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

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  1. 1.Core idea: The solution of differential equation by the method of separation of variables
  2. 2.Relates to other subtopics in Differential Equations
  3. 3.Order and degree identification
  4. 4.Separation of variables
  5. 5.Master The solution of differential equation by the method of separation of variables definitions and standard results
  6. 6.Solve 20 timed MCQs for The solution of differential equation by the method of separation of variables

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⋅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
Q2MathsUnit 9: Differential Equations
(1−xy+x2y2)dx=x2dy\left(1-x y+x^{2} y^{2}\right) d x=x^{2} d y
Q3MathsUnit 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:
Q4MathsUnit 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}}}
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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