-b.3.1.1 find the general solution of the second order y''+2y'-3y=0

[karush]

$\tiny{3.1.1}$
find the general solution of the second order differential equation.
$$y''+2y'-3y=0$$
assume that $y = e^{rt}$ then,
$$r^2+2r-3=0\implies (r+3)(r-1)=0$$
new stuff... so far..

 

[topsquark]

$\tiny{3.1.1}$
find the general solution of the second order differential equation.
$$y''+2y'-3y=0$$
assume that $y = e^{rt}$ then,
$$r^2+2r-3=0\implies (r+3)(r-1)=0$$
new stuff... so far..

Good so far. So r = -3 and r = 1. Therefore two linearly independent solutions would be \(\displaystyle y = e^{-3t}\) and \(\displaystyle y = e^{t}\).

One last little bit for you. We have two linearly independent solutions to a second order linear homogenous differential equation. How do you write down the general solution?

-Dan

 

[karush]

Good so far. So r = -3 and r = 1. Therefore two linearly independent solutions would be \(\displaystyle y = e^{-3t}\) and \(\displaystyle y = e^{t}\).

One last little bit for you. We have two linearly independent solutions to a second order linear homogenous differential equation. How do you write down the general solution?

-Dan

$$y=c_1e^{-3t}+c_2e^{t}$$

guess that's it?

what is linear independent solutions?

 

[topsquark]

$$y=c_1e^{-3t}+c_2e^{t}$$

guess that's it?

what is linear independent solutions?

Yes, that's it.

Let f(t) and g(t) be differentiable functions. Then they are called linearly dependent if there are nonzero constants c1 and c2 with \(\displaystyle c_1 f(t) + c_2 g(t) = 0\) for all t. If they are not linearly dependent then they are linearly independent.

-Dan