Solve: $(x+1)^3y''+3(x+1)^2y'+(x+1)y=6\log(x+1)$
Is my solution correct:
Answer given in the book : $y(x+1)=c_1+c_2\log(x+1)+\log3(x+1)$
For my later reference: link wolfram alpha
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$\begingroup$The $CF$ should be $(C_1+C_2z)e^{-z}$. The rest of your solution is correct and we finally find $$y(x)(x+1)=c_1+c_2\log(x+1)+\log^3(x+1).$$
$\endgroup$ 1 $\begingroup$Robert pointed out your mistake..
Here is another approach
$$(x+1)^3y''+3(x+1)^2y'+(x+1)y=6\log(x+1)$$ divide by $x+1$ $$(x+1)^2y''+3(x+1)y'+y=6\frac {\log(x+1)}{x+1}$$ on the left there is a derivative $$((x+1)^2y')'+((x+1)y)'=6\frac {\log(x+1)}{x+1}$$ Integrate $$(x+1)^2y'+(x+1)y=6\int \frac {\log(x+1)}{x+1} dx$$ $$(x+1)^2y'+(x+1)y=3\log^2(x+1)+K_1$$ Divide again by $x+1$ $$(x+1)y'+y=3\frac {\log^2(x+1)}{x+1}+K_1 \frac 1 {x+1}$$ $$((x+1)y)'=3\frac {\log^2(x+1)}{x+1}+K_1 \frac 1 {x+1}$$ Integrate again $$(x+1)y=3\int\frac {\log^2(x+1)}{x+1}+K_1 \int \frac {dx} {x+1}$$ $$(x+1)y=\log^3(x+1)+K_1 \ln |{x+1}|+K_2$$ $$ \boxed {y=\frac {\log^3(x+1)}{x+1}+K_1 \frac {\ln |{x+1}|}{x+1}+\frac {K_2}{x+1}}$$
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