Answer is E 7. Find the solution to the initial value problem dy da 6ry2(3ar2 +...
Find the solution to the initial value problem dy 6xy + y2 + (3x2 + 2xy + 2y) dx =0 y(1) = 3 OA x+y + 2xy2 + y2 + x = 31 OB. 6xy + 2y2 + x = 37 3x²y + 2x2y + x3 + 2x2 + 2y = 24 OD 3x2y + xy2 + y2 = 27 ОЕ xºy + x2y2 + y2 + x = 22
QUESTION 19 Find the solution to the initial value problem dy 6xy + y2 + (3x2 + 2xy + 2y) dc = 0 { wives y(1) = 3 ОА. 3x²y + xy² + y2 = 27 xºy + x²y2 + y2 + x = 22 Ос. 3.xạy + 2x^y + x3 + 2x2 + 2y = 24 x+y + 2xy2 + y2 + x = 31 OL 6xy + 2y2 + x = 37
Bb Take Test: MATH 216 SUMMER 2 X Bb Blackboard Collaborate Ultra - M X X c olearn.okan.edu.tr/webapps/assessment/take/launch.jsp?course_assessment_id=_9160_18_course_id=_11453_18_content_id=_118194_1&step=null K ... Remaining Time: 1 hour, 14 minutes, 34 seconds Question Completion Status QUESTION 19 4 points Save Aww. Find the solution to the initial value problem dy 6.xy + y2 + (3x2 + 2xy + 2y) de =0 y(1) = 3 ry+r+y2 + y² + x = 22 3x+y +2.c?y++2r2 + 2y = 24 6.ry + 2y2 + = 37 00....
Solve the initial value problem dy dx+2y-4e0y(O)2 The solution is y(x) Solve the initial value problem dy dx+2y-4e0y(O)2 The solution is y(x)
In problems 7 and 8 find the solution of the given initial value problem in explicit form: 7. sin 2.x dx + cos 3y dy = 0, y /2) = 1/3. 8. y' (1-22)/2 dy = arcsin x dx, y(0) = 1.
solution for all 4 please In Problems 1-3, solve the given DE or IVP (Initial-Value Problem). [First, you need to determine what type of DE it is. 1. (2xy + cos y) dx + (x2 – x sin y – 2y) dy = 0. 1 dy 2. + cos2 - 2.cy y(y + sin x), y(0) = 1. + y2 dc 3. [2xy cos (2²y) – sin x) dx + x2 cos (x²y) dy = 0. (1+y! x" y® is...
Previous Problem Problem List Next Problem (1 point) Find the solution to the initial-value problem dy = 2y+zez?, y(0) = 10 y = help (formulas) Preview My Answers Submit Answers Show me another
Solve the initial value problem. 9 dy 3 +5y 3 e 0, y(0)=7 dx The solution is y(x) =I
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...
Solve the initial value problem. 7 dy + 9y - 9 e-X = 0, y(0) = dx 8 The solution is y(x) =