StudyLabs

Lecture 06 - Jan 15

3 min read

Riccati Equation (Continued)

will reduce the eqn to a linear eqn in .

Then (I) →

Second Order Linear Homogeneous Equation

Theorem

Let be a set of functions defined on an interval , and let be linearly independent solutions on . Then, the Wronskian:

Proof

Given, are LD.

⟹ (not all zeros) s.t.

Since are differentiable, we have

For any , we must have all zeros before .

This gives matrix:

Misplaced &\phi_1 & \phi_2 & \ldots & \phi_k \\ \phi_1' & \phi_2' & \ldots & \phi_k' \\ \vdots & \vdots & & \vdots \\ \phi_1^{(k-1)} & \phi_2^{(k-1)} & \ldots & \phi_k^{(k-1)} \end{vmatrix} = W(\phi_1, \ldots, \phi_k) = 0$$ --- ### Proof (First Order Case) Suppose we have a first order DE: $$\Rightarrow y' + aq = 0, \text{ then we know that } ce^{-ax} \text{ is a soln.} \text{ where } c \text{ is the soln of } rya = 0$$ We know that if we differ. $e^{rx}$ any times, we get only a const. mult. of $e^{rx}$. This gives an indication that $e^{rx}$ could be a soln. of $L(y) = 0$ for some $r$. Consider, $$L(ce^{rx}) = (c_2^2 + a_1r + a_2) \cdot e^{rx} = P(r) \cdot e^{rx}$$ $e^{rx}$ could be a soln of $y' + a_1y + a_2y = 0$ ⟺ $P(r) = 0$. By Fundam. theorem, $P(r)$ should have two roots, say $r_1, r_2$. **Case 1:** Suppose $r_1 \neq r_2$ Then $e^{r_1y}$ & $e^{r_2x}$ are soln of $L(y) = 0$. **Case 2:** Suppose $r_1 = r_2 = r$ ⟹ $P(r) = 0, P'(r) = 0$ Consider, $\frac{\partial}{\partial t}L(e^{tx}) = L\left[\frac{\partial}{\partial t}(e^{tx})\right] = L(xe^{tx})$ $$L(e^{rx}) = P(r) \cdot e^{rx}$$ ⟹ $P(r) \cdot e^{rx} + P(r) \cdot xe^{rx} = L(cxe^{rx})$ Since $P'(r) = P(r) = 0$: $L(cxe^{rx}) = 0$ $$\therefore xe^{rx} \text{ is soln of eqn.}$$ --- ## Theorem If $x, r_1$ are distinct roots of $P(r)$, then the funcs: $$\phi_1(x) = e^{r_1x}, \quad \phi_2(x) = e^{r_2x}$$ and gen. soln is $\phi = c_1e^{r_1x} + c_2e^{r_2x}$ If $x$ is repeated soln of $P(r)$, then $$\phi_1(x) = e^{rx}, \quad \phi_2(x) = xe^{rx}$$ and gen. soln is $\phi = (c_1 + c_2x)e^{rx}$

Graph View