Answer to Question #238374 in Differential Equations for Juwel

Question

d^2y/dx^2-6dy/dx+9y=1+x+x^2

Answer 1

$\frac{d^{2}y}{dx^2}-6\frac{dy}{dx}+9y=1+x+x^{2}$

P(s)=s²-6s+9

=(s-3)²

Since we have two identical roots,the homogeneous solution will have the form ;

y_h=C₁e^3x + C₂xe^3x

y_p=Ax²+Bx+C

y'_p=2Ax+B

y''_p=2A

9Ax²+(9B-12A)x+(2A-6B+9C)=x²+x+1

9A=1 $\implies$ A=$\frac{1}{9}$

9B-12A=1 $\implies B=\frac{7}{27}$

2A-6B+9C=1$\implies C=\frac{7}{27}$

So y_p=$\frac{1}{9}x^{2}+\frac{7}{27}x+\frac{7}{27}$

$\therefore$ The general solution is:

$y=C_1e^{3x}+C_2xe^{3x}+\frac{1}{9}x^{2}+\frac{7}{27}x+\frac{7}{27}$