Chapter 8: Problem 25

Find a particular solution satisfying the given conditions. \(3 x^{2} y d x+x^{3} d y=0, \quad y=2\) when \(x=1\).

Short Answer

Expert verified

y = \frac{2}{x^3}.

Step by step solution

- Identify the differential equation

The given differential equation is: \[3 x^{2} y dx + x^{3} dy = 0\]

- Rewrite in separable form

Rewrite the equation to separate the variables: \[\frac{dy}{dx} = -\frac{3 x^2 y}{x^3} = -\frac{3 y}{x}\]

- Integrate both sides

Integrate both sides of the differential equation: \[\frac{dy}{y} = -\frac{3}{x} dx\] \[\text{Integrating gives: } \ \text{ln}|y| = -3 \text{ln}|x| + C\]

- Solve the integral

Simplify the integrated form: \[\text{ln} y = \text{ln} C - 3 \text{ln} x\] \[\text{ln} y = \text{ln} \frac{C}{x^3}\] \[\text{y} = \frac{C}{x^3}\]

- Apply initial conditions

Use the given initial conditions to find C: When \(x = 1\), \(y = 2\): \[2 = \frac{C}{1^3} \ C = 2\]

- Write the particular solution

Substitute \(C\) back into the equation: \[y = \frac{2}{x^3}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

separable differential equations

Separable differential equations are a type of differential equation that can be written as the product of a function of the independent variable (e.g., x) and a function of the dependent variable (e.g., y). In simpler terms, this type of equation allows us to separate the variables so that each side of the equation only contains one variable. This property makes them easier to integrate.

For example, consider the original differential equation given: \[3 x^{2} y dx + x^{3} dy = 0\] We can rewrite this in a way that separates the variables y and x. Dividing each term by \(dx\) and rearranging terms, we obtain:

\[ \frac{dy}{dx} = -\frac{3 x^2 y}{x^3} = -\frac{3 y}{x} \]
This key step allows us to then integrate each side independently.

initial conditions

Initial conditions are specific values given for the solution of a differential equation at a certain point. These conditions help us determine the particular solution of a differential equation, and are essential in solving real-world problems where specific constraints are applied.

In our exercise, the initial conditions provided were: \( y = 2\) when \( x = 1\).
We use these values to find the exact value of the integration constant (C), which allows us to narrow down from the general solution to the specific one that fits these conditions.

integration in differential equations

Integration is a crucial step in solving differential equations. It involves finding the antiderivative of a function, which helps us determine the relationship between the variables involved.

In our problem, after separating the variables, we integrate both sides of the separated equation:
\[ \frac{dy}{y} = -\frac{3}{x} dx \]
Integrating both sides, we obtain:
\[ \text{ln}|y| = -3 \text{ln}|x| + C \]
We can simplify this further to:
\[ \text{ln} y = \text{ln} C - 3 \text{ln} x \]
Exponentiating both sides to remove the logarithms gives us:
\[ y = \frac{C}{x^3} \]
We then apply the initial conditions to find the value of C, giving us the particular solution to the differential equation.