BB, respectively, such that 1OA+1OB=1,\frac{1}{O A}+\frac{1}{O B}=1, where OO is the origin. The equation of such a curve passing through (5,4)(\mathbf{5}, \mathbf{4}) is (x−1)2+(x-1)^{2}+ (y−1)2=25(y-1)^{2}=25 Reason OA=x+ydydx\boldsymbol{O A}=\boldsymbol{x}+\boldsymbol{y} \frac{\boldsymbol{d} \boldsymbol{y}}{\boldsymbol{d} \boldsymbol{x}} and OB=x+ydydxdydx\boldsymbol{O} \boldsymbol{B}=\frac{\boldsymbol{x}+\boldsymbol{y} \frac{d \boldsymbol{y}}{d \boldsymbol{x}}}{\frac{d \boldsymbol{y}}{d \boldsymbol{x}}}

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