Using variation of parameters to derive a general solution?

[Eclair_de_XII]

Homework Statement

"By choosing the lower limit of integration in Eq. (28) in the text as the initial point $t_0$, show that $Y(t)$ becomes

$Y(t)=\int_{t_0}^t(\frac{y_1(s)y_2(t)-y_t(t)y_2(s)}{y_1(s)y_2'(s)-y_1'(s)y_2(s)})g(s)ds$

Show that $Y(t)$ is a solution of the initial value problem:

$L[y]=g(t)$
$y(t_0)=0$
$y'(t_0)=0$"

Homework Equations

Equation 28: $Y(t)=-y_1(t)\int_{t_0}^t\frac{y_2(s)g(s)}{W(y_1,y_2)(s)}ds+y_2(t)\int_{t_0}^t\frac{y_1(s)g(s)}{W(y_1,y_2)(s)}ds$

The Attempt at a Solution

My main method of trying to solve this would be to differentiate $Y(t)$ twice and substituting in the derivatives into the general equation $y''+p(t)y'+q(t)y=g(t)$. So...

$Y(t)=\int_{t_0}^t(\frac{y_1(s)y_2(t)-y_t(t)y_2(s)}{y_1(s)y_2'(s)-y_1'(s)y_2(s)})g(s)ds$
$\frac{d}{dt}Y(t)=(-y_1'(t)\int_{t_0}^t\frac{y_2(s)g(s)}{W(y_1,y_2)(s)}ds-y_1(t)\frac{y_2(t)g(t)}{W(y_1,y_2)(t)})+(y_2'(t)\int_{t_0}^t\frac{y_1(s)g(s)}{W(y_1,y_2)(t)}ds+y_2(t)\frac{y_1(t)g(t)}{W(y_1,y_2)(t)})$

I feel like there should be an easier way to do this, because I don't really wish to calculate the second derivative of $Y(t)$. Perhaps I should do something with the $L[y]$ operator? I don't exactly know how it works.

 

[LCKurtz]

Homework Statement

"By choosing the lower limit of integration in Eq. (28) in the text as the initial point $t_0$, show that $Y(t)$ becomes

$Y(t)=\int_{t_0}^t(\frac{y_1(s)y_2(t)-y_t(t)y_2(s)}{y_1(s)y_2'(s)-y_1'(s)y_2(s)})g(s)ds$

Show that $Y(t)$ is a solution of the initial value problem:

$L[y]=g(t)$
$y(t_0)=0$
$y'(t_0)=0$"

Homework Equations

Equation 28: $Y(t)=-y_1(t)\int_{t_0}^t\frac{y_2(s)g(s)}{W(y_1,y_2)(s)}ds+y_2(t)\int_{t_0}^t\frac{y_1(s)g(s)}{W(y_1,y_2)(s)}ds$

The Attempt at a Solution

My main method of trying to solve this would be to differentiate $Y(t)$ twice and substituting in the derivatives into the general equation $y''+p(t)y'+q(t)y=g(t)$. So...

$Y(t)=\int_{t_0}^t(\frac{y_1(s)y_2(t)-y_t(t)y_2(s)}{y_1(s)y_2'(s)-y_1'(s)y_2(s)})g(s)ds$
$\frac{d}{dt}Y(t)=(-y_1'(t)\int_{t_0}^t\frac{y_2(s)g(s)}{W(y_1,y_2)(s)}ds-y_1(t)\frac{y_2(t)g(t)}{W(y_1,y_2)(t)})+(y_2'(t)\int_{t_0}^t\frac{y_1(s)g(s)}{W(y_1,y_2)(t)}ds+y_2(t)\frac{y_1(t)g(t)}{W(y_1,y_2)(t)})$

I feel like there should be an easier way to do this, because I don't really wish to calculate the second derivative of $Y(t)$. Perhaps I should do something with the $L[y]$ operator? I don't exactly know how it works.

Think about Leibnitz rule:$$
\frac d {dt}\int_{t_0}^t f(s,t)~ds = f(t,t) + \int_{t_0}^t f_t(s,t)ds$$Try applying that to$$
Y(t)=\int_{t_0}^t(\frac{y_1(s)y_2(t)-y_1(t)y_2(s)}{y_1(s)y_2'(s)-y_1'(s)y_2(s)})g(s)ds$$Try applying that directly to your integral without taking anything out of the integral. I think you will see it works easier than what you did and you will notice immediately that part of your answer cancels out. Not that your work is incorrect but I think you will find it seems less messy and won't be so terrified about taking another derivative.

 

[Eclair_de_XII]

$\int_{t_0}^tf_t(s,t)ds$

Do you mean I take the partial derivative with respect to $t$? Anyway, thanks. This simplified the problem very much.