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First order first degree differential equations

02

Chapter First order first degree differential equations Highlights: 2.1 First order first degree differential equations 2.2 Types of first order first degree differential equations 2.3 Section-A (Variable separable) 2.4 Section-B (Homogeneous) 2.5 Section-C (Reducible to homogeneous form)

2.6 Section-D (Exact) 2.7 Section-E (Reducible to exact form) 2.8 Section-F (Linear) 2.9 Section-G (Reducible to linear form) 2.10 Exercise

2.1 First order first degree differential equations: The differential equations whose both order and degree are one are called first order and first degree differential equations. In this chapter, solution techniques of first order first degree differential equations are discussed. Example (2.1): The differential equations

dy  y  0 , cos(x+y)dy = dx, dx

and (x2 + y2)dx = 2xydy are first order first degree differential equations. 2.2 Types of first order first degree differential equations: According to the solution criterions, there are seven types first order first degree differential equations. (a) Variable separable, (b) Homogeneous, (c) Reducible to homogeneous form, (d) Exact, (e) Reducible to exact form, (f) Linear and (g) Reducible to linear form. . 13

First order first degree differential equations

2.3 Section-A Variable separable first order first degree differential equations 2.A.1 Variable separable differential equations: If a differential equation can be expressed in the form f1(x) dx = f2(y) dy, then it is called variable separable differential equation. Variables can be separated by directly or by the method of substitution which are discussed by some done examples.

dy y0 dx dy dy dy Solution: Given that = – y ; Or, = – dx ;  y  0 ; Or, dx dx y dy Or,  = –  dx [Integrating on both sides] y Or, ln y = – x + c Example (2.A.1): Solve the differential equation

Example (2.A.2): Solve x2(1 + y)dy + y2(x – 1)dx = 0 Solution: Given that x2(1 + y)dy + y2(x – 1)dx = 0

1  y  1 1 1 1  x 1 dy =  2 dx ; Or, 2 dy + dy = 2 dx – dx 2  y x x y  x   y 

Or, 

Taking integration, we have

1

y

2

Or, – y

 x

1

1

1

1

 y dy =  x

dy +

+ ln y = – x

–y

1

dx –

2

1

1

 x dx

– ln x + c

+ ln xy = c

Example (2.A.3): Solve

dy + dx

[Answer]

(1  y ) =0 (1 x 2 ) 2

dy (1  y 2 ) + = 0; Or, Solution: Given that dx (1  x 2 ) 14

dy 1 y

2

+

dx 1  x2

=0

First order first degree differential equations 1 1  sin  y + sin  x = c [Integrating]

[Answer]

dy x(2 ln x  1) = dx sin y  y cos y dy x(2 ln x  1) = Solution: Given that dx sin y  y cos y Example (2.A.4): Solve

Or, (sin y + y cos y)dy = x(2 ln x + 1) dx



Or, sin ydy +

 y cos ydy = 2 x ln xdx +  xdx [Taking integration]





Or, sin ydy + y sin y – sin ydy = 2 lnx

x2 –  xdx + 2

So, y sin y = x2 lnx + c [Canceling unnecessary terms]

 xdx [Answer]

dx dx = a(x2 + ) dy dy dx dx = a(x2 + ) Solution: Given that x – y dy dy Example (2.A.5): Solve x – y

Or, x dy – y dx = ax2 dy + a dx ; Or,

Or, (x – ax2)dy = (a + y) dx ;

a  dx dy dy 1 = ; Or, =  dx ay a  y  x 1  ax  x(1  ax)

Or, ln (a + y) = ln x – ln (1 – ax) + ln k ; [Taking c = ln k] Or, ln {(1 – ax)(a + y)} = ln (kx) So, (1 – ax)(a + y) = kx [Answer] Example (2.A.6): Solve sec2x tany dx + sec2y tanx dy = 0 Solution: Given that sec2x tany dx + sec2y tanx dy = 0 Or,

sec 2 x  tan x dx +

sec 2 y  tan y dy = 0 [Integrating]

Or, ln(tan x) + ln(tan y) = ln k ; Or, ln(tan x. tan y) = ln k ; So, tan x tan y = k [Answer] Example (2.A.7): Solve

dy = ex – y + x2 e – y dx 15

First order first degree differential equations

Solution: Given that Or, ey = ex +

1 3 x +c 3

dy = e – y (ex + x2) ; Or, ey dy = (ex + x2)dx dx [Integrating]

[Answer]

Example (2.A.8): Solve x cos2y dx = y cos2x dy Solution: Given that Or,

 x sec

2

xdx =

x y dx = dy 2 cos x cos 2 y

 y sec

2

ydy [Taking integration on both sides]



