Homogeneous Equations with Constant Coefficients | Mathematical Physics | University Notes | Physics Notes | B.Sc. Physics Notes by Study Buddy Notes

Homogeneous Equations with Constant Coefficients

A homogeneous linear differential equation with constant coefficients has the form:

$$a_n \frac{d^n y}{dx^n} + a_{n-1} \frac{d^{n-1} y}{dx^{n-1}} + \cdots + a_1 \frac{dy}{dx} + a_0 y = 0$$

where $a_0, a_1, \ldots, a_n$ are constants and $y$ is the unknown function of $x$. The general solution to such equations can be derived by solving the associated characteristic equation.

Step 1: Formulate the Characteristic Equation

For simplicity, consider a second-order homogeneous equation with constant coefficients:

$$a \frac{d^2 y}{dx^2} + b \frac{dy}{dx} + c y = 0$$

To solve this, we assume a solution of the form $y = e^{mx}$, where $m$ is a constant to be determined.

1. Substitute $y=e^{m\mathrm{x, }}$$\frac{dy}{dx} = m e^{mx}$, and $\frac{d^2 y}{dx^2} = m^2 e^{mx}$  into the equation:

   $$a m^2 e^{mx} + b m e^{mx} + c e^{mx} = 0$$ 

2. Factor out $e^{mx}$ (since $e^{mx}\mathrm{\ne }\mathrm{0):}$

   $$(a m^2 + b m + c) e^{mx} = 0$$
   

3. This results in the characteristic equation:

   $$a m^2 + b m + c = 0$$
   

Step 2: Solve the Characteristic Equation

The characteristic equation $a m^2 + b m + c = 0$ is a quadratic equation in $m$, which can be solved using the quadratic formula:

$$m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

The nature of the roots $m_1$ and $m_2$ of this equation determines the form of the general solution.

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Case 1: Distinct Real Roots

If $b^2 - 4ac > 0$, the roots $m_1$ and $m_2$ are real and distinct. In this case, the general solution of the differential equation is:

$$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$$

where $C_1$ and $C_2$ are constants determined by initial conditions.

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Case 2: Repeated Real Roots

If $b^2 - 4ac = 0$, the characteristic equation has a repeated root $m = -\frac{b}{2a}$. The general solution for this case is:

$$y = (C_1 + C_2 x) e^{mx}$$

Here, $C_1$ and $C_2$ are constants. The term $x e^{mx}$ arises because the root is repeated, which means that $e^{mx}$ alone would not be sufficient to provide a general solution.

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Case 3: Complex Roots

If $b^2 - 4ac < 0$, the roots are complex, and we can write them as $m=\alpha \pm i\mathrm{\beta ,\; where }$$\alpha = -\frac{b}{2a}$ and $\beta = \frac{\sqrt{4ac - b^2}}{2a}$.

In this case, the general solution is given by:

$$y = e^{\alpha x} (C_1 \cos(\beta x) + C_2 \sin(\beta x))$$

This form comes from Euler’s formula, $e^{i\beta x}=\cos (\beta x)+i\sin (\beta x)\; ,\; and\; ensures\; that\; the\; solution\; remains\; real-valued.$

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Examples

Example 1: Solve $y'' - 3y' + 2y = 0$

1. Form the characteristic equation:

   $$m^2 - 3m + 2 = 0$$
   

2. Solve for $m$:

   $$m = \frac{3 \pm \sqrt{9 - 8}}{2} = \frac{3 \pm 1}{2}$$ $$m_1 = 2, \quad m_2 = 1$$
   

3. The general solution is:

   $$y = C_1 e^{2x} + C_2 e^{x}$$
   

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Example 2: Solve $y^{\mathrm{\prime }\mathrm{\prime }}+4y^{\mathrm{\prime }}+4y=0$

1. Form the characteristic equation:

   $$m^2 + 4m + 4 = 0$$
   

2. Solve for $m$:

   $$m = \frac{-4 \pm \sqrt{16 - 16}}{2} = -2$$
   

   Here, we have a repeated root $m = -2$
   

3. The general solution is:

   $$y = (C_1 + C_2 x) e^{-2x}$$
   

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Example 3: Solve $y'' + y = 0$

1. Form the characteristic equation:

   $$m^2 + 1 = 0$$
   

2. Solve for $m$:

   $$m = \pm i$$
   

   This results in complex roots $m = 0 \pm i$, so $\alpha = 0$ and $\beta =1.$

3. The general solution is:

   $$y = C_1 \cos x + C_2 \sin x$$
   

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Summary of Solution Types

• Distinct Real Roots $m_{\mathrm{1\; and }}$$m_2$: $y=C_1{e}^{m_{1}x}+C_2{e}^{m_{2}x}$
• Repeated Real Root $\mathrm{m: }$$y=(C_1+C_{2}x)e^{mx}$
• Complex Roots $\alpha \pm i \beta$: $y=e^{\alpha x}(C_1\cos (\beta x)+C_2\sin (\beta x))$

This approach provides a structured method to solve second-order homogeneous differential equations with constant coefficients, applicable to higher-order equations by extending the characteristic polynomial method.