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JEE Main Mock Test Series 2020 & Previous Year Papers

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Illustration 1: The differential equation dy/dx = √(1-y2)/y determines a family of circle with:
1. variable radii and  affixed centre at (0, 1)
2. variable radii and  affixed centre at (0, -1)
3. fixed radius 1 and variable centres along the x-axis
4. fixed radius 1 and variable centres along the y-axis
Solution: The given equation is
dy/dx = √(1-y2)/y
Taking the terms of y and x on separate sides
y/√(1-y2)dy = dx
Integrating both sides, we get
∫y/√(1-y2) dy = ∫dx
-√(1-y2) = x + c
Hence, we get x2 +y2 + 2cx + c2 - 1 =0
or 1- y2 = (x + c)2
This clearly shows that this differential equation represents a circle of fixed radius 1 and variable centres along x –axis.

Illustration 2: Solve (D+ 4) y = x sin2x.
Solution:The C.F can be easily obtained as C.F.  = c1cos 2x + c2 sin2x
P.I. = 1/ (D2+4). x sin 2x
= {x- 1/(D2+4) .2D}. 1/(D2+4). sin 2x
= {x- 1/ (D2+4). 2D}{-x/4 cos 2x}
= -x2/4 cos 2x + ½.1/(D2+4) (cos 2x- 2x sin 2x)
= -x2/4 .cos 2x + ½ 1/(D2+4) cos 2x – 1/(D2+4) x sin 2x
= -x2/8 cos 2x +1/16 x sin 2x
So, y= c1cos 2x + c2 sin 2x – x2/ 8 cos 2x + 1/16 x sin 2x.

Illustration 3: Solve cos (x+y+1) dx- dy = 0
Solution: Rearranging the terms it can easily be reduced to variable separable.
So, dy / dx= cos (x+y+1)
If we substitute t = x+y+1, we get
dt/ dx = 1+ dy/dx
So, dy/dx = (dt/dx)-1.
Hence the equation becomes, (dt / dx) -1 = cos t
So, dt / dx = 1+ cos t
Now taking the terms of t on one side and of x on the other
dt / (1+cos t) = dx ……(1)
For integrating (1+cos t), first we write it as 2 cos2t/2 using the formula cos 2x = 2 cos2x-1.
Now integrating dt/ 2cos2 (t/2) i.e. ½ sec (t/2) gives tan t/2. So, using this in equation
(1) we get
tan t/2 – x = c
Hence, tan [(x+y+1)/2] –x = c.

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