# Thread: integration -- getting constants

1. ## integration -- getting constants

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

2. $\displaystyle \displaystyle { y_1=\frac{1}{t} }$

$\displaystyle \displaystyle { y_2=y_1 \; \nu }$

$\displaystyle \displaystyle { \nu=\frac{2}{5}c t^{5/2}+k }$

$\displaystyle \displaystyle { y_2=t^{-1} \nu=\frac{2}{5}c t^{3/2}+\frac{k}{t} }$

$\displaystyle \displaystyle { y=c_1 y_1+c_2 y_2=c_1 \frac{1}{t} +c_2\frac{2}{5}ct^{3/2}+\frac{kc_2}{t}=\frac{1}{t}(c_1+kc_2)+c_2 \frac{2}{5} c t^{3/2}=c_1' \frac{1}{t}+c_2't^{3/2}. }$

3. Originally Posted by rlkmg
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
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.