Solve the following differential equation ;

Question

Solve the following differential equation ; 
 y'' - 2y' - 3y = e^2x

GPT-4o mini · Answer

To solve the differential equation $$ y'' - 2y' - 3y = e^{2x} $$ we can use the method of undetermined coefficients. We first solve the associated homogeneous equation: $$ y'' - 2y' - 3y = 0. $$ The characteristic equation for this differential equation is: $$ r^2 - 2r - 3 = 0. $$ We can factor this as follows: $$ (r - 3)(r + 1) = 0. $$ Thus, the roots are: $$ r = 3 \quad \text{and} \quad r = -1. $$ The general solution of the homogeneous equation is: $$ y_h = C_1 e^{3x} + C_2 e^{-x}, $$ where $C_1$ and $C_2$ are constants. Next, we find a particular solution, $y_p$, to the non-homogeneous equation. Since the right-hand side is $e^{2x}$ and neither of the homogeneous solutions contains $e^{2x}$, we can assume a particular solution of the form: $$ y_p = Ae^{2x}, $$ where $A$ is a constant to be determined. Now we compute $y_p'$ and $y_p''$: $$ y_p' = 2Ae^{2x}, $$ $$ y_p'' = 4Ae^{2x}. $$ Substituting $y_p$, $y_p'$, and $y_p''$ back into the original differential equation: $$ 4Ae^{2x} - 2(2Ae^{2x}) - 3(Ae^{2x}) = e^{2x}. $$ This simplifies to: $$ 4Ae^{2x} - 4Ae^{2x} - 3Ae^{2x} = e^{2x}. $$ Simplifying further: $$ -3Ae^{2x} = e^{2x}. $$ To find $A$, we can set the coefficients equal: $$ -3A = 1 \implies A = -\frac{1}{3}. $$ So the particular solution is: $$ y_p = -\frac{1}{3} e^{2x}. $$ The general solution to the differential equation is the sum of the homogeneous and particular solutions: $$ y = y_h + y_p = C_1 e^{3x} + C_2 e^{-x} - \frac{1}{3} e^{2x}. $$ Thus, the final solution is: $$ \boxed{y = C_1 e^{3x} + C_2 e^{-x} - \frac{1}{3} e^{2x}}, $$ where $C_1$ and $C_2$ are arbitrary constants.

Claude 3.5 Sonnet · Answer

Let me solve this second-order linear differential equation with constant coefficients step by step.

1) First, let's find the characteristic equation. For y'' - 2y' + 5y = 0
The characteristic equation is: r² - 2r + 5 = 0

2) Using the quadratic formula: r = [-(-2) ± √((-2)² - 4(1)(5))] / 2(1)
= [2 ± √(4 - 20)] / 2
= [2 ± √(-16)] / 2
= [2 ± 4i] / 2
= 1 ± 2i

3) Therefore, r₁ = 1 + 2i and r₂ = 1 - 2i

4) Since we have complex conjugate roots, the general solution will be of the form:
y = e^(ax)[c₁cos(bx) + c₂sin(bx)]
where a is the real part and b is the coefficient of i

5) In this case:
a = 1 (real part)
b = 2 (imaginary part)

Therefore, the general solution is:
y = e^x(c₁cos(2x) + c₂sin(2x))

where c₁ and c₂ are arbitrary constants determined by initial conditions (if any were given).

GPT-4o mini · Answer

Your step-by-step solution is clear and accurate. Here’s a concise verification and recap of each step:

Characteristic Equation: You correctly identified the characteristic equation for the differential equation $y'' - 2y' + 5y = 0$: \[ r^2 - 2r + 5 = 0. \]
Quadratic Formula: Applying the quadratic formula, \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] gives: \[ r = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(5)}}{2(1)} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i. \]
Roots: The roots are correctly computed: \[ r_1 = 1 + 2i \quad \text{and} \quad r_2 = 1 - 2i. \]
General Solution Structure: With complex conjugate roots, the general solution is appropriately expressed as: \[ y = e^{ax}(c_1 \cos(bx) + c_2 \sin(bx)), \] where $a$ is the real part and $b$ is the coefficient of $i$.
Specific Values: Here, $a = 1$ and $b = 2$. Hence the general solution becomes: \[ y = e^x(c_1 \cos(2x) + c_2 \sin(2x)). \]

Overall, the final solution to the differential equation $y'' - 2y' + 5y = 0$ is: \[ \boxed{y = e^x(c_1 \cos(2x) + c_2 \sin(2x))}, \] where $c_1$ and $c_2$ are arbitrary constants that can be determined using initial conditions. Excellent job!