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

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The discussion revolves around solving the second-order homogeneous ordinary differential equation (ODE) y'' - 2y' - 15y = 0. By substituting y = e^(mx), the characteristic equation m^2 - 2m - 15 = 0 is derived, yielding roots m1 = 5 and m2 = -3. The general solution is expressed as y(x) = c1e^(m1x) + c2e^(m2x), where c1 and c2 are constants. The linearity of differentiation confirms that this solution satisfies the original ODE, demonstrating the validity of the approach. This method effectively illustrates the process of solving second-order homogeneous ODEs using characteristic equations.
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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.
 
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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.
 
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