solve-y-2ay-1-a-2-y-x-e-ax-a-real-

Question Number 90969 by abdomathmax last updated on 27/Apr/20

s o l v e y^{^{″}} - 2 {a y}^{^{'}} + (1 + a^{2}) y = x + e^{a x}

a r e a l

Commented by niroj last updated on 27/Apr/20

y^{^{″}} - {2 ay}^{^{'}} + (1 + a^{2}) y = x + e^{ax}

(D^{2} - 2 aD + 1 + a^{2}) y = x + e^{ax}

A . E .,

m^{2} - 2 am + (1 + a^{2}) = 0

m = \frac{2 a \bar{+} \sqrt{{4 a}^{2} - 4 (1 + a^{2})}}{2}

= \frac{2 a \bar{+} \sqrt{{4 a}^{2} - 4 - {4 a}^{2}}}{2} = \frac{2 a \bar{+} \sqrt{- 4}}{2}

m = a \bar{+} i

CF = e^{ax} (C_{1} \cos x + C_{2} sinx)

PI = \frac{x + e^{ax}}{D^{2} - 2 a D + (1 + a^{2})}

= \frac{1}{1 + a^{2}} [\frac{1}{1 + \frac{D^{2} - 2 aD}{1 + a^{2}}}] x + \frac{e^{ax}}{a^{2} - {2 a}^{2} + 1 + a^{2}}

= \frac{1}{1 + a^{2}} {(1 + \frac{D^{2} - 2 aD}{1 + a^{2}})}^{- 1} x + \frac{e^{ax}}{1}

= \frac{1}{1 + a^{2}} [1 - (\frac{D^{2} - 2 aD}{1 + a^{2}}) + {(\frac{D^{2} - 2 aD}{1 + a^{2}})}^{2} - \dots] x + e^{ax}

= \frac{1}{1 + a^{2}} [1 + \frac{2 aD}{1 + a^{2}}] x + e^{ax}

= \frac{1}{1 + a^{2}} [x + \frac{2 a}{1 + a^{2}} Dx] + e^{ax}

= \frac{1}{1 + a^{2}} [x + \frac{2 a}{1 + a^{2}}] + e^{ax}

y = CF + PI

y = e^{ax} (C_{1} cosx + C_{2} sinx) + \frac{x}{1 + a^{2}} + \frac{2 a}{{(1 + a^{2})}^{2}} + e^{ax} / / .

Commented by mathmax by abdo last updated on 28/Apr/20

t h a n k y o u s i r .

Commented by niroj last updated on 03/May/20

��