1 Homogeneous equation

A function f(x; y) is said to be homogeneous function in x and y of degree n if the sumof power of x and y in each term is same and equal to n.

Consider Mdx+Ndy = 0 (i) where M and N are both homogeneousfunctions of x and y of same degree. Then dy

dx= M

N= (y=x)

(y=x)= f( y

x).

Step I: Putting y = vx so that dydx

= v+x dvdx; where v is a function of x only. Therefore,

(i) becomes v + x dvdx

= f(v) =) dvf(v)v =

dxx=) R dv

f(v)v =R

dxx(Method of separation

of variable).

Step II: Find v as a function of x and then replace v by yxand rearranging. We shall

get our required result.

Example: 2xy dydx

= y2 x2

=) dydx

= y2x22xy

=) dydx

=( yx)21

2( yx)

=) v + x dvdx

= v212v

=) x dvdx

= 1+v22v

=) 2vdv1+v2

+ dxx= 0 (using separation of variable)

=) cx = x2 + y2:

2

Consider a dierential equation of the form

dy

dx=

ax+ by + c

Ax+By + C (i)

where a; b; c; A;B;C are constants.

a) If aB bA 6= 0; thenStep I: Find the intersecting point of the straight lines ax+ by + c = 0 and

Ax+By + C = 0, say, (h; k):

Step II: Substitute x = X + h and y = Y + k so that dx = dX, dy = dY and

dYdX

= aX+bYAX+BY

. (Homogeneous equation and proceed as 1 )

b) If aB bA = 0; then in this case, the above two lines are parallel and they are of the formax+ by + c = 0 and p(ax+ by) +C = 0: Then substitute ax+ by = V and then use the

method of separation of variable.

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3 Leibnitz's linear equation:

Consider dydx

+ Py = Q (i), where P and Q are the functions of x or constants.Step I: Find the IF: e

RPdx.

Step II: Multiply both side by IF and then rearranging ddx(y:e

RPdx) = Q:e

RPdx and then

integrating.

4 Bernoulli's equation:

Consider dydx

+ Py = Qyn (i), where P and Q are the functions of x or constants.Step I: Rewrite equation (i) of the form yn dy

dx+ Py1n = Q.

Step II: Putting z = y1n so that dzdx

= (1 n)yn dydx

and then rearranging.

Step III: The dierential equation become of the form dzdx

+ Pz(1 n) = Q(1 n), which isa leibnitz's linear equation and proceed as 3.

Step IV: Find z and then replace z by y1n and rearranging. We shall get our required result.

5 Method of variation of parameters

Consider a linear dierential equation dydx

+ Py = Q (i), where P and Q are thefunctions of x or constants.

Step I: Find the general solution of the reduced equation dydx

+ Py = 0 (ii) andthe solution is of the form y = C:e

RP dx, where C is an arbitrary constant.

Step II: In this method, we assume that y = v:eRP dx is a solution of equation (i), where

Q 6= 0: Then dydx

= eRP dx dv

dx P ve

RP dx.

Step III:Equation (i) becomes dvdx

= QeRP dx and then by method of separation, we shall

get out required result.

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Problem Sheet:

Homogeneous Equations

Solve:

1. (x2 y2) dy = xy dx.

2. x dy y dx =p(x2 + y2) dx.3. (x4 2xy3)dydx + (y4 2x3y) = 0.

4. x sin(yx) dy = (y sin(yx) x)dx.

5. x2 dy + (xy + 2y2)dx = 0:

6. (x3 + y3)dx = xy2 dy:

7. (1 + ex=y)dx+ ex=y (1 x=y)dy = 0:

Solve:

1. dydx =2x+9y206x+2y10 .

2. dydx =3x4y23x4y3 .

3. dydx =x+2y32x+y3 .

4. dydx =xyx+1 .

5. (2x+ 3y 5)dy + (3x+ 2y 5)dx = 0.

6. (3y + 2x+ 4)dx = (4x+ 6y + 5)dy.

7. (3y 7x+ 7)dx+ (7y 3x+ 3)dy = 0.

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Method of variation of parameters

Solve by the method of variation of parameters:

1. dydx 5y = sinx.

2. dydx y tanx = cosx.

3. dydx + 2xy = 4x.

4. dydx + 6y = 18e3x.

5. dxdt +1tx = t

2.

6. Find a solution of the dierential equation dydx5y = 0 in the form y = y1:Hence solve dydx 5y = sin x by the substitution y = vy1:

7. Find a solution of the dierential equation dydx y tanx = 0 in the formy = y1(x); say. Hence solve

dydx y tanx = cos x by the substitution

y = v(x)y1(x):

Bernoulli's Equation :

Solve :

1. xy dydx = y3ex2

.

2. (x2y3 + 2xy)dy = dx:

3. y(2xy + ex)dx exdy = 0.

4. dydx +1xy = x

2y6.

5. xdydx + y = y2x3 cosx.

6. cosx dydx y sinx+ y2 = 0.

7. y + 2dydx = y3(x 1).

8. x2y x3 dydx = y4 cosx.

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