Exercise 1.1 (Introduction to differential equations):

State the order and linearity:

43

3

22

2

2) 0 (order 3, non-linear)

5) 1 (order 2, non-linear)

7) ( sin ) (cos ) 2 (order 3, lin

d y dyx y

dxdx

d y dy

dxdx

y yθ θ

− + =

= +

′′′ ′− = ear)

Determine the linearity of the differential equation:

10) ( ) 0uudv v uv ue du+ + − =

If v is the dependent variable:

( ) 0 (1 )u udv dvu v uv ue u u v uedu du

+ + − = ⇒ + + =

(linear in v)

If u is the dependent variable:

( ) 0u duu v uv uedv

+ + − = (non-linear in u)

Verify that the indicated function is a solution of the given DE:

3

3

3 3

3 3 3 3

3 3

13) 6 13 0 ; cos 2

Let: cos 2

3 cos 2 2 sin 2

9 cos 2 6 sin 2 6 sin 2 4 cos 2

5 cos 2 12 sin 2

x

x

x x

x x x x

x x

y y y y e x

y e x

y e x e x

y e x e x e x e x

e x e x

′′ ′− + = =

=

′ = −

′′ = − − −

= −

3 3 3 3 3

substitute , , into the DE:

6 13 0

5 cos 2 12 sin 2 6(3 cos 2 2 sin 2 ) 13( cos 2 ) 0

x x x x x

y y y

y y y

e x e x e x e x e x

′ ′′

′′ ′− + =

− − − + =

3

0 0 (verified)

Therefore cos 2 is a solution to the DE: 6 13 0xy e x y y y

=

′′ ′= − + =

2

14) tan ; (cos ) ln(sec tan )

Let: (cos ) ln(sec tan )

cos sin ln(sec tan ) (sec tan sec )

sec tan

sin ln(sec tan ) 1

y y x y x x x

y x x x

xy x x x x c x

x x

x x x

′′ + = = − +

= − +

′ = + − ++

= + −

2sin cos ln(sec tan ) (sec tan sec )

sec tan

cos ln(sec tan ) tan

substitute , , into the DE:

tan

cos ln(sec tan ) tan (cos ) ln(sec

xy x x x x c x

x x

x x x x

y y y

y y x

x x x x x

′′ = + + ++

= + +

′ ′′

′′ + =

+ + − tan ) tan

tan tan (verified)

Therefore (cos ) ln(sec tan ) is a solution to the DE: tan

x x x

x x

y x x x y y x

+ =

=

′′= − + + =

3 1/2

1/2

3/2

3

3/2

18) 2 cos ; (1 sin )

Let: (1 sin )

1 (1 sin ) ( cos )

2

substitute , into the DE:

2 cos

1 2( (1 sin ) ( cos )) (

2

y y x y x

y x

y x x

y y

y y x

x x

−

−

−

−

′ = = −

= −

′ = − − −

′

′ =

− − − = 1/2 3

3/2 3/2

1/2 3

(1 sin ) ) cos

(1 sin ) cos (1 sin ) cos (verified)

Therefore (1 sin ) is a solution to the DE: 2 cos

x x

x x x x

y x y y x

−

− −

−

−

− = −

′= − =

2 2 2

2 2

2

2

20) 2 ( ) 0; 2 1

From: 2 1

differentiate form: 4 2 2 0

4 2( ) 0

xydx x y dy x y y

x y y

xydx x dy ydy

xydx x y dy

+ − = − + =

− + =

− − + =

− − − =

2

2 2 2

( 2) : 2 ( ) 0 (verified)

Therefore 2 1 is a solution to the DE: 2 ( ) 0

xydx x y dy

x y y xydx x y dy

÷ − ⇒ + − =

− + = + − =

2 119) =( 1)(1 2 ); ln

1

2 1 From: ln ln(2 1) ln( 1)

1

2 1 differentiate wrt to : 1

(2 1) ( 1)

dX XX X t

dt X

Xt X X t

X

dX dXt

X dt X dt

− − − = −

− = ⇒ − − − = −

⋅ − ⋅ =− −

2 1 1

(2 1) ( 1)

2( 1) (2 1) 1

(2 1)( 1)

1

(2 1)( 1)

dX

dt X X

dX X X

dt X X

dX

dt X X

− = − −

− − −= − −

− − −

1

( 1)(2 1) or =( 1)(1 2 ) (verified)

2 1 Therefore ln is a solution to the DE: =( 1)(1 2 )

1

dX dXX X X X

dt dt

X dXt X X

X dt

=

= − − − − −

− = − − −

2 2 2

2 2 2

2 2 2

2 2 2 2 2

2 2 2

0 1

0 1

0 1

0 0 1

22) 2 1;

From:

( ) ( )

