On this page below, the person chooses the constants to be anything they want, but can you actually do that?
Pauls Online Notes : Differential Equations - Reduction of Order
On this page below, the person chooses the constants to be anything they want, but can you actually do that?
Pauls Online Notes : Differential Equations - Reduction of Order
$\displaystyle
\displaystyle {
y_1=\frac{1}{t}
}
$
$\displaystyle
\displaystyle {
y_2=y_1 \; \nu
}
$
$\displaystyle
\displaystyle { \nu=\frac{2}{5}c t^{5/2}+k
}
$
$\displaystyle
\displaystyle {
y_2=t^{-1} \nu=\frac{2}{5}c t^{3/2}+\frac{k}{t}
}
$
$\displaystyle
\displaystyle {
y=c_1 y_1+c_2 y_2=c_1 \frac{1}{t} +c_2\frac{2}{5}ct^{3/2}+\frac{kc_2}{t}=\frac{1}{t}(c_1+kc_2)+c_2 \frac{2}{5} c t^{3/2}=c_1' \frac{1}{t}+c_2't^{3/2}.
}
$
Dear rlkmg,
c and k are arbitary constants which came up as a result of integration. Note that if you have a function $\displaystyle f(x)$ which is differentiable, then $\displaystyle \frac{d}{dx}\left[f(x)+C\right]=g(x)$ where C is any constant value. Therefore $\displaystyle \int{g(x)}dx=f(x)+C$
Hope you get the idea.