Question

可降价的高阶微分方程 求解过程 xy'+1=y'^2

Answer 1

xy'+1=y'^2
y'^2-xy-1=0
(y'-x/2)^2=1+x^2/4
y'=x/2+√(1+x^2/4) 或 y'=x/2-√(1+x^2/4)
dy/dx=x/2+√(1+x^2/4) dy/dx=x/2-√(1+x^2/4)
通解
y=x^2/4+(1/8)x√(1+x^2/4)+(1/4)ln|x/2+√(1+x^2/4)|+C
y=x^2/4-(1/8)x√(1+x^2/4)-(1/4)ln|x/2+√(1+x^2/4)|+C

∫√(1+x^2/4)dx
x/2=tanu
=(1/2)∫secu^3du
=(1/2)∫secudtanu=(1/2)secutanu-(1/2)∫tanu^2 secudu
=(1/2)secutanu-(1/2)∫secu^3du+(1/2)∫secudu
∫secu^3du=(1/2)secutanu+(1/2)ln|secu+tanu|
(1/2)∫secu^3du=(1/4)secutanu+(1/4)ln|secu+tanu|
∫√(1+x^2/4)dx=(1/8)x√(1+x^2/4)+(1/4)ln|x/2+√(1+x^2/4)|