Math, asked by seshamchamundeswari9, 7 days ago

(D-2)^2y=8(e^2x+sin2x+y^2)

Answers

Answered by ansarizara478

Answer:

What is the particular integral of (D^2-4D+4) y=8 (e^2x+sin2x + x^2)?

Given diffrential equation is

=> (D^ 2 - 4D + 4) y = 8( e^ 2x + sin2x + x^ 2)

Particular integral = 8e^ 2x/(D^ 2- 4D +4) +8sin2x/(D^ 2 - 4D + 4) + 8x^2/(D^ 2 - 4D + 4)

P. I= y1 + y2 + y3. .(1)

y1= 8e^2x/(D^ 2 - 4D + 4)

Put D= 2

=> y1 = 8e^2x/(4 - 8 + 4)

=> y1= 8e^2x/0

Denominator is zero so we will differentiate the denominator.

=> y1 = 8xe^2x/(2D - 4)

Again, when we put D = 2 then denominator is become zero. Hennce we will differentiate again

=> y1 = 8e^ 2x/ 2

=> y1 = 4e^ 2x

Now y2 = 8sin2x/(D^ 2 - 4D + 4)

=> y2 = 8sin2x/(- 4 - 4D + 4)

=> y2 = 8sin2x/( - 4D)

=> y2 = - 2[1/D(sin2x) ]

=> y2 = - 2( - cos2x/2 )

=> y2 = cos2x

At Last, y3 = 8x^2 /(D^ 2 - 4D + 4)

=> y3 = 8x^2/{ 4 (1- D +D^2/4) }

=> y3= 4x^2/( 1 - D + D^2/4)

=> y3 = 4x^2[ 1 - D + D^2/4]^-1

=> y3 =4x^2[ 1- (D- D^2/4) ]^-1

=> y3 = 4x^2 [ 1 +( D - D^ 2/4) + ( D - D^2/4) ^2 ….…and so on]

=> y3 = [ 4x^2 + 4D(x^2) - 4D^2(x^2/4) ]

=> y3 = 4x^2 + 8x - 2

Hence, PI= y1 + y2 + y3

=> PI= 4e^2x+ cos2x +4x^2+ 8x - 2

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