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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
    • 3 years ago
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    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
    • 3 years ago
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    oh thank you! silly me....hihi

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