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（x+1）y"+y'=ln(x通解+1)
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（x+1）y"+y'=ln(x通解+1)
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答案和解析
令 z = y' 则 z' = y''
So (x + 1)y'' + y' = ln(x + 1)
--> (x + 1)z' + z = ln(x + 1)
--> (x + 1)z' + z(x + 1) = ln(x + 1)
--> [z(x + 1)]' = ln(x + 1)
--> z(x + 1) = ∫ ln(x + 1) dx
--> z(x + 1) = xln(x + 1) - ∫ (x + 1 - 1)/(x + 1) dx
--> z(x + 1) = xln(x + 1) - ∫ [1 - 1/(x + 1)] dx
--> z(x + 1) = xln(x + 1) - x + ln|x + 1| + C
--> z(x + 1) = - x + (x + 1)ln|x + 1| + C
--> z = (C - x)/(x + 1) + ln|x + 1|
--> y' = (C - x)/(x + 1) + ln|x + 1|
--> y' = C1*ln|x + 1| - 2x + xln|x + 1| + C2
So (x + 1)y'' + y' = ln(x + 1)
--> (x + 1)z' + z = ln(x + 1)
--> (x + 1)z' + z(x + 1) = ln(x + 1)
--> [z(x + 1)]' = ln(x + 1)
--> z(x + 1) = ∫ ln(x + 1) dx
--> z(x + 1) = xln(x + 1) - ∫ (x + 1 - 1)/(x + 1) dx
--> z(x + 1) = xln(x + 1) - ∫ [1 - 1/(x + 1)] dx
--> z(x + 1) = xln(x + 1) - x + ln|x + 1| + C
--> z(x + 1) = - x + (x + 1)ln|x + 1| + C
--> z = (C - x)/(x + 1) + ln|x + 1|
--> y' = (C - x)/(x + 1) + ln|x + 1|
--> y' = C1*ln|x + 1| - 2x + xln|x + 1| + C2
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