For the Linear, 2nd Order ODE $$\frac{d^2y}{dx^2} + P(x)\frac{dy}{dx} + Q(x)y = F(x)\text{,}$$ I've been asked to show that the equation's particular solution can be written $$y_p(x) = y_2(x)\int^{x}\frac{y_1(s)F(s)}{W(y_1(s),y_2(s))}ds - y_1(x)\int^{x}\frac{y_2(s)F(s)}{W(y_1(s),y_2(s))}ds\text{.}$$
It was given to start with $y_p(x) = y_1(x)v(x)$ and develop a 1st order ODE for $v'(x)$.
I started out by computing the 1st and 2nd order derivatives of $y_p(x)$ and substituting that into $\frac{d^2y}{dx^2} + P(x)\frac{dy}{dx} + Q(x)y = F(x)\text{.}$ This yields $$y_1\frac{d^2v}{dx^2} + 2\frac{dy_1}{dx}\frac{dv}{dx} + \frac{d^2y_1}{dx^2}v + P(x)y_1\frac{dv}{dx} + P(x)\frac{dy_1}{dx}v + Q(x)y_1v = F(x)\text{.}$$ I then multiplied both sides by $y_1$ to get $$y_1^2\frac{d^2v}{dx} + 2\frac{dy_1}{dx}\frac{dv}{dx}y_1 +vy_1\left[\frac{d^2y_1}{dx} + P(x)\frac{dy_1}{dx} + Q(x)y_1\right] + P(x)y_1^2\frac{dv}{dx} = y_1F(x)\text{.}$$ Since $y_1$ is a solution of the homogeneous equation $\left(Ly=0\right)$, the expression inside the brackets vanishes, and what's left can be written as $$\frac{d}{dx}\left[y_1^2\frac{dv}{dx}\right] + P(x)y_1^2\frac{dv}{dx} = y_1F(x)\text{.}$$ Now I multiplied both sides by the integrating factor $e^{\int^{x}P(s)ds}$, leading to $$\frac{d}{dx}\left[y_1^2\frac{dv}{dx}e^{\int^{x}P(s)ds}\right] = y_1F(x)e^{\int^{x}P(s)ds}\text{.}$$ Notice that $e^{\int^{x}P(s)ds} = \frac{1}{W(y_1(x),y_2(x))}\text{,}$ so the equation becomes $$\frac{d}{dx}\left[\frac{y_1^2}{W(y_1(x),y_2(x))}\frac{dv}{dx}\right] = \frac{y_1F(x)}{W(y_1(x),y_2(x))}$$ Integrating both sides: $$\frac{y_1^2}{W(y_1(x),y_2(x))}\frac{dv}{dx} = \int^{x}\frac{y_1(s)F(s)}{W(y_1(s),y_2(s))}ds$$ or $$\frac{dv}{dx} = \frac{W(y_1(x),y_2(x))}{y_1^2}\int^{x}\frac{y_1(s)F(s)}{W(y_1(s),y_2(s))}ds = \frac{d}{dx}\left[\frac{y_2}{y_1}\right]\int^{x}\frac{y_1(s)F(s)}{W(y_1(s),y_2(s))}ds$$ This is beginning to look very similar to $y_p(x)$, but after integrating, I would need the equation to be $$v = \frac{y_2(x)}{y_1(x)}\int^{x}\frac{y_1(s)F(s)}{W(y_1(s),y_2(s))}ds - \int^{x}\frac{y_2(s)F(s)}{W(y_1(s),y_2(s))}ds$$ How do I make this leap? What am I missing?