2 2

2

x t xx

x t xx

x t xx

x t x t xx x

x x x

dyxy y e e dt c e

dx

y e e dt c e

d dy e e dt c e

dx dx

de e dt xe e dt c xe

dx

e e xe

− −

− −

− −

− − −

− −

∫

∫

∫

∫ ∫

+ = = +

= +

′ = +

= − −

= ⋅ −2 2 2 2 2

2 2 2 2 2 2

0 01 1

0 01 1

2 1 2 2

substitute into the DE:

2 1 1 2 2 2 ( ) 1

t x x t xx x

x t x x t xx x

e dt c xe xe e dt c xe

y

dyxy xe e dt c xe x e e dt c e

dx

− − −

− − − −

∫ ∫

∫ ∫

− = − −

′

+ = ⇒ − − + + =

2 2 2

0 1

1 1 (verified)

Therefore is a solution to the DE: 2 1x t xx dyy e e dt c e xy

dx

− −∫

=

= + + =

Exercise 1.2 (Initial Value Problems)

Given 1

1

(1 )x

yc e

−=

+is a one-parameter family of solutions of the DE: 2y y y′ = − , find a

solution of the first-order IVP consisting this DE and the given initial condition:

1

(0)1

1 1

1) (0) 1/ 3 0, 1/ 3

1 substitute into the general solution

(1 )

1 1

3 (1 )

(1 ) 3 4

1 solution for the IVP is:

(1 4 )

2) ( 1) 2

x

x

y x y

yc e

c e

c c

ye

y x

−

−

−

= − ⇒ = = −

=+

− =+

+ = − ⇒ = −

=−

− = ⇒

1

1 1( 1)1

(1 )

1, 2

1 substitute into the general solution

(1 )

1 1 1 2 1

2 2(1 )

1 solution for the IVP is:

[1 0.5 ]

x

x

y

yc e

c e cec e

ye

−

− −

− +

= − =

=+

= ⇒ + = ⇒ = −+

=−

Given 1 2cos sinx c t c t= + is a two-parameter family of solutions of the DE: 0x x′′ + = , find a

solution of the second-order IVP consisting this DE and the given initial conditions:

1 2 1 2

1 2 1 2

9) ( / 6) 1/ 2, ( / 6) 0

Given: cos sin , sin cos

( / 6) 1/ 2 / 6, 1/ 2

1 1 3 1 cos sin ..... (i)

2 6 6 2 2 2

( / 6) 0 / 6, 0

x x

x c t c t x c t c t

x t x

c c c c

x t x

π π

π π

π π

π π

′= =

′= + = − +

= ⇒ = =

= + ⇒ = +

′ = ⇒ = =

1 2 1 2

1 2

1 3 0 sin cos 0 ..... (ii)

6 6 2 2

solve (i) and (ii): 3 / 4, 1/ 4

3 1 solution for the IVP is: cos sin

4 4

c c c c

c c

x t t

π π= − + ⇒ = − +

= =

= +

1 2 1 2

1 2 1 2

10) ( / 4) 2, ( / 4) 2 2

Given: cos sin , sin cos

( / 4) 2 / 4, 2

1 1 2 cos sin 2 ..... (i)

4 4 2 2

( / 4) 2 2 / 4, 2 2

x x

x c t c t x c t c t

x t x

c c c c

x t x

π π

π π

π π

π π

′= =

′= + = − +

= ⇒ = =

= + ⇒ = +

′ = ⇒ = =

1 2 1 2

1 2

1 1 2 2 sin cos 2 2 ..... (ii)

4 4 2 2

solve (i) and (ii): 3, 1

solution for the IVP is: 3cos sin

c c c c

c c

x t t

π π= − + ⇒ = − +

= = −

= −

Given 1 2x xy c e c e−= + is a two-parameter family of solutions of the DE: 0y y′′ − = , find a

solution of the second-order IVP consisting this DE and the given initial conditions:

1 2 1 2

1 11 2

1 11 2

21 2

12) (1) 0, (1)

Given: ,

(1) 0 0 ..... (i)

(1) ..... (ii)

1 1 solve (i) and (ii): ,

2 2

sol

x x x x

y y e

y c e c e y c e c e

y c e c e

y e e c e c e

c c e

− −

−

−

′= =

′= + = −

= ⇒ = +

′ = ⇒ = −

= = −

21ution for the IVP is: ( )

2

x xy e e −= −

1 2 1 2

11 2

11 2

1 2

13) ( 1) 5, ( 1) 5

Given: ,

( 1) 5 5 ..... (i)

( 1) 5 5 ..... (ii)

solve (i) and (ii): 0, 5 /

x x x x

y y

y c e c e y c e c e

y c e c e

y c e c e

c c e

− −

−

−

′− = − = −

′= + = −

− = ⇒ = +

′ − = − ⇒ − = −

= =

(1 )solution for the IVP is: 5 xy e− +=