Or, x tan x – tan xdx = y tan y –

 tan ydy

Or, x tan x – ln sec x = y tan y – ln sec y + c Example (2.A.9): Solve Solution: Given that

a x

a x

[Integrating by parts] [Answer]

dy +x=0 dx

dy = – x ; Or, dy = – dx

x dx ; a x

a xa dx [Taking integration] a x dx Or,  dy =   a  x dx + a ax 3 2 [Answer] Or, y =  (a  x ) 2 + 2a a  x + c 3 Or,

 dy =  

Example (2.A.10): Solve y(1 + xy)dx + x(1 – xy)dy = 0 Solution: Given that y(1 + xy)dx + x(1 – xy)dy = 0

dx dy  )=0 x y

Or, y dx + xdy + xy(y dx – x dy) = 0 ; Or, d(xy) + x2y2(

d ( xy ) dx dy 1  +( ) = 0; Or,  + ln x – ln y = c [Integrating] 2 xy x y ( xy ) x 1 So, ln  =c [Answer] y xy

Or,

16

First order first degree differential equations

Example (2.A.11): Solve Solution: Given that

dy = (4x + y + 1)2 dx

dy = (4x + y + 1)2 ........... (1) dx

Let 4x + y + 1 = z

dy dz = [Differentiating with respect to x] dx dx dy dz Or, = –4 dx dx dz dz dz – 4 = z2 ; Or, = z2 + 4 ; Or, 2 = dx ; From (1), we have dx dx z  22 dz Or,  2 =  dx [Integrating on both sides] z  22 1  z Or, tan 1 = x + c [c is integrating constant] 2 2 1 1  tan 1 (4x  y  1) = x + c [Putting value of z] [Answer] 2 2 So, 4 +

Example (2.A.12): Solve (x + y)2

dy = 2x + 2y + 5 dx

dy = 2x + 2y + 5 ............... (1) dx dy dy dz dz Let x + y = z ; So, 1 + = ; Or, = –1 dx dx dx dx dz From (1), we have z2( – 1) = 2z + 5 ; dx z 2 dz dz Or, z2 = z2 + 2z + 5 ; Or, 2 = dx ; dx z  2z  5 z 2 dz Or,  2 = dx [Integrating on both sides] z  2z  5  Solution: Given that (x + y)2

17

First order first degree differential equations

z2  2 z  5  2 z  2  3 dz =  dx  z 2  2z 5 dz z 2  2z  5 2z  2 dz   2 Or,  2 =  dx dz – 3  z  2z  5 z  2z  5 (z  1) 2  2 2 3 1 z  1 Or, z – ln(z2 + 2z + 5) – tan =x+c 2 2 3 1 x  y  1 =x+c Or, x + y – ln{(x + y)2 + 2(x + y) + 5} – tan 2 2 3 1 x  y  1 Or, y = ln{(x + y)2 + 2(x + y) + 5} + tan + c [Answer] 2 2 Or,

Example (2.A.13): Solve cos(x + y)dy = dx Solution: Given that cos(x + y)dy = dx ; Or, cos(x + y)

dy = 1 ....... (1) dx

dy dy dz dz = ; Or, = –1 dx dx dx dx dz dz From (1), we have cos z ( – 1) = 1 ; Or, – 1 = sec z ; dx dx dz cos zdz 1   Or, = dx ; Or, = dx ; Or, 1  dz = dx ; sec z  1 1  cos z  1  cos z  z 1 z Or, dz – sec 2 dz = dz ; Or, z – tan = x – c [Integrating] 2 2 2 x y = x – c [Substituting value of z] Or, x + y – tan 2 x y So, y + c = tan [Answer] 2 Let

x + y = z ; So, 1 +

   x Example (2.A.14): Solve 1  e y  dx  e y 1   dy = 0 x

 

x

 

18



y

First order first degree differential equations

x x   x y  y   Solution: Given that 1  e dx  e 1  dy = 0 ;    y   x x e y (1  ) y dx =  ....... (1) Or, x dy 1 e y

dx dv =v+y dy dy dv ev (1  v) dv ev (1  v) From (1), we have v + y =  ; Or, y = –v  v v dy dy 1e 1e dv dy v  ev 1ev  dv ; Or, y = ; Or, = v dy y v  ev 1 e Let x = vy 

Or, ln y = – ln (v + ev) + ln c [Integrating] Or, ln{y(v + ev)} = ln c ; Or, y(v + ev)} = c

x y

Or, y (  e x

So, x + y e

y

x

y

) = c [Putting v =

=c

x ] y

[Answer]

