For the next three questions, consider the differential equation: y" – 2y/ + y = e* arctan r If the complementary function of this equation is y. = ce+czxe², then by the method of variation of parameters, we can assume form of its particular solution, y, = u1Y1 + uzY2 . Determine uj. If the complementary function of this equation is y. = ce"+cre", then by the method of variation of parameters, we can assume form of its particular solution, y, = u1Yı + uzy2 . Determine uz. Determine the particular solution of the given DE.

Q: Find the differential of the function. y = e³x dy = =

A: We have given function  y=e3x dydx=ddxe3x dydx=e3x·ddx3x dydx=e3x·3 dydx=3e3x dy=3e3x dx

Q: (бх + у?)dх + y(2х — Зу)dy %3D 0

A: We have to solve given problem:

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Q: y'' – 37y" + 316y' – 780y = (e6* + e-5x)2

A:

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A: We can find the solution.