Question

For the differential equation \(xy\frac{dy}{dx} = (x+2)(y+2)\), find the curve passing through the point \((1, -1)\).

Hint:

When separating variables, if you have a fraction where the degree of the numerator is equal to or greater than the degree of the denominator (like \(\frac{y}{y+2}\)), perform algebraic division (or a simple manipulation as shown) to simplify it before integrating.

Answer 1

Step 1: Understanding the Concept:
This is a first-order differential equation. It can be solved using the method of separation of variables. After finding the general solution (containing an arbitrary constant C), we use the given point \((1, -1)\) to find the specific value of C, which gives the particular solution or the equation of the specific curve.
Step 2: Key Formula or Approach:
1. Separate the variables by moving all terms involving \(y\) to one side with \(dy\) and all terms involving \(x\) to the other side with \(dx\).
2. Integrate both sides of the separated equation.
3. Use the initial condition (the point \((1, -1)\)) to solve for the constant of integration, C.
4. Substitute C back into the general solution to get the equation of the curve.
Step 3: Detailed Explanation:
The given differential equation is: \[ xy\frac{dy}{dx} = (x+2)(y+2) \] 1. Separate the variables: \[ \frac{y}{y+2} dy = \frac{x+2}{x} dx \] The left side can be simplified for easier integration: \(\frac{y}{y+2} = \frac{y+2-2}{y+2} = 1 - \frac{2}{y+2}\).
The right side can also be simplified: \(\frac{x+2}{x} = 1 + \frac{2}{x}\).
So the separated equation becomes: \[ \left(1 - \frac{2}{y+2}\right) dy = \left(1 + \frac{2}{x}\right) dx \] 2. Integrate both sides: \[ \int \left(1 - \frac{2}{y+2}\right) dy = \int \left(1 + \frac{2}{x}\right) dx \] \[ y - 2\ln|y+2| = x + 2\ln|x| + C \] This is the general solution.
3. Use the initial condition \((1, -1)\):
Substitute \(x=1\) and \(y=-1\) into the general solution to find C. \[ -1 - 2\ln|-1+2| = 1 + 2\ln|1| + C \] \[ -1 - 2\ln(1) = 1 + 2(0) + C \] Since \(\ln(1) = 0\): \[ -1 - 2(0) = 1 + 0 + C \] \[ -1 = 1 + C \implies C = -2 \] 4. Write the particular solution:
Substitute \(C = -2\) back into the general solution: \[ y - 2\ln|y+2| = x + 2\ln|x| - 2 \] This can be rearranged as: \[ y - x + 2 = 2\ln|x| + 2\ln|y+2| \] \[ y - x + 2 = 2\ln|x(y+2)| \] Step 4: Final Answer:
The equation of the curve passing through the point \((1, -1)\) is \(y - 2\ln|y+2| = x + 2\ln|x| - 2\).