Problem 38
Question
Solve by separating variables. \(3 y^{2} \frac{d y}{d x}=5 x\)
Step-by-Step Solution
Verified Answer
Separate variables, integrate both sides, solve for \(y\): \(y(x) = -\frac{1}{\frac{5}{6}x^2 + C}\).
1Step 1: Rewrite the Equation
Given the equation is \(3 y^{2} \frac{d y}{d x}=5 x\). Our goal is to separate variables, which means we want all terms involving \(y\) on one side of the equation and all terms involving \(x\) on the other side. Begin by dividing both sides of the equation by \(y^2\) and multiply by \(dx\) to get: \[ \frac{1}{y^2} \, dy = \frac{5}{3} x \, dx \]
2Step 2: Integrate Both Sides
Now that we have the variables separated, we can integrate both sides with respect to their respective variables. The left side is integrated with respect to \(y\) and the right side with respect to \(x\). \[ \int \frac{1}{y^2} \, dy = \int \frac{5}{3} x \, dx \]
3Step 3: Solve the Integrals
Evaluate the integrals. The integral of \(\frac{1}{y^2}\) is \(-\frac{1}{y}\), and the integral of \(\frac{5}{3} x\) is \(\frac{5}{6} x^2\). Don't forget the integration constant, which can be added to either side. \[ -\frac{1}{y} = \frac{5}{6}x^2 + C \]
4Step 4: Solve for y
To express \(y\) explicitly, rearrange the equation to solve for \(y\). Start by solving for \(-\frac{1}{y}\):\[ -\frac{1}{y} = \frac{5}{6}x^2 + C \]Taking the reciprocal of both sides gives:\[ \frac{1}{y} = -\left(\frac{5}{6}x^2 + C\right) \]Hence, \[ y = -\frac{1}{\frac{5}{6}x^2 + C} \]
5Step 5: Final Solution
The solution to the differential equation, expressed in terms of \(y\), is:\[ y(x) = -\frac{1}{\frac{5}{6}x^2 + C} \]
Key Concepts
Understanding Differential EquationsIntegration Techniques for Solving EquationsSolving for y
Understanding Differential Equations
Differential equations play a crucial role in modeling various phenomena in real-world applications, such as physics, biology, and economics. These equations involve derivatives, which measure how a function changes as its input changes. A differential equation expresses a relationship between a function and its derivatives.
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In this context, the given differential equation is \(3 y^{2} \frac{d y}{d x}=5 x\). Here, we have an equation relating \(y\), its derivative \(\frac{dy}{dx}\), and the variable \(x\). Solving this equation means finding a function \(y(x)\) that satisfies this relationship.
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In many cases, differential equations can be tricky to solve directly. However, the separation of variables is a technique that allows us to tackle a broad class of differential equations. By rearranging the terms, we aim to separate variables so they appear on different sides of the equation.
\
In this context, the given differential equation is \(3 y^{2} \frac{d y}{d x}=5 x\). Here, we have an equation relating \(y\), its derivative \(\frac{dy}{dx}\), and the variable \(x\). Solving this equation means finding a function \(y(x)\) that satisfies this relationship.
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In many cases, differential equations can be tricky to solve directly. However, the separation of variables is a technique that allows us to tackle a broad class of differential equations. By rearranging the terms, we aim to separate variables so they appear on different sides of the equation.
Integration Techniques for Solving Equations
Integration techniques are pivotal in solving differential equations once the variables are separated. In the given equation, \( \frac{1}{y^2} \, dy = \frac{5}{3} x \, dx \), each side can be integrated separately.
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For integrating \( \frac{1}{y^2} \, dy \), recognize that this is equivalent to integrating \( y^{-2} \, dy \). The antiderivative of \( y^{-2} \) is \( -y^{-1} \), which simplifies to \( -\frac{1}{y} \).
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On the other side, integrating \( \frac{5}{3} x \, dx \) requires us to find the antiderivative of a polynomial term. The general rule for integrating \( x^n \) is to add one to the exponent and divide by the new exponent. Thus, the integral of \( x \) is \( \frac{1}{2}x^2 \). Multiplying by \( \frac{5}{3} \) gives us \( \frac{5}{6}x^2 \). Remember to always include the constant of integration, \( C \), which represents any constant that can be added to the function without affecting the derivative.
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For integrating \( \frac{1}{y^2} \, dy \), recognize that this is equivalent to integrating \( y^{-2} \, dy \). The antiderivative of \( y^{-2} \) is \( -y^{-1} \), which simplifies to \( -\frac{1}{y} \).
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On the other side, integrating \( \frac{5}{3} x \, dx \) requires us to find the antiderivative of a polynomial term. The general rule for integrating \( x^n \) is to add one to the exponent and divide by the new exponent. Thus, the integral of \( x \) is \( \frac{1}{2}x^2 \). Multiplying by \( \frac{5}{3} \) gives us \( \frac{5}{6}x^2 \). Remember to always include the constant of integration, \( C \), which represents any constant that can be added to the function without affecting the derivative.
Solving for y
Once the integration is complete, our main goal is to express the solution explicitly as \( y = f(x) \). The integration step gives us the equation \( -\frac{1}{y} = \frac{5}{6}x^2 + C \). Our task now is to solve for \( y \).
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Begin by isolating \( -\frac{1}{y} \) on one side, which it already is. To find \( y \), we take the reciprocal of both sides of the equation. This operation converts \( -\frac{1}{y} \) into \( y \). Thus:
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This expression gives us \( y \) in terms of \( x \), effectively solving the original differential equation. It's important to recognize how the constant \( C \) affects the solution. Different values of \( C \) will correspond to different solutions, representing a family of curves rather than just a single function.
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Begin by isolating \( -\frac{1}{y} \) on one side, which it already is. To find \( y \), we take the reciprocal of both sides of the equation. This operation converts \( -\frac{1}{y} \) into \( y \). Thus:
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- \( \frac{1}{y} = -\left(\frac{5}{6}x^2 + C\right) \)
- Reciprocate both sides to solve for \( y \)
- \( y = -\frac{1}{\frac{5}{6}x^2 + C} \)
This expression gives us \( y \) in terms of \( x \), effectively solving the original differential equation. It's important to recognize how the constant \( C \) affects the solution. Different values of \( C \) will correspond to different solutions, representing a family of curves rather than just a single function.
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