3.8 Q-5

Question Statement

Solve the following differential equation:

$$\frac{dy}{dx} = \frac{y}{x^2}, \quad (y > 0)$$

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Background and Explanation

This is a first-order differential equation that can be solved using separation of variables. We will separate the terms involving $y$ and $x$, integrate each side, and solve for $y$.

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Solution

1. Separate the variables:

   Start with the given equation:

$$\frac{dy}{dx} = \frac{y}{x^2}$$

Rearranging the terms to separate $y$ and $x$:

$$\frac{1}{y} , dy = \frac{1}{x^2} , dx$$

2. Integrate both sides:

   Now, integrate both sides. The integral of $\frac{1}{y} , dy$ is $\ln |y|$, and the integral of $\frac{1}{x^2} , dx$ is $-\frac{1}{x}$:

$$\int \frac{1}{y} , dy = \int \frac{1}{x^2} , dx$$

This gives:

$$\ln y = -\frac{1}{x} + C_1$$

3. Solve for $y$:

   To solve for $y$, exponentiate both sides:

$$y = e^{-\frac{1}{x} + C_1}$$

Use the properties of exponents to separate the terms:

$$y = e^{-\frac{1}{x}} \cdot e^{C_1}$$

Since $e^{C_1}$ is just a constant, we can write it as $C$. Therefore, the final solution is:

$$y = C e^{-\frac{1}{x}}$$

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Key Formulas or Methods Used

• Separation of variables: Rearranged the equation to separate terms involving $y$ and $x$.
• Integration: Integrated both sides with respect to their respective variables.
• Exponential function: Used the exponential function to solve for $y$.

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Summary of Steps

1. Separate the variables: $\frac{1}{y} , dy = \frac{1}{x^2} , dx$
2. Integrate both sides: $\ln y = -\frac{1}{x} + C_1$
3. Solve for $y$: $y = C e^{-\frac{1}{x}}$