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## MathSofiya 3 years ago $y''+6y'+9y=1+x$ $r_1=r_2=-3$ $y_c=c_1e^{-3x}+c_2xe^{-3x}$ $y_p(x)=Ax+B$ $y_p'(x)=A$ $y_p''(x)=0$ $0+(9A)x+(6A+B)=1+x$ $A=\frac19$ $B=\frac13$ $y_p(x)=\frac19x+\frac13$ $y(x)=c_1e^{-3x}+c_2e^{-3x}+\frac19x+\frac13$ how did I do?

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1. dumbcow

pretty good...i only see 1 thing it should be 0+(9A)x +(6A+9B) = 1+x 9(Ax +B) = (9A)x + 9B it only changes the B value to 1/27

2. MathSofiya

oh thank you! silly me....hihi

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