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Online Practice: Ordinary differential equations, their order and degree for JEE

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Sharpen your mathematics preparation with interactive online practice for Ordinary differential equations, their order and degree. Goodmarks offers 9+ JEE-style MCQs mapped to the official syllabus, each with detailed explanations so you learn from every attempt.

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Part of Differential Equations in JEE Main Mathematics.

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Q1MathsUnit 9: Differential Equations

The order, degree of the differential equation satisfying the relation

\sqrt{1+x^{2}}+\sqrt{1+y^{2}}=\lambda(x \sqrt{1+y^{2}})

\left.y \sqrt{1}+x^{2}\right)

Q2MathsUnit 9: Differential Equations

The family of curves represented by

\frac{d y_{1}}{d x}=\frac{x^{2}+x+1}{y^{2}+y+1}

and the family represented by

\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,

\left(\frac{d^{2} y}{d x^{2}}\right)^{3}=\sin y+3 x \quad

are

Q4MathsUnit 9: Differential Equations

The order and degree of the differential equation.

\left(\frac{d^{2} y}{d x^{2}}\right)^{3}+\left(\frac{d y}{d x}\right)=\int y d x

are respectively.

Q5MathsUnit 9: Differential Equations

Find the order and degree of

(1+

\left.\boldsymbol{y}^{\prime}\right)^{1 / 2}=\boldsymbol{y}^{\prime \prime}

Q6MathsUnit 9: Differential Equations

Assertion The order of the differential equation, of which

\boldsymbol{x} \boldsymbol{y}=\boldsymbol{c} \boldsymbol{e}^{\boldsymbol{x}}+\boldsymbol{b} \boldsymbol{e}^{-\boldsymbol{x}}+\boldsymbol{x}^{\boldsymbol{2}}

is a solution, is 2 Reason The differential equation is

x \frac{d^{2} y}{d x^{2}}+

2 \frac{d y}{d x}-x y+x^{2}-2=0

Q7MathsUnit 9: Differential Equations

The differential equation corresponding to

x y=c^{2},

where

c

is an arbitrary constant, is:

Q8MathsUnit 9: Differential Equations

Order and degree of

\left(x^{2}+2 x\right) y_{2}^{2}+\left(x^{2}-2\right) y_{1}^{3}-2(x+

\mathbf{3}) \boldsymbol{y}=\mathbf{0}

are:

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