Answer to Question #140898 in Differential Equations for Victor

Question

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

Answer 1

Given the D.E $y'''+y''=8x^2$

The auxiliary equation is of the form $m^3+m^2=0$

Equivalently we have $m^2(m+1)=0$ .......(1)

Solving (1) we have $m=-1$ or $m=0$ (twice)

Then the complimentary function is of the form : $Y_c(x)=A_1e^{-x}+A_2x+A_3$

Where $A_1,A_2,A_3$ are constants.

Assuming the general form of the particular integral of the RHS of the DE:

$Y_p = Ax^4+Bx^3+cx^2$

$Y'_{p} = 4Ax^3+3Bx^2+2cx\\ Y''_{p} = 12Ax^2+6Bx+2c\\ Y'''_{p} = 24Ax+6B\\ \\Then\\ Y'''_p+Y''_p=12Ax^2+6Bx+24Ax+6B+2c=8x^2$

By the comparison of coefficients we have

$12A=8 ;$ $6B+24A=0;$ $6B+2c= 0$

Solving the equations simultaneously we have:

$A=\frac{2}{3};B=-\frac{8}{3};c=8$

Therefore, $Y_p=Y_p = \frac{2}{3}x^4-\frac{8}{3}x^3+8x^2$

The general solution of the DE is now of the form:

$Y(x)=Y_c(x)+Y_p(x)= (A_1e^{-x}+A_2x+A_3)+\frac{2}{3}x^4-\frac{8}{3}x^3+8x^2$

$Y(x)=A_1e^{-x}+\frac{2}{3}x^4-\frac{8}{3}x^3+8x^2+A_2x+A_3$