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The reducible higher order differential equation XY '+ 1 = y' ^ 2
xy'+1=y'^2
y'^2-xy-1=0
(y'-x/2)^2=1+x^2/4
Y '= x / 2 + √ (1 + x ^ 2 / 4) or y' = x / 2 - √ (1 + x ^ 2 / 4)
dy/dx=x/2+√(1+x^2/4) dy/dx=x/2-√(1+x^2/4)
general solution
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)|
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