求一道高数题17.
3个回答
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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^2/(y-1)
p = 0 或 dp/dy = p/(y-1)
p = dy/dx = 0, 得 y = C;
dp/dy = p/(y-1) ,得 dp/p = dy/(y-1)
lnp = ln(y-1)+lnC1, p = C1(y-1) = dy/dx,
C1dx = dy/(y-1), ln(y-1) = C1x + lnC2
y-1 = C2e^(C1x)
设 y' = p 则 y'' = dp/dx = (dp/dy)(dy/dx) = pdp/dy
微分方程化为 pdp/dy = p^2/(y-1)
p = 0 或 dp/dy = p/(y-1)
p = dy/dx = 0, 得 y = C;
dp/dy = p/(y-1) ,得 dp/p = dy/(y-1)
lnp = ln(y-1)+lnC1, p = C1(y-1) = dy/dx,
C1dx = dy/(y-1), ln(y-1) = C1x + lnC2
y-1 = C2e^(C1x)
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