Math Problem Statement

Solve it

Apply variable separation method to find the solution to: dy - (y - 1)^2 dx = 0

Solution

The differential equation provided is:

dy(y1)2dx=0dy - (y - 1)^2 dx = 0

This can be rewritten as:

dy=(y1)2dxdy = (y - 1)^2 dx

To solve this using the method of separation of variables, follow these steps:

  1. Separate the variables yy and xx: dy(y1)2=dx\frac{dy}{(y - 1)^2} = dx

  2. Integrate both sides. On the left side, we integrate with respect to yy, and on the right side, with respect to xx:

dy(y1)2=dx\int \frac{dy}{(y - 1)^2} = \int dx

The integral on the left side is:

dy(y1)2=1y1\int \frac{dy}{(y - 1)^2} = -\frac{1}{y - 1}

The integral on the right side is:

dx=x+C\int dx = x + C

where CC is the constant of integration.

  1. Now, equate both integrals:

1y1=x+C-\frac{1}{y - 1} = x + C

  1. Solve for yy:

1y1=(x+C)\frac{1}{y - 1} = -(x + C)

y1=1(x+C)=1Cxy - 1 = \frac{1}{-(x + C)} = \frac{1}{C - x}

y=1+1Cxy = 1 + \frac{1}{C - x}

Thus, the general solution to the differential equation is:

y=1+1Cxy = 1 + \frac{1}{C - x}

Let me know if you'd like any further details or have any questions.

Here are 5 related questions for practice:

  1. How do you solve a differential equation using the integrating factor method?
  2. What are the common types of first-order differential equations?
  3. How do you interpret a separable differential equation graphically?
  4. Can this differential equation be solved using another method like exact equations?
  5. How can initial conditions be applied to find a specific solution to this equation?

Tip: Always check whether a differential equation is separable before trying more advanced methods, as separation of variables often simplifies the problem considerably.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Separation of Variables
Differential Equations

Formulas

dy = (y - 1)^2 dx
∫ dy/(y - 1)^2 = ∫ dx

Theorems

Integration by Substitution

Suitable Grade Level

College-level Calculus