Solving the Differential Equation 2y'' - 5y' - 3y = 0 with Initial Conditions y(0) = 1, y'(0) = 10

Introduction

Differential equations are a powerful tool in mathematics, with applications in various fields, including physics, engineering, and economics. In this article, we will focus on solving a second-order linear differential equation with constant coefficients and given initial conditions.

The Differential Equation

The differential equation we need to solve is:

2y'' - 5y' - 3y = 0

where y'' represents the second derivative of y with respect to the independent variable, and y' represents the first derivative of y.

Initial Conditions

The initial conditions for this problem are:

y(0) = 1
y'(0) = 10

This means that the value of the function y at x = 0 is 1, and the value of the first derivative of y at x = 0 is 10.

Step 1: Find the Characteristic Equation

To solve the differential equation, we need to find the characteristic equation. The characteristic equation is a quadratic equation that is derived from the differential equation by substituting the expression r^2 for y'', r for y', and 0 for y.

The characteristic equation for the given differential equation is:

2r^2 - 5r - 3 = 0

Step 2: Solve the Characteristic Equation

To solve the characteristic equation, we can use the quadratic formula:

r = (-b ± √(b^2 - 4ac)) / (2a)

where a = 2, b = -5, and c = -3.

Substituting these values, we get:

r = (-(-5) ± √((-5)^2 - 4(2)(-3))) / (2(2))
r = (5 ± √(25 + 24)) / 4
r = (5 ± √49) / 4
r = (5 ± 7) / 4

This gives us two roots:

r1 = 3
r2 = -2

Step 3: Find the General Solution

The general solution of the differential equation is given by:

y = C1e^(3x) + C2e^(-2x)

where C1 and C2 are arbitrary constants that will be determined using the initial conditions.

Step 4: Determine the Values of C1 and C2

To find the values of C1 and C2, we need to substitute the initial conditions into the general solution.

Substituting y(0) = 1 and y'(0) = 10 into the general solution, we get:

1 = C1 + C2
10 = 3C1 - 2C2

Solving this system of equations, we get:

C1 = 2
C2 = -1

Step 5: Write the Final Solution

Substituting the values of C1 and C2 into the general solution, we get the final solution:

y = 2e^(3x) - e^(-2x)

This is the solution to the differential equation 2y'' - 5y' - 3y = 0 with the initial conditions y(0) = 1 and y'(0) = 10.