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Find the general solution of the differential equation y ′ + ycosx = (LNX) e-sinx
The given equation is a first order linear differential equation, and P (x) = cosx, q (x) = (LNX) e-sinx, so the general solution of the original equation is y = e − & nbsp; P (x) DX [& nbsp; Q (x) e & nbsp; P (x) dxpdx + C] = e − & nbsp; cosxdx [& nbsp; (LNX) e − sinxe − & nbsp; cosxdxdx + C] = e-sinx (& nbsp; lnxdx + C) = e-sinx (xlnx-x + C)
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