Second Solution of Homogeneous Linear Second-Order ODE
Theorem 1. Let $$y'' + P(x)y' + Q(x)y = 0$$ be a homogeneous second-order linear ODE on an interval $I$, and let $y_1$ be a known solution of the ODE on $I$. Then the second solution $y_2$ is $$y_2(x) = y_1(x) \int \frac{\exp \left[ - \int P(x) \, dx \right]}{y^2_1(x)} \, dx,$$ and the general solution is $$y = C_1 y_1(x) + C_2 y_2(x) \\ = C_1 y_1(x) + C_2 y_1(x) \int \frac{\exp \left[ - \int P(x) \, dx \right]}{y^2_1(x)} \, dx$$ where each $C_i$ is a constant.
Proof. Note that the given ODE is linear, which means that it has $2$ linearly independent solutions $y_1$ and $y_2$. Let $y_2(x) = u(x) y_1(x)$ where $u$ is not a constant on $I$. Then we have $$y'_2 = uy_1' + u'y_1 \\ \Longrightarrow y_2'' = uy_1'' + 2u'y_1' + u''y_1 \\ \Longrightarrow y_2'' + Py_2' + Qy_2 = (uy_1'' + 2u'y_1' + u''y_1) + P(uy_1' + u'y_1) + Q(uy_1) \\ = u(y_1'' + Py_1' + Qy_1) + y_1u'' + (2y_1' + Py_1)u' \\ y_1u'' + (2y_1' + Py_1)u' = 0$$ Let $w = u'$. Then we have $$y_1u'' + (2y_1' + Py_1)u' = 0 \\ \Longrightarrow y_1w' + (2y_1' + Py_1)w = 0 \\ \Longrightarrow \frac{dw}{w} = - \frac{2y_1' + Py_1}{y_1} dx = - \left( 2 \frac{y_1'}{y_1} + P \right) dx \\ \Longrightarrow \ln |w| = - 2 \ln |y_1| - \int P(x) \, dx + C \\ \Longrightarrow \ln |wy_1^2| = - \int P(x) \, dx + C \\ \Longrightarrow wy_1^2 = C_1 \exp \left[ - \int P(x) \, dx \right] \\ \Longrightarrow w = \frac{C_1 \exp \left[ - \int P(x) \, dx \right]}{y_1^2} = u' = \frac{du}{dx} \\ \Longrightarrow u = C_1 \int \frac{\exp \left[ - \int P(x) \, dx \right]}{y_1^2} dx + C_2.$$ Take $C_1 = 1$ and $C_2 = 0$. Then $$u = \int \frac{\exp \left[ - \int P(x) \, dx \right]}{y_1^2} dx \\ \Longrightarrow y_2(x) = y_1(x) \int \frac{\exp \left[ - \int P(x) \, dx \right]}{y_1^2} dx.$$ Thus a general solution of the ODE is $$y = C_1y_1(x) + C_2y_1(x) \int \frac{\exp \left[ - \int P(x) \, dx \right]}{y_1^2} dx. \blacksquare$$
선형 동차 2계 ODE의 어떤 한 해가 주어졌을 때, 이 해를 가지고 위와 같은 치환을 통해 다른 선형독립인 해를 구할 수 있다.