Question

求一道高数题18.

Answer 1

该微分方程属于缺 x 型，即缺自变量型。
设 y' = p 则 y'' = dp/dx = (dp/dy)(dy/dx) = pdp/dy
微分方程化为 pdp/dy = p^3+p
p = 0 或 dp/dy = 1+p^2
p = dy/dx = 0, 得 y = C；
dp/dy = 1+p^2 ，得 dp/(1+p^2) = dy, arctanp = y+C1
p = dy/dx = tan(y+C1) = sin(y+C1)/cos(y+C1),
cos(y+C1)dy/sin(y+C1) = dx
ln[sin(y+C1)] = x+ lnC2, sin(y+C1) = C2e^x.

Answer 2

这个是分布微分法，将正弦函数换到微分符号的后面，然后运用方法来一部部分布积分。

Answer 3

Answer 4