Answer to Question #285964 in Differential Equations for hector12

Question

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 

Answer 1

$$16y'''+9y''=1458 x^{ k−1} (2x − 1)$$

Corresponding homogeneous differential equation

$$16y'''+9y''=0$$

Characteristic (auxiliary) equation

$$16r^3+9r^2=0$$

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

$$r_1=r_2=0, r_3=-\dfrac{9}{16}$$

The general solution of the homogeneous differential equation is

$$y=c_1+c_2x+c_3e^{-(9/16)x}$$

The particular solution of the non homogeneous differential equation is

$$y_p=Ax^4+Bx^3+Cx^2$$

$$y_p'=4Ax^3+3Bx^2+2Cx$$

$$y_p''=12Ax^2+6Bx+2C$$

$$y_p'''=24Ax+6B$$

Substitute

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

$$=1458 x^{ k−1} (2x − 1)$$

$$x^2=x^{k-1+1}$$

$$108A=2916$$

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

$$96B+18C=0$$

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

(b)

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