{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

WrittenHW06-Solutions

# WrittenHW06-Solutions - Math 216(Section 50 1(3.51 Let us...

This preview shows pages 1–3. Sign up to view the full content.

Math 216 (Section 50) Written Homework #6 – Solutions Fall 2007 1 ( 3.5–1 ) Let us assume y p ( x ) = A e 3 x . Plugging this into the equation, we have 9 A e 3 x + 16 A e 3 x = e 3 x = 25 A e 3 x = e 3 x = 25 A = 1 = A = 1 25 . Therefore y p ( x ) = 1 25 e 3 x . 2 ( 3.5–3 ) Let us assume y p ( x ) = A cos(3 x ) + B sin(3 x ) . Plugging this into the equation, we have [ - 9 A cos(3 x ) - 9 B sin(3 x )] - [ - 3 A sin(3 x ) + 3 B cos(3 x )] - 6 [ A cos(3 x ) + B sin(3 x )] = 2 sin(3 x ) = ( - 15 A - 3 B ) cos(3 x ) + (3 A - 15 B ) sin(3 x ) = 2 sin(3 x ) = ⇒ - 15 A - 3 B = 0 and 3 A - 15 B = 2 Solving them gives A = 1 39 , B = - 5 39 . Therefore y p ( x ) = 1 39 cos(3 x ) - 5 39 sin(3 x ) = 1 39 [cos(3 x ) - 5 sin(3 x )] 3 ( 3.5–5 ) Since sin 2 x = 1 - cos 2 x 2 = 1 2 - 1 2 cos 2 x. We can assume y p ( x ) = A cos 2 x + B sin 2 x + C. Plugging this into the equation, we have - 4 A cos 2 x - 4 B sin 2 x - 2 A sin 2 x + 2 B cos 2 x + A cos 2 x + B sin 2 x + C = 1 2 - 1 2 cos 2 x = ( - 3 A + 2 B ) cos 2 x + ( - 2 A - 3 B ) sin 2 x + C = 1 2 - 1 2 cos 2 x = ⇒ - 3 A + 2 B = - 1 2 , - 2 A - 3 B = 0 , C = 1 2 . Solving the equations for A and B gives A = 3 26 , B = - 1 13 . Therefore y p ( x ) = 3 26 cos 2 x - 1 13 sin 2 x + 1 2 = 1 26 (3 cos 2 x - 2 sin 2 x + 13) . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4 ( 3.5–36 ) Let us first find the complementary solution. The characteristic equation is r 4 - 4 r 2 = 0 = r 2 ( r 2 - 4) = 0 = r 2 ( r + 2)( r - 2) = 0 = r = 0 , ± 2 , where r = 0 has multiplicity 2. So the complementary solution can be written as follows: y c ( x ) = c 1 + c 2 x + c 3 e 2 x + c 4 e - 2 x . Let us next find the particular solution y p ( x ). Assume y p ( x ) = x 2 ( A + Bx + Cx 2 ) = Ax 2 + Bx 3 + Cx 4 . Plugging into the given differential equation, ( - 8 A + 24 C ) - 24 Bx - 48 Cx 2 = x 2 . Hence - 8 A + 24 C = 0 , B = 0 , - 48 C = 1 = A = - 1 16 , B = 0 , C = - 1 48 . Thus y p ( x ) = - 1 16 x 2 - 1 48 x 4 Now the solution for the initial value problem can be written as follows: y ( x ) = c 1 + c 2 x + c 3 e 2 x + c 4 e - 2 x - 1 16 x 2 - 1 48 x 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

WrittenHW06-Solutions - Math 216(Section 50 1(3.51 Let us...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online