The power series solution of the Initial-Value Problem (IVP) (a + 1 yul + zy +...
Question 7 3 pts The solution of the Initial-Value Problem (IVP) zy! - 2y = 4(x - 2) y(1) = 4 y (1) = -1 is 1 23 +22 -3 +3 +2.3 -2.0.4 1 Y L 22 - 2.0 + 4 2 None of them 0 4 2.- - 2 + 1 y = 2 Question 8 3 pts The power series solution of the Initial-Value Problem (IVP) (22 +1)yll + xy + 2xy = 0 y(0) = 2 is...
The power series solution of the Initial-Value Problem (IVP) (x² + 1)yl + xy + 2xy = 0 y(0) = 2 is given by y(0) = 3 4 13 325 2 y=2(1 + + :). 2 + + 3 20 6 2 2125 y= 2 + 3x + +. 6 2 4 23 3.25 y = 32 =3(< + + -) +2 (1 + + :) 3 20 6 2 7.23 21.25 y= 2 + x + + + +......
The power series solution of the Initial-Value Problem (IVP) (x² + 1)yl + xy + 2xy = 0 y(0) = 2 is given by y(0) = 3 4 13 325 2 y=2(1 + + :). 2 + + 3 20 6 2 2125 y= 2 + 3x + +. 6 2 4 23 3.25 y = 32 =3(< + + -) +2 (1 + + :) 3 20 6 2 7.23 21.25 y= 2 + x + + + +......
Question 8 3 pts The power series solution of the Initial-Value Problem (IVP) (x2 + 1)yll + xyl + 2xy = 0 y(0) = 2 is given by y (0) = 3 23 3x5 = 2 1 + + ...)+(2-* + + ...) + 3 20 None of them 7x3 21.25 y= 2 + 3x + +... 6 2 4 --- 3(-one -+...) +2(1-**+..) 7274 y= 2 + x + +...
3 pts Question 7 is The solution of the initial-Value Problem (IVPI 38 - 2y = 4(x - 2) v(I) = 4 W(1) = -1 4 +-9 + 1 O 1 +? - 24 0 417 +-+3 None of them o 1 +- 2c + 4 D Question 8 3 pts The power series solution of the initial Value Problem (IVP) (x+1) + xy + 2xy = 0 (0) = 2 (0) = 3 given by None of them 2125...
The solution of the Initial-Value Problem (IVP) ((2 + y)dz - edy=0 (1) = 0 is given by Oy=ze?-1-1 O = em ? None of them Oy= (x + y) lns Oy= 2ln(0+ y)
The solution of the Initial-Value Problem (IVP) (x + y)dx - xdy = 0 ((1) = 0 is given by y = fer-1 - 1 0 None of them Oy= x ln(x + y) y=x Inc Oy= (x + y) Inc
Question 1 3 pts The solution of the Initial-Value Problem (IVP) Į (x + y)dx – xdy = 0 1 y(1) = 0 is given by Oy= (x + y) In x None of them Oy= xel-1-1 O y = x ln(x + y) Oy= x In x
The solution of the Initial-Value Problem (IVP) ( z* yn – 2y = 4(x - 2) y(1) = 4 Y (1) = -1 is None of them 1 yang +82-2+3 1 y = +23 - 2x + 4 Oy - 1/4 +2² - 22+4 4 y= + x2 – 22+1
Question 7 3 pts The solution of the Initial-Value Problem (IVP) x? yll – 2y = 4(x - 2) y(1) = 4 yl(1) = -1 is 1 y = + x3 - 2x + 4 22 None of them 4 y = + x2 23 + 1 2 1 O Y + x2 – 2x + 4 2 O y = *+2-- + x2 - x + 3 23