Answer to Question #199971 in Differential Equations for sania khan

Question

solve the undetermined coefficient y''+7y' =42x²+5x+1

Answer 1

Let us solve the differential equation $y''+7y' =42x^2+5x+1.$

The characteristic equation $k^2+7k=0$ has the solutions $k_1=0$ and $k_2=-7.$ Therefore, the general solution of the differential equation is of the form $y(x)=C_1+C_2e^{-7x}+y_p.$ Since $k=0$

is a simple root of the characteristic equation, $y_p=x(ax^2+bx+c)=ax^3+bx^2+cx.$

$y_p'=3ax^2+2bx+c,\ \ y_p''=6ax+2b.$

It follows that $6ax+2b+7(3ax^2+2bx+c)=21ax^2+(6a+14b)x+(2b+7c)=42x^2+5x+1.$

Consequently, $21a=42, \ 6a+14b=5, \ 2b+7c=1.$ Therefore, $a=2,\ b=-\frac{1}{2},\ c=\frac{2}{7}.$

We concude that the general solution of the differential equation is of the form $y(x)=C_1+C_2e^{-7x}+2x^3-\frac{1}{2}x^2+\frac{2}{7}x.$