Finn den generelle løsningen av y′′−3y′+2y=e3xy'' - 3y' + 2y = e^{3x}y′′−3y′+2y=e3x.
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Homogen: r2−3r+2=(r−1)(r−2)=0r^2 - 3r + 2 = (r-1)(r-2) = 0r2−3r+2=(r−1)(r−2)=0, yh=C1ex+C2e2xy_h = C_1 e^x + C_2 e^{2x}yh=C1ex+C2e2x. Partikulær: yp=12e3xy_p = \tfrac{1}{2}e^{3x}yp=21e3x. Generell: y=C1ex+C2e2x+12e3xy = C_1 e^x + C_2 e^{2x} + \tfrac{1}{2}e^{3x}y=C1ex+C2e2x+21e3x.
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