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Online Practice: Differential Equations for JEE

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Sharpen your mathematics preparation with interactive online practice for Differential Equations. Goodmarks offers 14+ JEE-style MCQs mapped to the official syllabus, each with detailed explanations so you learn from every attempt.

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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=siny+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 tanysec2xdx+tanxsec2ydy=0\tan y \cdot \sec ^{2} x d x+\tan x \cdot \sec ^{2} y d y=0 is
Q5MathsUnit 9: Differential Equations
(1xy+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 (x1)2+(x-1)^{2}+ (y1)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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How do I practice Differential Equations online for JEE?

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How many Differential Equations questions are available?

Goodmarks currently has 14+ JEE-aligned MCQs for Differential Equations, with more added regularly.

Is this aligned with the JEE Main syllabus?

Yes. All Mathematics questions on Goodmarks are organised by official JEE Main units and subtopics, including Ordinary differential equations, their order and degree, The solution of differential equation by the method of separation of variables, Solution of a homogeneous and linear differential equation of the type dy/dx + p(x)y = q(x).

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