Please help me answer the following questions by following the same format that the examples show. The examples are going to be attached to the exercises.

First Exercises: Numbers 8, 18, and 30

\fSolve the initial-value problem. 13. y' = x2 + 2x - 3; )(0) = 4 14. y' = 2x, y(1) = 7 15. y' = (3x + 1; y(0) = 2 16. y' = sin x - x, y(0) = 3 17. f'(x) = x23 - x,f(1) = -6 18. f'(x) = 1/x - 2x + x1/2; f(4) = 2 19. y' - xV/ x2 + 1; y(0) = 3 20. y' = xe*; y(0) = 1 21. y' = x3/(x* + 1)2; y(1) - -2 22. y' = sin x/(2 + cos x); y( 7) = 3 23. y' = xex; y(0) = 2 24. y' = x sin x; y(0) = 3 25. y' = In x; y(1) = 2 26. y' = x In x, y(1) = 0 27. y' = x sin(x2); y(0) = 3 28. y' = exVex + 3; )(0) = 1 Solve the higher-order initial-value problem. 29. f"(x) = 2;f(0) = 3 and f'(0) = 4\fSolve the differential equation. Be sure to check for possible constant solutions. If necessary, write your 27. y' answer implicitly. 28. y' 1. ay = 4xy 2. 3/2 ay = 5X dx dx 29. y' 3. X 4. y' = x2y3 30. y' 2y 31. So xVy2 + 1 6. y' ex+1 is y ey 32. So dy x2 sec y 7. 8. ay = exy Vex + 4 dx (x 3 + 1 ) 3 dx all 9. + 10.2 dy = 12er y3 APP X 33. U 1 1. y' = x cosy 12. y' = ytys a) Vx ex+ 2y 13. y' = 14. y' sin y + cos y ey+ 1 tw b) 15. ay 16. dy _sinLy de ( 12 + 1 ) (1 4 + 1 ) di 34. C in 17. dy = 3x2(y - 2)2 18. dy y2 - 1 W dx dx dr Solve the initial-value problem. If necessary, write your answer implicitly. 19/ 3y 2 dy dx - 2x = 0; y(2) = 5\fFind : dz , dz, dz d x dy dx tz-3) and dz dy 10 1-5 ) 1 ) 2 = 2 X - 3xy dx d2 = 2- 34-7 62 (- 21 - 3 ) - 2 - 3(-37 = 111 dz = 0 - 3x = [- 3x => dz dy dy (0 1 - 5) = - 3 (0 ) = 10] 3 ) 2 = 3x 2 - 2 x y + y x ( ) ) lol d2 = 3 (2 x ) - 2y to d x L7 6X - 24 / 1- 2 1-3) 6(- 27 - 2 ( - 3 ) = 1- 6) (2 = 0 - 2X ( 1) + 1 dy = - 2 x + 1 10 1 - 5 ) - 2(0) + 1 = 1] 5 ) 2 = 2X - 3y no constant so don't plug in 42 = 2 - 0 = 12 - Fx ( - 2 , -4 ) = 12 dx dy dz = 0 - 3 = 3 -7 fy (- 3,-2) =131\f1 ) f ( x , y ) = * 2 + x y + y ? - y fx = 2 x + y fy = x + 24 - 1 f x x = 2 FY X = 1 fxy = 1 FYY = 2 D/= 2 ( 2) - [132 / D - (x x fry - [ xy ] = 370 fxx = 270 7.2 goes at the end . Minimum- ( -3131-3) 2 X 14 2 0 Solve for fx = 2x + y= 0 -> y=- 2x - 2 7 - 2y Solup fy = x+ zy-1=0=7 x+2(-2)-1=0 x - 4 x - 1=0 y = - 2 x - 3X = 1 y= - 2 ( -3 ) 3 - 3 X = - Y = 3 For Z use the original Equation z = f ( - 1 /3 , 2 / 3 ) WI - = ( 3 ) 2 + ( 3) ( 3 ) + (3)2- 27 ) f ( X ly ) = x 2 + y 2 - 2x + ly - 2 fy - 2 4 + 1 fx = 2 x - 2 fxx = 2 FVX = 0 fxy = 0 fy = 2 D = 2 ( 2 ) - [0] 2 = 470 f x X = 2 7 0 Relative Minimum : (17-21 - 2 ) 2 x - 4-0 + 12 +2 24 + 4= 0 fx = 2 x - 2 = 0 = > X= 1 fy = 2 4 + 4 = 0 = 7 1= - 2 y = - 2 2 = f ( 1 1 - 2 ) = ( 1)2 + ( - 2)2- 2 ( 1 ) + 4 ( - 2) - 2 2= - 7 a f ( x y ) = * 2 + y 2 + 2 x - Lly fx = 2 X +2 fy = 24 - 4 fxx = 2 fyx = 0 fxy = 0 fyy = 2 D = 2 ( 2 ) - [012 = 470 fxy = 270 Minimum : ( -1,21-5)(x = 2 X 42 = 7 - 1 2 x +2=0 24-up0 -2 - 2 2x- -2 X = - 1 1/ 2 2 2 = (- 1)2+ (272+ 2(-1)-4(2) = -5 1 1 ) f (x ,y ) = 4 x 2 - 42 fx = 8x fy = - 24 fx x - 8 f yx = 0 FX y = 0 Fyy = - 2 1) = 8 ( 2 ) - [032 =16 fXXX = 870 0 4 - 16 Saddle paint (0,0,0) fx = 8x = >0 8) 8 2 X 2 0 fy = 24 = > 0 Y = 2 Z = (0, 0) = 1(0)2-(0)2(7.3) Maximum- Minimum Problems finding the Relative Extrema for Z = F ( x Ly ) Theorem. Second Derivative test D = f x x fyy - [ f xy] ? a greats If DJO and fxx 70 : Relative Minimum * IF D 70 and fox co : Relative Minimum If D ZO : saddle point ex D = O . No information