Compute the wronskian they are fundamental set solutions and the general solution what claimed and the

Math 3301	Homework Set 5 – Solutions	10 Points

2. (2 pts)

g x ( )		=	x	−2	e	x	−		x	−3 e		x	+		cx		−3		. The guess for the particular and its
g x ( )		( )		=	c 1 e			−10		t	+	c t 2		e		−10		t
3. (2 pts) I’ll leave it to you to verify that	y c	( )		=	c 1 e			−10		t	+	c t 2		e		−10		t

derivatives are,

100	At	3	+	(	60	A	+	100	B t )	2	+	(	6	A	+	40	B	+	100 C t )	+	2	B	+	20 C	+	100	D	=	50 t	3	−

Setting coefficients equal and solving gives,

The general solution is then,

y t ( )	=	c 1 e	−10	t	+	c t 2 e		−10	t	+	1	t	3	−		3		t	2	+	1	t	−
y t ( )	=	c 1 e	−10	t	+	c t 2 e		−10	t	+	2	t	3	−		10		t	2	+	20	t	−
6. (3 pts) I’ll leave it to you to verify that						y c	( )		=	c 1 e +					c 2		e	8	t	. The guess for the particular solution and

Plugging this into the differential equation and simplifying gives,

e	3 t		(	−11	A	−	3	B	)	cos	( ) (	3	A	−	11 B	)	sin	( )		=	5 e	3 t	cos	( )

26 cos ( )−26 3 sin ( )

7. (3 pts) I’ll leave it to you to verify that cy ( ) = c 1 cos 2 ( t ) + c 2 sin 2 ( t ) . The guess for the particular

	Homework Set 5 – Solutions	10 Points

Plugging this into the differential equation and simplifying gives,

Now apply the initial conditions.

The actual solution is then,

y t ( )	= −	33		)	−		sin 2 ( t	) (	5 t	2	−	3 t	+	13	)	e	6

general solution is ( )	=	c 1 e	r t 1	+	c 2	e	r t 2	.		−	1	t									−	1	t
general solution is ( )	=	c 1 e	r t 1	+		( )				−	1	t									−	1	t
4. I’ll leave it to you to verify that					y c	( )		=	c 1 e		2		cos	(	5	t	)	+	c 2	e		2		sin	(	5	t	)	. The guess for the

Plugging this into the differential equation and simplifying gives,

Math 3301	(	−37	A	−	9	B	)											10 Points
Math 3301	(	−37	A	−	9	B	)	cos 3 ( ) (	9	A	−	37	B	)	sin 3 ( )	=		10 Points

y t ( )	=	c 1 e	−4	t	+	c 2	e	7	t	+	18	cos 3 ( )	−	74	sin 3 ( )
y t ( )	=	c 1 e	−4	t	+	c 2	e	7	t	+	145	cos 3 ( )	−	145	sin 3 ( )