Problem 4. Verify that the differential equation is exact then solve it! (4x + 2y)dx +...
Solve the following exact differential equation with initial value. (5x + 4y)dx + (4x - 8y3)dy = 0, y(0) = 2
Solve the exact differential equation (4x*y+sinx)dx+(x4-y)dy=0.
4. Solve the exact differential equation. (1-2xy)dx + (4y3 - x2)dy 0 4. Solve the exact differential equation. (1-2xy)dx + (4y3 - x2)dy 0
Dif equations 4 4. a) Determine whether the following differential equation is exact. (x + 2y) dx + (2x - y)dy = 0 b) Find the general solution using the method of exact differentials.
No 4. Solve the differential equation dy dx . Solve the initial value problem: y" + 3y' + 2y 10 cosx, y(0) 1,y'(0) 0
(15 points) In this problem we consider an equation in differential form M dx + N dy = 0. (- (4xy2 + 4y)) dx +(- (4x²y + 4x))dy = 0 Find My N. If the problem is exact find a function F(x, y) whose differential, dF(x, y) is the left hand side of the differential equation. That is, level curves F(t, y) = C, give implicit general solutions to the differential equation. If the equation is not exact, enter NE...
In this problem we consider an equation in differential form M dx + N dy = 0. (4x4 + 2y) dx +(- (2x + y2))dy = 0 Find My Nx = = If the problem is exact find a function F(x, y) whose differential, dF(x, y) is the left hand side of the differential equation. That is, level curves F(x, y) solutions to the differential equation. C, give implicit general If the equation is not exact, enter NE otherwise find...
can someone solve this differential equation Which of the following is an exact differential equation ? Select one: a. 3xdy + (x − 2) dx = 0 b. x'ydx – y’xdy = 0 c. 2xydx + (2 + x²) dy 50 d. (2x² + 1) dx – xydy = 0
correct me if wrong Verify by substitution that y is the solution for the differential equation Y' + 4y = 1.4 The solution is y = Cel-4x) + 0.35 Where the initial Condition (IC) is given by: y(x=0) = 2 y = dy +4y=114 Cdx) dx dy +4y= 1.46dx) dy +4(2)=11444 dy +8=1,4dx
Solve the equation. (2x)dx + (2y - 4x^y 'dy =0 by multiplying by the integrating factor. An implicit solution in the form F(x,y)=C is = C, where C is an arbitrary constant, and (Type an expression using x and y as the variables.) the solution y = 0 was lost the solution x = 0 was lost no solutions were lost