Solving 2nd Order Homogeneous ODE - Joe's Question on Yahoo Answers

[MarkFL]

Here is the question:

Differential Calculus Problem?

Consider the differential equation given by y''-2y'-15y=0

part 1) Set y=e^mx, determine the values of m which make y satisfy the above equation

part 2) Let m1 and m2 be the two values found in part (a). Show that for any constants c1 and c2 the function y=c1e^m1x+c2e^m2x also satisfies the above equation.

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello joe,

We are given the 2nd order homogeneous ODE:

$$y''-2y'-15y=0$$

Part 1.) Letting:

$$y=e^{mx}$$

We can see that:

$$\frac{d^ny}{dx^n}=m^ne^{mx}$$ where $n\in\mathbb{N}$

Hence, substituting this function into the ODE, the ODE becomes:

$$m^2e^{mx}-2me^{mx}-15e^{mx}=0$$

Divide through by $$e^{mx}\ne0$$ and we have:

$$m^2-2m-15=0$$

This is what is referred to as the characteristic or auxiliary equation. Factoring, we obtain:

$$(m-5)(m+3)=0$$

And so we find the characteristic roots are:

$$m=-3,\,5$$

These are the two values of $m$ for which $y=e^{mx}$ is a solution to the given ODE.

Part 2) Let:

$$y(x)=c_1e^{m_1x}+c_2e^{m_2x}$$

As before, given the linearity of differentiation, we can see:

$$y^{(n)}(x)=c_1m_1^ne^{m_1x}+c_2m_2^ne^{m_2x}$$

Hence, substituting this function into the ODE, we obtain:

$$\left(c_1m_1^2e^{m_1x}+c_2m_2^2e^{m_2x} \right)-2\left(c_1m_1e^{m_1x}+c_2m_2e^{m_2x} \right)-15\left(c_1e^{m_1x}+c_2e^{m_2x} \right)=0$$

Arrange on like terms:

$$\left(c_1m_1^2e^{m_1x}-2c_1m_1e^{m_1x}-15c_1e^{m_1x} \right)+\left(c_2m_2^2e^{m_2x}-2c_2m_2e^{m_2x}-15c_2e^{m_2x} \right)=0$$

Factor both expressions:

$$c_1e^{m_1x}\left(m_1^2-2m_1-15 \right)+c_2e^{m_2x}\left(m_2^2-2m_2-15 \right)=0$$

Now, since we have defined $m_1$ and $m_2$ wot be the two roots of the quadratic characteristic equation $m^2-2m-15=0$, we must therefore have:

$$c_1e^{m_1x}\cdot0+c_2e^{m_2x}\cdot0=0$$

$$0=0$$

This shows that the function:

$$y(x)=c_1e^{m_1x}+c_2e^{m_2x}$$

is the general solution to the given ODE.