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U. + 2y + y + 1 -e: y(0) = 0, y'(o) - 2 In Problems 31-36, determine the form of a particular solution for the differential equation. Do not solve. 31. y" + y = sin : + i cos + + 10' 32. y" - y = 2+ + te? + 1221 x" - x' - 2x = e' cos - + cost y" + 5y' + 6y = sin t - cos 2t 35. y" –...
(d) The line integral [(x+y?)dx + (x2 + 2xy)dy, where the positively oriented curve C is the boundary of the region in the first quadrant determined by the graphs of x=0, y=x2 and y=1, can be converted to A 2xdydx 0 0 BJ 2 xdxdy 0 0 С -2x)dyda 00 D none of the above (e) Consider finding the maximum and minimum values of the function f(x, y) = x + y2 - 4x + 4y subject to the constraint...
Use the reduction of order method to solve the following problem given one of the solution y1. (a) (x^2 - 1)y'' -2xy' +2y = 0 ,y1=x (b) (2x+1)y''-4(x+1)y'+4y=0 ,y1=e^2x (c) (x^2-2x+2)y'' - x^2 y'+x^2 y =0, y1=x (d) Prove that if 1+p+q=0 than y=e^x is a solution of y''+p(x)y'+q(x)y=0, use this fact to solve (x-1)y'' - xy' +y =0
use the fact that y=x is a solution of the homogeneous equation x^2y''-2xy'+2y= 0 to completely solve thee differential equation x^2y''-2xy'+2y= x^2
c. exact with solution ZXY+y+C d. exact with solution 2xy +y=c e. not exact lemy 8. The solution of (x +2y )dx + ydy = 0 is a. Inx+ln(y+x)=C b. 1((v+x)/x)=C c. 1(y+x)+x/(x+x)=C d. In(y+x)+x/(y+x) +c e. it cannot be solved
Solve by the Method of Undetermined Coefficients. 1. " - 3y' - 4y = 3e2x (ans. y = C1e4x + cze* - e2x) 2. " - 4y = 4e3x (ans. y = C1 e - 2x + C2 e 2x + 4/5 e3x) 3. 2y" + 3y' + y = x2 + 3 sin x (ans. y = ci e-* + C2 e-x/2 + x2 - 6x + 14 - 3/10 sin x- 9/10 cos x) 4. Y" + y'...
[10] 2. Solve the initial value problem (x^ + 3y^ + c)2 + 2xy + y = 0, g(1) = 1.
15. (2xy + y^2 ) dx + (2xy + x^2 − 2x 2y^2 − 2xy^3 ) dy = 0
7. Given that y(x) = sin 2x is a particular solution to y" + 2y + 4y - 4 cos 2x = 0, find the general solution.
Question 10 Find the differential equation of the given family y =C + 2 a)xy 3y +6=0 b) 2y-e-0 c)y+2y1=0 y+2y-1-0 d) y+2xy 1-0 f) None of the above. Question 11