H4_1130

# H4_1130 - f x To do that use the following results in...

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HW 4 Solutions 3.1.27 a) If y = y e is a constant solution, then ( y e ) 0 = 0 , ( y e ) y 00 = 0 . Plug these into ay 00 + by 0 + cy = d . i.e. a 0 + b 0 + cy e = d . Thus, y e = d/c is the only constant solution. b) Let Y = y - y e . Then Y + y e = y is a solution of the diﬀerential equation ay 00 + by 0 + cy = d . i.e. a ( Y + y e ) 00 + b ( Y + y e ) 0 + c ( Y + y e ) = d This implies, aY 00 + bY 0 + c ( Y + y e ) = d After plugging in y e = d/c , we will get the following diﬀerential equation for Y: aY 00 + bY 0 + c ( Y + d/c ) = d which is the same as aY 00 + bY 0 + cY = 0 3.2.28 P ( x ) y 00 + Q ( x ) y 0 + R ( x ) y = 0. is EXACT if it can be rewritten as[ P ( x ) y 0 ] 0 + [ f ( x ) y ] 0 = 0. Set them equal : P ( x ) y 00 + Q ( x ) y 0 + R ( x ) y = [ P ( x ) y 0 ] 0 + [ f ( x ) y ] 0 = P ( x ) y 00 + P 0 ( x ) y 0 + f ( x ) 0 y + f ( x ) y 0 = P ( x ) y 00 + ( P 0 ( x ) + f ( x )) y 0 + f 0 ( x ) y Equate the coeﬃcients: Q ( x ) = P 0 ( x ) + f ( x ) and R ( x ) = f 0 ( x ) To eliminate f(x), we need to take the derivative of the ﬁrst equation and get Q 0 ( x ) = P 00 ( x ) + f 0 ( x ). Then use the second equation ( replace f’ by R(x) ) to get the desired equality: Q 0 ( x ) = P 00 ( x ) + R ( x ) or P 00 ( x ) - Q 0 ( x ) + R ( x ) = 0

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3.2.29 y 00 + xy 0 + y = 0 P ( x ) = 1 , Q ( x ) = x, R ( x ) = 1 Check exactness condition P 00 ( x ) - Q 0 ( x ) + R ( x ) = 0: Since P 00 ( x ) = 0, Q 0 ( x ) = 1 and R ( x ) = 1 the condition is immediately satisﬁed . (0-1+1=0 ). So, this diﬀerential equation is exact. Now ﬁnd
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Unformatted text preview: f ( x ). To do that use the following results in previous question Q ( x ) = P ( x ) + f ( x ) and R ( x ) = f ( x ) The ﬁrst equation implies that f ( x ) = Q ( x )-P ( x ) = x-0 = x . Hence, the diﬀerential equation can be written as [ y ] + [ xy ] = 0 which is the same as [ y + xy ] = 0 Integrating this equation once, we get y + xy = c 1 Iit is now a ﬁrst order linear diﬀerential equation. Use integrating factor method to solve this equation for y. μ = e R xdx = e x 2 / 2 then ( e x 2 / 2 y ) = c 1 e x 2 / 2 . Integrating this you get, e x 2 / 2 y = R c 1 e x 2 / 2 dx + c 2 Thus the solution is y = 1 e x 2 / 2 Z c 1 e x 2 / 2 dx + c 2...
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H4_1130 - f x To do that use the following results in...

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