Problem 14

Question

Solve the differential equations in Exercises $9-22$ $$\sqrt{2 x y} \frac{d y}{d x}=1$$

Step-by-Step Solution

Verified

Answer

The solution to the differential equation is $ y = \frac{1}{4} \left( \frac{2x^{3/2}}{3} + C \right)^2 $.

1Step 1: Separate Variables

The given differential equation is $ \sqrt{2xy} \frac{dy}{dx} = 1 $. First, separate the variables by rewriting the equation as: $ \sqrt{2xy} \, dy = dx $. This allows us to integrate each side with respect to its own variable.

2Step 2: Prepare for Integration

To integrate both sides, manipulate the equation to make it easier to integrate. We have $ \sqrt{2x} \sqrt{y} \, dy = dx $. Divide each side by $ \sqrt{y} $: $ \sqrt{2x} \, dy = \frac{dx}{\sqrt{y}} $. Converting $ \sqrt{y} $ to exponential form gives us $ y^{1/2} $, so $ \frac{dy}{\sqrt{y}} $ becomes $ y^{-1/2} \, dy $.

3Step 3: Integrate Both Sides

Integrate both sides of the equation. The left side integrates as follows: $ \int y^{-1/2} \, dy = 2y^{1/2} $. The right side is $ \int \sqrt{2x} \, dx $, which can be simplified using substitution or recognizing it as a standard form, giving $ \frac{2x^{3/2}}{3} $ after integration.

4Step 4: Combine and Simplify the Expression

Combine the results of the integrations: $ 2y^{1/2} = \frac{2x^{3/2}}{3} + C $, where $ C $ is the constant of integration. Simplifying for $ y $, square both sides: $ 4y = \left( \frac{2x^{3/2}}{3} + C \right)^2 $.

5Step 5: Solve for y

Isolate $ y $ by dividing by 4, resulting in $ y = \frac{1}{4} \left( \frac{2x^{3/2}}{3} + C \right)^2 $. This represents the general solution to the differential equation.

Key Concepts

Separation of VariablesIntegrationGeneral Solution

Separation of Variables

Separation of variables is a common method used to solve simple differential equations. This technique involves manipulating the equation so that each variable appears with its derivative, effectively separating them onto different sides of the equation. For example, in the given differential equation $ \sqrt{2xy} \frac{dy}{dx} = 1 $, we first aim to separate $ x $ and $ y $.
We can do this by rearranging the equation as $ \sqrt{2xy} \, dy = dx $. Now the equation is set up for integration, with $ dy $ terms on one side, and $ dx $ terms on the other side.
This process is crucial because it transforms the problem into two simpler integration problems, one for each variable.

Integration

Integration is the process of finding the integral, used here to revert a derivative back to its original function. Once we have the separated differential equation $ \sqrt{2x} \sqrt{y} \, dy = dx $, the next step is to integrate each side.

On the left side, we manipulate the terms $ \sqrt{y} $ by writing it in exponential form as $ y^{-1/2} $. This leads to $ \, y^{-1/2} \, dy $ which integrates to $ 2y^{1/2} $.
On the right side, we have $ \sqrt{2x} \, dx $, which integrates to $ \frac{2x^{3/2}}{3} $.

The art of integration often involves rewriting functions in forms that are recognizable as basic integration rules, thus simplifying the integration process.
Both sides are now translated into a simpler form that allows us to find the solution involving indefinite integrals and the constant of integration $ C $.

General Solution

The general solution to a differential equation includes an arbitrary constant since integration usually results in an indefinite integral. After integrating, we combined the results $ 2y^{1/2} = \frac{2x^{3/2}}{3} + C $. This equation defines a family of curves, differing by the value of $ C $, which makes it a general solution.
To simplify this further for $ y $, we isolate it on one side by squaring both sides: $ 4y = \left( \frac{2x^{3/2}}{3} + C \right)^2 $. Finally, dividing by 4 gives $ y = \frac{1}{4} \left( \frac{2x^{3/2}}{3} + C \right)^2 $.

This form represents all potential solutions, capturing the relationships between $ x $ and $ y $ in the context of the differential equation.
The constant $ C $ can be determined if an initial condition is given.

Thus, the general solution provides a comprehensive view of the behavior described by the differential equation.

Problem 14

Other exercises in this chapter

Problem 14

Find the limits in Exercises $13-20 .$ (If in doubt, look at the function's graph.) $$ \lim _{x \rightarrow-1^{+}} \cos ^{-1} x $$

View solution

Problem 14

In Exercises $13-24,$ find the derivative of $y$ with respect to the appropriate variable. $$y=\frac{1}{2} \sinh (2 x+1)$$

View solution

Problem 14

Use l'Hopital's rule to find the limits in Exercises $7-50$ . $$ \lim _{t \rightarrow 0} \frac{\sin 5 t}{2 t} $$

View solution

Problem 14

Find the derivative of $y$ with respect to $x, t,$ or $\theta,$ as appropriate. \begin{equation}y=\left(9 x^{2}-6 x+2\right) e^{3 x}\end{equation}

View solution