高数--微分方程 求通解

Question

如题所述

Answer 1

整理后,均可化为一阶线性方程.
一阶线性方程: y' +yP(x) = Q(x)的通解为:
y = [e^(-∫Pdx)]*{ ∫Q*[e^(∫Pdx)]dx +C}

1.dy/dx = y/(x+y), 改写为: dx/dy = x/y +1, dx/dy -x/y =1.
(将x看作是y的函数) :有P=-1/y, Q =1.
-∫Pdy =lny+c1, (可 取c1 =0), [e^(-∫Pdy)]* =y, (对数的性质)
按公式,有: x = y*{∫1*[e^(∫-1/ydy)] dy+C} =y*{∫[e^(-lny)]dy +C} =y*{∫1/y)dy +C}
= y*(lny +C)
即,通解为:x = y *(lny +C).
2.变形为: y' -2xy/(x^2 +1) = x^2 +1.
这里:P=-2x/(x^2 +1), Q = x^2 +1.
-∫Pdx)] = ln(x^2 +1) +c1 (取c1 =0). e^(-∫Pdx)] = x^2 +1 (对数的性质)
e^(∫Pdx)] = 1/(x^2 +1).
故得通解: y = (x^2 +1)* ∫Q*[e^(∫Pdx)]dx +C}
= (x^2 +1)* {∫(x^2 +1)*[1/(x^2 +1)]dx +C}
=(x^2 +1)* { ∫1*dx +C}
=(x^2 +1)*(x+C)
即,通解为: y = (x^2 +1)*(x+C)