Assignment 6 Solutions

# Assignment 6 Solutions - MATH 138 Calculus 2, Solutions to...

This preview shows pages 1–2. Sign up to view the full content.

MATH 138 Calculus 2, Solutions to Assignment 6 1: (a) Verify that y = x sin x is a solution of the DE y ( y 00 + y ) = x sin 2 x . Solution: We have y 0 = sin x + x cos x and y 00 = cos x + cos x - x sin x = 2 cos x - x sin x and so y ( y 00 + y ) = ( x sin x )(2 cos x - x sin x + x sin x ) = ( x sin x )(2 cos x ) = x (2 sin x cos x ) = x sin 2 x . (b) Find all the solutions of the form y = ax 2 + bx + c to the DE ( y 0 ( x ) ) 2 + 4 x = 3 y ( x ) + x 2 + 1. Solution: For y = ax 2 + bx + c we have y 0 = 2 ax + b , so ( y 0 ( x ) ) 2 + 4 x = 3 y ( x ) + x 2 + 1 ⇐⇒ ( y 0 ( x ) ) 2 + 4 x - 3 y ( x ) - x 2 - 1 = 0 ⇐⇒ (2 ax + b ) 2 + 4 x - 3( ax 2 + bx + c ) - x 2 - 1 = 0 ⇐⇒ (4 a 2 - 3 a - 1) x 2 + (4 ab + 4 - 3 b ) x + ( b 2 - 3 c - 1) = 0 ⇐⇒ 4 a 2 - 3 a - 1 = 0 , 4 ab + 4 = 3 b , and b 2 = 3 c + 1 From 4 a 2 - 3 a - 1 = 0 we get (4 a + 1)( a - 1) = 0 and so a = - 1 4 or a = 1. When = - 1 4 , the equation 4 ab + 4 = 3 b gives - 1 + 4 = 3 b so b = 1, and then the equation b 2 = 3 c + 1 gives 1 = 3 c + 1 so c = 0. When a = 1, 4 ab + 4 = 3 b gives 4 b + 4 = 3 b so b = - 4 and then b 2 = 3 c + 1 gives 16 = 3 c + 1 so c = 5. Thus there are two solutions, and they are y = - 1 4 x 2 + x and y = x 2 - 4 x + 5. 2:

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## Assignment 6 Solutions - MATH 138 Calculus 2, Solutions to...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online