Mathematics_Today_-_October_2016

On squaring, we get

dy

dx

dy

dx

2

2

(^22)
⎛ 1
⎝⎜
⎞
⎠⎟
=+⎛⎝⎜ ⎞⎠⎟
Clearly, order of differential equation is 2 and
degree is also 2.

3. The general equation of the family of all straight
   lines is given by y = mx + c, where m and c are
   parameters.

Now,ymxc dy

dx

=+⇒ =m ⇒ dy=

dx

2

2 0

So, the required differential equation is dy

dx

2

2 =^0

4. We h a v e dy
   dx x

y e

x

x

+ ⋅ =

1 −^2

This is of the form

dy

dx

+=Py Q, where P

x

=^1

andQ e

x

x

=

− 2

.

I.F.== =∫

∫

ee ePdx x

(^1) dx 2 x
.

5. We h a v eypx ap b=+^22 +^2

⇒−ypx ap b=+^222

⇒ (y – px)^2 = a^2 p^2 + b^2 [On squaring both sides]

⇒ y^2 + x^2 p^2 – 2xyp = a^2 p^2 + b^2

⇒ (x^2 – a^2 )p^2 – 2xyp+ (y^2 – b^2 ) = 0

⇒−()xa⎝⎜⎛dy⎟⎞⎠ −⋅⎝⎜⎛ ⎟⎠⎞+()− =

dx

xy dy

dx

(^22) yb
(^222)
20
Clearly, it is a differential equation of order 1 and
degree 2.

6. We h a v e y = Aex + Be–x + x^2 ...(1)
   Differentiating (1) w.r.t. x, we get
   dy
   dx

=Aexx−Be− + 2 x ...(2)

Again differentiating (2) w.r.t. x, we get

dy

dx

AexxBe

2

2 =+ +^2

−

...(3)

() ( ) 13 2

2

2

−⇒−y dy=^2 −

dx

x

or, dy

dx

yx

2

2

− +^2 − 20 = , which is the required

differential equation.

7. We h a v e log dy
   dx

⎛⎝⎜ ⎞⎠⎟=+ax by

⇒ dy==+ ⋅

dx

eeeax by ax by ⇒^1 =

e

bydy e dxax

⇒ ∫edy edx−by =∫ ax

⇒

−

=+

e−

b

e

a

C

by ax

⇒ ae–by + beax = K, where K = –abC.

This is the required solution of the given differential

equation.

8. Given, (1 + xy)ydx + (1– xy)xdy = 0
   ⇒ (ydx + xdy) + xy(ydx – xdy) = 0
   ⇒ d(xy) = xy(xdy – ydx)
   ⇒ dxy= −
   xy

xdy ydx

xy

()

22 ⇒ = −

dxy

xy

dy

y

dx

x

()

()^2

Integrating, we get −^1 = − +

xy

logyxclog

⇒−^1 = ⎛⎝⎜ ⎞⎠⎟+

xy

y

x

log c, which is the required

solution.

9. We h a v e y = ae^3 x + be–x
   ∴ dy= − −
   dx

3 ae^3 xxbe and dy

dx

aexxbe

2

2

=+ 9 3 −

Now, dy

dx

dy

dx

yaebexx

2

2

−− 239 =+()^3 −

–2(3ae^3 x^ – be–x) – 3(ae^3 x + be–x) = 0

Hence, y = ae^3 x^ + be–x is a solution of the differential

equation dy

dx

dy

dx

y

2

2 −−^230 =

10. We h a v e dy
    dx x

+= >^1 ye xx;( 0 )

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

dx

+=Py Q

Where, P

x

==^1 andQex

∴ I.F.== ==ee ex∫Pdx ∫x

(^1) dx logx
∴ The solution is given by
yx xe dx C
=+∫ x
or yx = ex(x – 1) + C
This is the required solution of the given differential
equation.

11. We have (1 + e^2 x)dy + ex(1 + y^2 )dx = 0
    ⇒
    
    
    • 
      
    
    
    
    

+

+

dy =

y

e

e

dx

x

1122 x^0