3.8 Q-7

Question Statement

Solve the differential equation:

$$x , dy + y(x - 1) , dx = 0$$

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Background and Explanation

This is a first-order linear differential equation. To solve it, we will use the method of separation of variables. The goal is to separate the terms involving $y$ on one side and the terms involving $x$ on the other side. Once we have separated the variables, we can integrate both sides of the equation to find the general solution.

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Solution

1. Rewrite the equation:

   Start by isolating terms involving $dy$ on one side and $dx$ on the other:

$$x , dy + y(x - 1) , dx = 0$$

Rearrange it as:

$$x , dy = -y(x - 1) , dx$$

2. Separate the variables:

   Now, we want to get all terms involving $y$ on one side and terms involving $x$ on the other side. Divide both sides by $y$ and $x$:

$$\frac{1}{y} , dy = -\left( 1 - \frac{1}{x} \right) , dx$$

The equation is now separated with $y$ terms on the left and $x$ terms on the right.

3. Integrate both sides:

   We can now integrate both sides. Start with the left-hand side, which is the integral of $\frac{1}{y} , dy$:

$$\int \frac{1}{y} , dy = \ln |y|$$

On the right-hand side, we have two terms to integrate:

$$\int -1 , dx = -x \quad \text{and} \quad \int \frac{1}{x} , dx = \ln |x|$$

Thus, we have:

$$\ln |y| = \ln |x| - x + \ln |C|$$

Here, $C$ is the constant of integration.

4. Simplify the equation:

   Combine the logarithmic terms on the right-hand side:

$$\ln |y| = \ln |C \cdot x| - x$$

Now exponentiate both sides to remove the logarithms:

$$\frac{y}{Cx} = e^{-x}$$

Multiply both sides by $Cx$ to solve for $y$:

$$y = Cx e^{-x}$$

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Key Formulas or Methods Used

• Separation of Variables: Rearranged the equation to separate terms involving $y$ and $x$.
• Integration of Common Functions: Integrated $\frac{1}{y}$, $-1$, and $\frac{1}{x}$ with respect to their respective variables.
• Logarithmic Properties: Used properties of logarithms to combine terms.
• Exponentiation: Exponentiated both sides of the equation to eliminate the natural logarithms.

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Summary of Steps

1. Rearrange the given equation: $x , dy = -y(x - 1) , dx$.
2. Separate the variables: $\frac{1}{y} , dy = -\left( 1 - \frac{1}{x} \right) , dx$.
3. Integrate both sides:
   • $\int \frac{1}{y} , dy = \ln |y|$
   • $\int -1 , dx = -x$ and $\int \frac{1}{x} , dx = \ln |x|$
4. Combine the results: $\ln |y| = \ln |Cx| - x$.
5. Exponentiate both sides and solve for $y$: $y = Cx e^{-x}$.