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Solve (x+3)²y"-4(x+3)y'+6y = log(x+3)

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$(x+3)^2 y^{\prime \prime}-4(x+3) y^{\prime}+6 y=\log (x+3)$

$t^2 y^{\prime \prime}-4 t y^{\prime}+6 y=0 \] \[ m^2-5 m+6=0 \Rightarrow m=2,3$

$y_n=C_1 t^2+C_2 t^3=C_1(x+3)^2+C_2(x+3)^3$

$y_p=\frac{1}{6} \log (x+3)$

$y=C_1(x+3)^2+C_2(x+3)^3+\frac{1}{6} \log (x+3)$

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Divya Sharma

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