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Chapter 10: Problem 6

Solve the following differential equations: $$\frac{d y}{d t}=\frac{t^{2} y^{2}}{t^{3}+8}$$

Short Answer

Expert verified
Separated variables, integrated both sides, expressed y in terms of t.

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \(\frac{d y}{d t}=\frac{t^{2} y^{2}}{t^{3}+8}\). This is a first-order ordinary differential equation and appears to be separable.
02

Separate the Variables

Separate the variables by moving all terms involving \(y\) to one side and all terms involving \(t\) to the other side: \(\frac{1}{y^2} \frac{d y}{d t}=\frac{t^{2}}{t^{3}+8}\). Now, rewrite this as: \(\frac{1}{y^2} dy=\frac{t^2}{t^3+8} dt\).
03

Integrate Both Sides

Integrate both sides of the equation: \(\begin{aligned} &\text{\textbf{Left side:}} \int \frac{1}{y^2} dy = -\frac{1}{y} + C_1\ \ &\text{\textbf{Right side:}} \int \frac{t^2}{t^3+8} dt = \frac{1}{3} \frac{t^3 + 8}{\frac{3}{8}} dt = \frac{1}{3} \frac{(t^3+8)}{(t^3+8)} + C_2 \end{aligned}\). Combine the constants: \( -\frac{1}{y} = \frac{1}{3} \text{ln}|t^3+8| + C \).
04

Solve for y

To isolate \(y\), solve the equation: \(-\frac{1}{y} = \frac{1}{3} \text{ln}|t^3+8| + C\) => \(y = -\frac{1}{\frac{1}{3} \text{ln}|t^3+8| + C}\).

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Key Concepts

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

Separable Differential Equation
A differential equation is considered separable if you can rewrite it so that all the terms involving one variable (and its differential) are on one side, and all the terms involving the other variable (and its differential) are on the other side. This allows you to integrate each side independently.
In our given problem, the equation is \(\frac{d y}{d t}=\frac{t^{2} y^{2}}{t^{3}+8}\). By moving the terms involving \(y\) to one side and the terms involving \(t\) to the other, we achieve:
\(\frac{1}{y^2} dy=\frac{t^2}{t^3+8} dt\).
This separated form is crucial for the next step, which is integration.
Integration
Once a differential equation is separated, the next step is to integrate both sides. Integration is the reverse process of differentiation. Here we perform the integrals on both sides:
\text{Left side: } \int \frac{1}{y^2} dy = -\frac{1}{y} + C_1\
This is a basic integral that assumes knowledge of the power rule.
\text{Right side: } \int \frac{t^2}{t^3+8} dt\
This requires recognizing the denominator's influence on differentiating the numerator, possibly involving partial fractions or a simple substitution. The result is:
\-\frac{1}{y} = \frac{1}{3} \text{ln}|t^3+8| + C\
We combined constants into one term, \(C\).
First-Order Ordinary Differential Equation
A first-order ordinary differential equation (ODE) involves the first derivative of the unknown function but no higher derivatives. In our example, \(\frac{d y}{d t}=\frac{t^{2} y^{2}}{t^{3}+8}\), the highest derivative is \(\frac{d y}{d t}\), indicating it is a first-order ODE. This type of equation is simpler than higher-order ODEs and often solved via separation of variables or integrating factors. After solving for \(y\), we get:
\(y = -\frac{1}{\frac{1}{3} \text{ln}|t^3+8| + C}\).
This final expression shows the solution explicitly in terms of \(t\).

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Most popular questions from this chapter

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