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

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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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