# Proof of Variation of Parameters for Linear, 2nd Order ODE

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?

Not missing anything. Just integrate by parts. Since $$\frac{dv}{dx} = \frac{d}{dx}\left[\frac{y_2}{y_1}\right]\int^{x}\frac{y_1(s)F(s)}{W(y_1(s),y_2(s))}ds$$ you have $$v=\int^x\left\{\frac{d}{dx}\left[\frac{y_2}{y_1}\right]\int^{x}\frac{y_1(s)F(s)}{W(y_1(s),y_2(s))}ds\right\}\,dx=\frac{y_2(x)}{y_1(x)}\int^{x}\frac{y_1(s)F(s)}{W(y_1(s),y_2(s))}ds$$ $$-\int^x \frac{y_2(x)}{y_1(x)}\frac{y_1(x)F(x)}{W(y_1(x),y_2(x))}\,dx$$ $$=\frac{y_2(x)}{y_1(x)}\int^{x}\frac{y_1(s)F(s)}{W(y_1(s),y_2(s))}ds - \int^{x}\frac{y_2(s)F(s)}{W(y_1(s),y_2(s))}ds$$