if asked to find all solutions of a BVP in the form that $y''=ky, y(0)=y_0, y(1)=y_1$ for $k\gt0$, how to solve problems like this?
I know some of the models: $y=acost+bsint$, $c_1e^t+c_2e^t$, $ksint$, but given only these info, how do we approach this problem? Thanks!
$$\frac{d^2y}{dx^2} - ky = 0$$
Then the auxilliary equation gives us:
$$\lambda^2 - k = 0 \implies \lambda = \pm \sqrt{k}$$
Since $k > 0$, these are real solutions
Then the general solution is given by:
$$y = c_{1}e^{\sqrt{k}x}+c_{2}e^{-\sqrt{k}x}$$
We are given that $y(0) = y_{0}$, $y(1)=y_{1}$
Plugging this in:
$$y_{0} = c_{1}+c_{2} \implies c_{1} = y_{0}-c_{2}$$
$$y_{1} = c_{1}e^{\sqrt{k}} +c_{2}e^{-\sqrt{k}}\implies y_{1} = (y_{0}-c_{2})e^{\sqrt{k}}+c_{2}e^{-\sqrt{k}}$$
Re-arranging in terms of $c_{2}$:
$$c_{2}(e^{-\sqrt{k}}-e^{\sqrt{k}})=y_{1}-y_{0}e^{\sqrt{k}}$$
$$\implies c_{2} = \frac{y_{1}-y_{0}e^{\sqrt{k}}}{e^{-\sqrt{k}}-e^{\sqrt{k}}}$$
Then $$c_{1} = \frac{y_{0}(e^{-\sqrt{k}}-e^{\sqrt{k}}) -y_{1}-y_{0}e^{\sqrt{k}}}{e^{-\sqrt{k}}-e^{\sqrt{k}}}$$
$$\implies c_{1} = \frac{y_{0}e^{-\sqrt{k}}-y_{1}}{e^{-\sqrt{k}}-e^{\sqrt{k}}} = -c_{2}$$
Then the general solution is given by:
$$y = \frac{y_{0}e^{-\sqrt{k}}-y_{1}}{e^{-\sqrt{k}}-e^{\sqrt{k}}}(e^{\sqrt{k}x}-e^{-\sqrt{k}x})$$