y' – 3y = e3t, y(0) = 1
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- Verify that 0t ду дх дх ду 2. 3 y 3 f (x, у) %3D 2х*у — х"у — хуSolve the equation xydx - (x + 2y)2dy = 0 A exy = c y³ (xy) B exy = cy³ (x-y) c) exy = c y³ (x+y) D exy = cy* (x-y)What is the mixed, second order partial derivative of this function. * f (x, y) = 2x² + y² fz. (2, y) = 2(2x? + y²)} - 4x*(2x² + y²)-} Option 1 fay(x, y) = –2xy (2æ² + y°)¯} O Option 2