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**Illustration 1: The differential equation dy/dx = âˆš(1-y ^{2})/y determines a family of circle with:**

dy/dx = âˆš(1-y

Taking the terms of y and x on separate sides

y/âˆš(1-y

Integrating both sides, we get

âˆ«y/âˆš(1-y

-âˆš(1-y

Hence, we get x

or 1- y

This clearly shows that this differential equation represents a circle of fixed radius 1 and variable centres along x â€“axis.

**Illustration 2: Solve (D ^{2 }+ 4) y = x sin2x.**

P.I. = 1/ (D

= {x- 1/(D

= {x- 1/ (D

= -x

= -x

= -x

So, y= c

**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 â€¦â€¦(1)

For integrating (1+cos t), first we write it as 2 cos2t/2 using the formula cos 2x = 2 cos^{2}x-1.

Now integrating dt/ 2cos^{2} (t/2) i.e. Â½ sec ^{2 }(t/2) gives tan t/2. So, using this in equation

(1) we get

tan t/2 â€“ x = c

Hence, tan [(x+y+1)/2] â€“x = c.

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