Question

y'''+y''=8x^2

Answer 1

Third-order linear ordinary differential equations.

$y'''+y''=8x^2$ (1)

 Homogeneous equation

$y'''+y''=0$ (2)

The characteristic equation is:

$\lambda^3+ \lambda^2=0$

$\lambda^2(\lambda+1)=0$

$\lambda_{1,2}=0,$

$\lambda+1=0$

$\lambda_3=-1$ .

The general solution of homogeneous equations (2) is

$y_h(x)= C_1+C_2x+ C_3\exp(-x)$

Then the general solution of nonhomogeneous equations is given by

$y = y_h + y_p .$

The particular solution is

$y_p =x^2*(ax^2+bx+c).$

To determine the unknown coefficients (a,b,c) , we can use the initial equation (1)

$y'_p =(ax^4+bx^3+cx^2)'= 4ax^3+3bx^2+2cx$

$y''_p =(4ax^3+3bx^2+2cx)'= 12ax^2+6bx+2c$ (3)

$y'''_p =(12ax^2+6bx+2c)'= 24ax+6b$ (4)

Substituting (3), (4) into equation (1) we obtain

$24ax+6b+ 12ax^2+6bx+2c=8x^2$ ,

$12ax^2+(24a+6b)x+(6b+2c)=8x^2$

Further, we equate the coefficients with equal degrees:

$12a=8\\ 24a+6b=0\\ 6b+2c=0$

From here:

$a =\frac{2}{3},b=-\frac{8}{3},c=8$

So, the general solution of nonhomogeneous equations (1) is

$y(x)= C_1+C_2x+ C_3\exp(-x)+\frac{2x^4}{3}-\frac{8x^3}{3}+8x^2$ .