Question #285964

For which values of A , B , C and k is the function yp(x) = x 2 (A x2 + B x + C) a particular solution of the equation 16 y 000 + 9 y 00 = 1458 x k−1 (2x − 1) ? (a) A = 27 , B = 165 , C = −880 , k = 2 (b) A = 27 , B = −219 , C = 1168 , k = 2 (c) A = 27 , B = −219 , C = 1168 , k = 4 (d) A = 54 , B = −411 , C = 2192 , k = 3 (e) A = 54 , B = −411 , C = 2192 , k = 4 


1
Expert's answer
2022-01-10T15:16:42-0500
16y+9y=1458xk1(2x1)16y'''+9y''=1458 x^{ k−1} (2x − 1)

Corresponding homogeneous differential equation


16y+9y=016y'''+9y''=0

Characteristic (auxiliary) equation


16r3+9r2=016r^3+9r^2=0

r2(16r+9)=0r^2(16r+9)=0

r1=r2=0,r3=916r_1=r_2=0, r_3=-\dfrac{9}{16}

The general solution of the homogeneous differential equation is


y=c1+c2x+c3e(9/16)xy=c_1+c_2x+c_3e^{-(9/16)x}


The particular solution of the non homogeneous differential equation is


yp=Ax4+Bx3+Cx2y_p=Ax^4+Bx^3+Cx^2

yp=4Ax3+3Bx2+2Cxy_p'=4Ax^3+3Bx^2+2Cx

yp=12Ax2+6Bx+2Cy_p''=12Ax^2+6Bx+2C

yp=24Ax+6By_p'''=24Ax+6B

Substitute


16(24Ax+6B)+9(12Ax2+6Bx+2C)16(24Ax+6B)+9(12Ax^2+6Bx+2C)

=1458xk1(2x1)=1458 x^{ k−1} (2x − 1)

x2=xk1+1x^2=x^{k-1+1}

108A=2916108A=2916

384A+54B=1458384A+54B=-1458

96B+18C=096B+18C=0

k=2,A=27,B=219,C=1168k=2, A=27, B=-219, C=1168

(b)


A=27,B=219,C=1168,k=2A=27, B=-219, C=1168, k=2




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