Question

Y''-2y'+y=(x^3)*(cos2x) using method of undetermined coefficients

Answer 1

here p=1-
q=2
r=1
1+p+q=0
y=e^x is part of cf
$let \: y = v {e}^{x} \\ \frac{dy}{dx} = v {e}^{x} + {e}^{x} \frac{dv}{dx} \\ \frac{ {d}^{2} y}{d {x}^{2} } = 2{e}^{x} \frac{dv}{dx} + v {e}^{x} + {e}^{x} \frac{ {d}^{2}v }{d {x}^{2} } \\ put \: in \: original \: eqn \\ (2 {e}^{x} \frac{dv}{dx} +v {e}^{x} + {e}^{x} \frac{ {d}^{2}v }{d {x}^{2} } ) - 2(v {e}^{x} + {e}^{x} \frac{dv}{dx} ) + v {e}^{x} = {x}^{3} \cos(2x) \\ \frac{ {d}^{2} v}{d {x}^{2} } ( {e}^{x} ) + \frac{dv}{dx} (2 {e}^{x} - 2 {e}^{x} ) - v(2 {e}^{x} - 2 {e}^{x} ) = {x}^{3} \cos(2x) \\ \\ \frac{ {d}^{2}v }{d {x}^{2} } ( {e}^{x} ) = {x}^{3} \cos(2x) \\ \frac{ {d}^{2}v }{d {x}^{2} } = {e}^{ - x} {x}^{3} \cos(2x) \\ \\ now \: u \: can \: solve \: it$
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