y" - y" - 2y = x² + cos(x)
Understand the Problem
The question is a second-order linear differential equation that involves the derivatives of a function y with respect to x. We need to find the general solution for this differential equation.
Answer
The general solution is $y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}$ for distinct real roots, $y(x) = (C_1 + C_2 x)e^{r x}$ for repeated roots, or $y(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x))$ for complex roots.
Answer for screen readers
The general solution depends on the nature of the roots derived from the characteristic equation.
If the roots are real and distinct, the solution is
$$ y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} $$
If the roots are real and repeated, the solution is
$$ y(x) = (C_1 + C_2 x)e^{r x} $$
If the roots are complex, the solution is
$$ y(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) $$
Steps to Solve
- Identify the Differential Equation
Write the second-order linear differential equation in standard form. Typically, it looks like this:
$$ a y'' + b y' + c y = 0 $$
where $y''$ is the second derivative of the function $y$ with respect to $x$, $y'$ is the first derivative, and $a$, $b$, and $c$ are constants.
- Characteristic Equation
Convert the differential equation into its characteristic equation by substituting $y = e^{rx}$ into the equation. This gives us the characteristic equation:
$$ a r^2 + b r + c = 0 $$
- Solve for Roots
Use the quadratic formula to find the roots $r$ of the characteristic equation, which are given by:
$$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$
- Determine the General Solution
Based on the nature of the roots:
- If the roots are real and distinct, the general solution is:
$$ y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} $$
- If the roots are real and repeated, the general solution is:
$$ y(x) = (C_1 + C_2 x)e^{r x} $$
- If the roots are complex conjugates, say $r = \alpha \pm i \beta$, the general solution is:
$$ y(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) $$
- Finalizing the General Solution
Substitute the calculated roots into the appropriate form of the general solution to complete the process.
The general solution depends on the nature of the roots derived from the characteristic equation.
If the roots are real and distinct, the solution is
$$ y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} $$
If the roots are real and repeated, the solution is
$$ y(x) = (C_1 + C_2 x)e^{r x} $$
If the roots are complex, the solution is
$$ y(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) $$
More Information
This process is known as solving second-order linear differential equations with constant coefficients. The method relies on finding roots of the associated characteristic polynomial, which tells us about the behavior of the solution.
Tips
- Forgetting to substitute $y = e^{rx}$ correctly in the differential equation.
- Not simplifying the characteristic equation properly, which can lead to errors in determining the roots.
- Misidentifying the nature of the roots (real vs. complex), which affects the form of the general solution.
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