Example (2.A.15): Solve (x – y)2

dy = a2 dx

dy = a 2 .............. (1) dx dy dy dz dz = ; Or, =1– Let, x – y = z ; 1 – dx dx dx dx dz From (1) we have, z2 (1 – ) = a 2 ; Or, (z2 – a 2)dx = z2dz dx z2 Or,  dx =  2 dz [Taking integration] z  a2 Solution: Given that (x – y)2

19

First order first degree differential equations

Or,

 dx =  dz + a  z 2

2

1 1 za dz ; Or, x = z + a2. ln +c 2 2a z  a a

a x y a ln + c [Putting value of z] [Answer] 2 x y a dy Example (2.A.16): Solve = sin(x + y) + cos(x + y) dx dy Solution: Given that = sin(x + y) + cos(x + y) .............. (1) dx dy dy dz dz Let x + y = z ; So, 1 + = ; Or, = –1 dx dx dx dx dz From (1), we have – 1 = sin z + cos z ; dx dz dz Or,  dx =  ; Or,  dx =  ; z z z 1  cos z  sin z 2 cos 2  2 sin cos 2 2 2 1 2 z sec dz 2 2 ; Or, x = ln(1  tan z ) + c Or,  dx =  z 2 1  tan 2 x y ) + c [Answer] So, x = ln(1  tan 2 x  y  a dy x ya Example (2.A.17): Solve ( ) = x  y  b dx x yb x  y  a dy x  y a Solution: Given that ( ) = .............. (1) x  y  b dx x  y b dy dy dz dz Let x + y = z ; So, 1 + = ; Or, = –1 dx dx dx dx dz (z  a )(z  b ) From (1), we have –1= ; (z  b )(z  a ) dx Or, y =

20

First order first degree differential equations

2( z 2  ab) z 2  ab  (b  a )z dz = 2 ; Or, dz = 2 dx z 2  ab dx z  ( b  a) z  ab b  a 2z )dz = 2 dx ; Or, (1 2 z 2  ab ba ln(z2 – ab) = 2x + c [Integrating] Or, z + 2

Or,

Or, (b – a) ln{(x + y)2 – ab} = 2(x – y) + k [Putting z = x + y and 2c = k]

xdx  ydy = Example (2.A.18): Solve xdy  ydx Solution: Given that

xdx  ydy = xdy  ydx

a2  x2  y2 x2  y 2

a2  x2  y 2 x2 y2

1 d ( x2  y2 ) a2  ( x2  y 2 ) Or, 2 = ; 2 2 xdy  ydx  x y x2( ) x2 1 2 2 d (x  y ) a2  ( x2  y 2 ) = x2 Or, 2 d (y / x ) x2  y2 1 d (x 2  y 2 ) a 2  ( x 2  y 2) Or, 2 = 1  ( y / x) 2 x 2  y 2 d ( y / x) 1 2 (x  y 2 )1 / 2 d (x 2  y 2 ) d ( y / x) Or, 2 = 1 (y / x )2 a 2  (x 2  y 2 ) Or,

d( x 2  y 2 ) a 2  ( x2  y2 ) 2

=

d ( y / x) 1 (y / x )2 21

First order first degree differential equations

Or, sin

1

x2  y2 y = tan 1 + c [Integrating] x a

[Answer]

2.A.2 Exercise: 1. Solve the following differential equations: (i) ydx + xdy = 0 [Ans: xy = c] (ii) 1 x 2 dy = 1 y 2 dx [Ans: sin–1y = sin–1x + c] (iii) (x + y)(dx – dy) = dx + dy [Ans: x – y = ln(x + y) + c] (iv) (x + y)dy + (x – y)dx = 0 [Ans: 2tan–1(y/x) + ln(x2 + y2) = c] (v)

1 dy + 1 = ex – y [Ans: ex + y = e2x + c] 2 dx

2. Solve using suitable substitutions: (i)

dy = dx

y  x [Ans: 2 y  x + 2 ln( y  x – 1) = x + c]

 dy    x  y [Ans: tan(x + y) – sec(x + y) = x + c]  dx  1 1  dy  (iii) cos    x  y [Ans: tan ( x  y) = x + c] 2  dx  dy 1 x  y = a 2 [Ans: y = a tan (iv) (x + y)2 + c] dx a 1

(ii) sin 

(v)

a2  x2  y 2 y xdx  ydy = [Ans: 2 tan 1 + 2 2 x x y xdy  ydx a 2  x 2  y 2 ] = c]

22...