(2x+3)+(2y−2)y′=0

4 min read Jun 16, 2024
(2x+3)+(2y−2)y′=0

Solving the Differential Equation (2x+3) + (2y-2)y' = 0

This article will guide you through solving the differential equation (2x+3) + (2y-2)y' = 0. We will use the method of separation of variables to find the general solution.

Understanding the Equation

The given equation is a first-order ordinary differential equation (ODE). It involves the derivative of a function y(x) with respect to x, denoted by y'.

Separation of Variables

  1. Rearrange the equation: We begin by rearranging the equation to separate the variables x and y:

    (2y-2)y' = -(2x+3)
    
  2. Divide both sides: Next, we divide both sides of the equation by (2y-2) and by -(2x+3) to isolate the y' term and the x term respectively:

    y' / (2y-2) = -1 / (2x+3)
    
  3. Integrate both sides: Now, we integrate both sides of the equation with respect to their respective variables:

    ∫ [1 / (2y-2)] dy = ∫ [-1 / (2x+3)] dx
    
  4. Solve the integrals: Evaluating the integrals, we obtain:

    (1/2) ln|2y-2| = -(1/2) ln|2x+3| + C
    

    Where C is the constant of integration.

  5. Simplify and solve for y: We can simplify the equation by multiplying both sides by 2 and combining the logarithms:

    ln|2y-2| + ln|2x+3| = 2C
    

    Using the property of logarithms that ln(a) + ln(b) = ln(a*b), we get:

    ln|(2y-2)(2x+3)| = 2C
    

    Exponentiating both sides and solving for y, we get:

    |(2y-2)(2x+3)| = e^(2C)
    

    Since e^(2C) is a constant, we can replace it with another constant K:

    (2y-2)(2x+3) = K
    

    Finally, solving for y, we get the general solution:

    y = (K + 2) / (4x + 6) + 1
    

General Solution

Therefore, the general solution to the differential equation (2x+3) + (2y-2)y' = 0 is:

y = (K + 2) / (4x + 6) + 1

Where K is an arbitrary constant.

Conclusion

We have successfully solved the differential equation (2x+3) + (2y-2)y' = 0 using the method of separation of variables. The general solution we obtained represents a family of curves, each determined by a specific value of the constant K. This solution can be used to analyze the behavior of the function y(x) and to find specific solutions that satisfy particular initial conditions.

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