Basis of a Kernel

Question

How would i find the basis of the kernel of the differential operator below

$$8y'' + 3y' + 7y$$

We know the equation was homogenous and i believe the basis is two dimensional

Answer 1

The key here is that for any $x_0, v_0 \in \mathbb{R}$, there exists a unique solution $x$ to the differential equation that satisfies $x(0) = x_0, x'(0) = v_0$.

If we let $L$ be the differential operator above, then this shows that the linear map $i: \ker L \to \mathbb{R}^2$ given by $i(y) = (y(0),y'(0))$ is bijective, and hence $\dim \ker L = \dim \mathbb{R}^2 = 2$.

Define the function $e_\alpha$ by $e_\alpha(t) = e^{\alpha t}$. Note that if $\alpha_1 \neq \alpha_2$, then the functions $e_{\alpha_1}, e_{\alpha_2}$ are linearly independent. To see this, suppose $\beta_1 e_{\alpha_1} + \beta_2 e_{\alpha_2} = 0$. By evaluating $e_\alpha(0), e'_\alpha(0)$ we have $\beta_1 + \beta_2 = 0$ and $\beta_1 \alpha_1 + \beta_2\alpha_2 = 0$ from which we have $\beta =0$.

Note that $L e_\alpha = (8 \alpha^2+3 \alpha+7) e_\alpha$, hence $L e_\alpha = 0$ iff $\alpha$ satisfies $8 \alpha^2+3 \alpha+7 = 0$.

Let $\alpha_1,\alpha_2$ be the two distinct values of $\alpha$ that satisfy the quadratic.

Then $\ker L = \operatorname{sp} \{ e_{\alpha_1}, e_{\alpha_2} \} $.

Answer 2

You will need to solve the differential equation: $$8y'' + 3y' + 7y = 0$$ and we can do this with the characteristic equation which can be found by assuming our solution is a power of an exponential: $y= e^{rx}$ which yields: $$(8r^2+3r+7)e^{rx} = 0$$ and so $r = \frac{-3 \pm i\sqrt{216}}{16}$

This leads to the two linearly independent solutions: $$y_1 = e^{(-3)x} \cdot\left(\cos\left(\frac{\sqrt{216}}{16}x\right) + i \sin\left(\frac{\sqrt{216}}{16}x\right)\right)$$ $$y_2 = e^{(-3)x} \cdot\left(\cos\left(\frac{\sqrt{216}}{16}x\right) - i \sin\left(\frac{\sqrt{216}}{16}x\right)\right)$$

In order to have two solutions with strictly real terms, we combine these as $z_1 = y_1+y_2$ and $z_2 =-i (y_1-y_2)$ to get:

$$z_1= 2e^{-3x} \cos\left(\frac{\sqrt{216}}{16}x\right)$$ and $$z_2 = 2e^{-3x} \sin\left(\frac{\sqrt{216}}{16}x\right)$$

Answer 3

Let $z_1, z_2$ the roots of the characteristic equation. Let $\Delta$ the differential operator, let $D$ be the derivation operator. Let $P = (X-z_1)(X-z_2)$.

Then as $z_1\neq z_2$, $\gcd(X-z_1, X-z_2) = 1 $ and some algebra theorem ensures that $$ \ker \Delta = \ker P(D) = \ker (D-z_1I) \oplus \ker (D-z_2I) $$

This leads to the result of Joel.