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The solution of differential equation by the method of separation of variables Concepts for JEE

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The solution of differential equation by the method of separation of variables is a core JEE Main Mathematics subtopic under Differential Equations. Master the definitions, standard results, and typical MCQ patterns tested in JEE Main and Advanced.

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Syllabus context

Part of Differential Equations in JEE Main Mathematics.

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Q1MathsUnit 9: Differential Equations

Solution of the differential equation

\tan y \cdot \sec ^{2} x d x+\tan x \cdot \sec ^{2} y d y=0

Q2MathsUnit 9: Differential Equations

\left(1-x y+x^{2} y^{2}\right) d x=x^{2} d y

Q3MathsUnit 9: Differential Equations

The normal at any point

\boldsymbol{P}(\boldsymbol{x}, \boldsymbol{y})

\mathbf{a}

curve meets the

x

-axis at

Q

and

N

is the foot of the ordinate at

P

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:

Q4MathsUnit 9: Differential Equations

Assertion A normal is drawn at a point

\boldsymbol{P}(\boldsymbol{x}, \boldsymbol{y})

\mathbf{a}

curve. It meets the

x

-axis and the

y

-axis in point

A

and

B

, respectively, such that

\frac{1}{O A}+\frac{1}{O B}=1,

where

O

is the origin. The equation of such a curve passing through

(\mathbf{5}, \mathbf{4})

(x-1)^{2}+

(y-1)^{2}=25

Reason

\boldsymbol{O A}=\boldsymbol{x}+\boldsymbol{y} \frac{\boldsymbol{d} \boldsymbol{y}}{\boldsymbol{d} \boldsymbol{x}}

and

\boldsymbol{O} \boldsymbol{B}=\frac{\boldsymbol{x}+\boldsymbol{y} \frac{d \boldsymbol{y}}{d \boldsymbol{x}}}{\frac{d \boldsymbol{y}}{d \boldsymbol{x}}}

Q5MathsUnit 9: Differential Equations

A certain radioactive material is known to decay at a rate proportional to the amount present. If after one hour it is observed that 10 percent of the material has decayed, find the half-life (period of time it takes for the amount of material to decrease by half) of the material (in hrs.)

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