Question

高等数学，如下图4道题目，求详细解答

Answer 1

望采纳。谢谢啦。

Answer 2

1(9) y' + y = e^x 是一阶线性微分方程， 通解
y = e^(-∫dx) [C + ∫e^x e^(∫dx) dx]
= e^(-x) {C + ∫e^(2x)dx] = e^(-x) {C + (1/2)e^(2x)]
= Ce^(-x) + (1/2)e^x
(13) dy/dx = 1/(x+y^2), 则 dx/dy = x+y^2,
dx/dy - x = y^2 是 x 对 y 的一阶线性微分方程，通解
x = e^(∫dy) [C + ∫y^2 e^(-∫dy) dy]
= e^y[C + ∫y^2 e^(-y) dy] = e^y[C - ∫y^2 de^(-y) ]
= e^y[C - y^2 e^(-y) + 2∫ye^(-y) dy]
= e^y[C - y^2 e^(-y) - 2∫yde^(-y) ]
= e^y[C - y^2 e^(-y) - 2ye^(-y) + 2∫e^(-y)dy ]
= e^y[C - y^2 e^(-y) - 2ye^(-y) - 2e^(-y) ].
2(3) y' - y = e^x
y = e^(∫dx) [C + ∫e^x e^(∫-dx) dx]
= e^x [C + ∫dx] = e^x(C + x),
y(0) = 1, 代入得 C = 1，则 y = (1+x)e^x.
(4) xy' - y = x^2e^x, x ≠0 时，y' -y/x = xe^x
y = e^(∫dx/x){C+∫xe^xe^(-∫dx/x)dx]
= x[C + ∫e^xdx] = x(C + e^x),
y(1) = 2 代入得 C = 2 - e，则 y = x(e^x - e + 2).