Mathematics_Today_-_October_2016

Integrating, we get

tan− +

+

(^1) ∫ 2 =
1
y dt
t
c
(Where t = ex)
⇒ tan–1y + tan–1t = c ⇒ tan–1y + tan–1(ex) = c
When x = 0, y = 1
⇒ tan–1 1 + tan–1(e^0 ) = c
or c=+=πππ
442
∴ tan−−^11 +=tan
2
yex π
This is the required solution of the given differential
equation.

12. We h a v e dy
    dx

+=2sinyx

This is a linear D.E. of the form

dy

dx

+=Py Q

Where, P = 2 and Q = sin x

Now, I.F.===eee∫∫Pdx^2 dx^2 x

∴ The solution is given by

yQdxc×=×I.F. ∫ I.F. +

⇒ ye^22 xx=+∫e sinx dx c

= − +

e^2 x xxc

5

( sin 2 cos )

⇒ yxxce=^1 − + −x

5

( sin 2 cos )^2

This is required solution of the given differential

equation.

13. Equation of a circle with centre (0, a) and radius a is
    x^2 + (y – a)^2 = a^2 or x^2 + y^2 = 2ay ...(1)
    where a is a parameter.
    Differentiating both sides of (1) w.r.t. x, we get

22 xy^2

dy

dx

ady

dx

ady

dx

xydy

dx

+= or =+

or a x

dy

dx

= y

⎛

⎝⎜

⎞

⎠⎟

+

...(2)

Putting the value of a from (2) in (1), we get

xy yx

dy

dx

yxyxy

dy

dx

(^22) += 2 222
⎛
⎝⎜
⎞
⎠⎟
⎧ +
⎨
⎪
⎩⎪
⎫
⎬
⎪
⎭⎪
− =
⎛
⎝⎜
⎞
⎠⎟
or ( )
or (xy)dy
dx
(^22) −− 20 xy=
This is the required differential equation.

14. Let P(x, y) be an arbitrary point on the given curve.
    The equation of the normal to the given curve at

(x, y) is Yy dy

dx

− =−−^1 ()Xx

It is given that the normal at every point passes

through a fixed point (α, β) (say).

Therefore, βα−y=−−dx

dy

()x

⇒ (x – α)dx + (y – β)dy = 0

Integrating both sides, we get

⇒−∫() ()xdxydyCαβ+∫ − =

⇒ ()x−α +()y−β =C

2 2

22

⇒ (x – α)^2 + (y – β)^2 = r^2 , where r^2 = 2C

Clearly, this equation represents a circle, having

centre at (α, β) and radius r.

15. We h a v e dy
    dx

+secxy⋅ =tanx ...(1)

This is a linear D.E. of the form dy

dx

+=Py Q.

Where, P = secx and Q = tanx

Now, I.F.==ee∫∫Pdx secxdx

= elog(secx + tanx) = secx + tanx

∴ The solution of (1) is given by

yx x⋅(sec +=tan ) ∫tan (secxx xdxc+ +tan )

=++∫(sec tanxxtan xdxc)

2

=+∫∫sec tanxxdx∫sec^2 xdx dxc−⋅^1 +

∴ y(secx + tanx) = secx + tanx – x + c

This is the required solution of the given differential

equation.

16. We have (x^3 + y^3 )dy – x^2 ydx = 0
    ⇒ =
    
    
    • 
      
    
    
    
    

dy

dx

xy

xy

2

33 ...(1)

This is a homogeneous differential equation.

Put yvx dy

dx

vxdv

dx

= ⇒ =+ ...(2)

Now, differential equation (1) becomes

vxdv

dx

v

v

+=

1 +^3

⇒ =

+

− =−

+

xdv

dx

v

v

v v

113 v

4

3

⇒^1 + += 0

3

4

v

v

dv dx

x

Integrating both sides, we get

(^110)
v^4 v dv
dx
x
⎡ +
⎣⎢
⎤
⎦⎥
∫∫+=
⇒−^1 ++ =
3 v^3
log | | log | |vxc