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Find general solutions (implicit if necessary, explicit if convenient) of the differential equations in Problems 1 through \(18 .\) Primes denote derivatives with respect to \(x.\) $$ \frac{d y}{d x}=3 \sqrt{x y} $$

Short Answer

Expert verified
The general solution is \( y = \left( x^{3/2} + \frac{C}{2} \right)^2 \).

Step by step solution

01

Separation of Variables

To solve the differential equation \( \frac{d y}{d x} = 3 \sqrt{x y} \), we'll start by rearranging it so each side contains only one variable. Divide both sides by \( \sqrt{y} \) and multiply by \( dx \) to separate variables:\[ \frac{d y}{\sqrt{y}} = 3 \sqrt{x} \, dx \]
02

Integrate Both Sides

Now, integrate both sides of the equation separately. The left side becomes:\[ \int \frac{1}{\sqrt{y}} \, dy = 2\sqrt{y} + C_1 \]And the right side becomes:\[ 3 \int \sqrt{x} \, dx = 3 \left( \frac{2}{3} x^{3/2} \right) + C_2 = 2 x^{3/2} + C_2 \]Thus, equating the two integrals gives us:\[ 2 \sqrt{y} = 2 x^{3/2} + C \] where \( C = C_2 - C_1 \).
03

Solve for \( y \)

Divide the entire equation by 2 to isolate \( \sqrt{y} \) on the left side:\[ \sqrt{y} = x^{3/2} + \frac{C}{2} \]Squaring both sides to solve for \( y \) gives:\[ y = \left( x^{3/2} + \frac{C}{2} \right)^2 \]

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Key Concepts

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

Separation of Variables
The technique of separation of variables is a powerful method used to solve certain types of differential equations. Specifically, it helps when you can express the equation in the form where all instances of one variable are on one side and the other variable is isolated on the opposite side.
This method hinges on the ability to "separate" the equation. For the given differential equation \( \frac{d y}{d x} = 3 \sqrt{x y} \), we can reorganize it so that each side of the equation has only one variable:
  • Divide both sides of the equation by \( \sqrt{y} \) to handle the dependency on \( y \).
  • Multiply through by \( dx \) to ensure all \( x \)-related terms are together. This gives us \( \frac{d y}{\sqrt{y}} = 3\sqrt{x} \, dx \).
The goal here is straightforward. By doing so, we can easily integrate each side with respect to its respective variable. This isolation of variables means we can process the integration separately, simplifying the problem significantly.
General Solutions
The concept of general solutions in differential equations refers to the expression that encompasses all possible solutions of the equation. In finding a general solution, we usually introduce an arbitrary constant, denoted as \( C \), to account for all potential initial conditions.
For the equation given, after separating variables, we arrive at the stage of integration. Integrating both sides:
  • For the \( y \)-side: \( \int \frac{1}{\sqrt{y}} \, dy = 2\sqrt{y} + C_1 \).
  • For the \( x \)-side: \( 3 \int \sqrt{x} \, dx = 2 x^{3/2} + C_2 \).
We set the two sides equal, which results as: \( 2\sqrt{y} = 2 x^{3/2} + C \).
This expression forms the implicit general solution of the differential equation. Here, \( C = C_2 - C_1 \) is the constant of integration that provides flexibility for any initial condition the equation might need to satisfy. This constant is essential in defining the specific curve of solutions that meet any required conditions.
Integration Techniques
Integration techniques are a collection of methods used to evaluate integrals, which are crucial in solving the separated differential equation. In our case, the integration process involves two integrals:
  • The left integral: \( \int \frac{1}{\sqrt{y}} \, dy \) is simplified by recognizing it as a standard integral form, resulting in \( 2\sqrt{y} + C_1 \).
  • The right integral: \( 3 \int \sqrt{x} \, dx \) also relies on standard integral formulas. Transforming \( \sqrt{x} \) into \( x^{1/2} \) allows us to use the power rule for integrals, yielding \( 2 x^{3/2} + C_2 \).
To solve such problems, recognizing patterns in integrals and applying them efficiently can significantly simplify your work. The constants \( C_1 \) and \( C_2 \) are crucial as they encompass the range of all possible solutions to this differential equation. Overall, integration techniques are indispensable tools in crafting precise and meaningful solutions to differential equations.

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Most popular questions from this chapter

Problems deal with a shallow reservoir that has a one-square-kilometer water surface and an average water depth of 2 meters. Initially it is filled with fresh water, but at time \(t=0\) water contaminated with a liquid pollutant begins flowing into the reservoir at the rate of 200 thousand cubic meters per month. The well-mixed water in the reservoir flows out at the same rate. Your first task is to find the amount \(x(t)\) of pollutant (in millions of liters) in the reservoir after 1 months. The incoming water has a pollutant concentration of \(c(t)=10\) liters per cubic meter \(\left(\mathrm{L} / \mathrm{m}^{3}\right)\). Verify that the graph of \(x(t)\) resembles the steadily rising curve in Fig. 1.5.9, which approaches asymptotically the graph of the equilibrium solution \(x(t)=20\) that corresponds to the reservoir's long-term pollutant content. How long does it take the pollutant concentration in the reservoir to reach \(10 \mathrm{~L} / \mathrm{m}^{3} ?\)

Verify that the given differential equation is exact; then solve it. $$ \left(x+\tan ^{-1} y\right) d x+\frac{x+y}{1+y^{2}} d y=0 $$

Suppose a uniform flexible cable is suspended between two points \((\pm L, H)\) at equal heights located symmetrically on either side of the \(x\) -axis (Fig. 1.4.12). Principles of physics can be used to show that the shape \(y=y(x)\) of the hanging cable satisfies the differential equation $$a \frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}$$ where the constant \(a=T / \rho\) is the ratio of the cable's tension \(T\) at its lowest point \(x=0\) (where \(y^{\prime}(0)=0\) ) and its (constant) linear density \(\rho\). If we substitute \(v=d y / d x\), \(d v / d x=d^{2} y / d x^{2}\) in this second-order differential equation, we get the first-order equation $$ a \frac{d v}{d x}=\sqrt{1+v^{2}} $$ Solve this differential equation for \(y^{\prime}(x)=v(x)=\) \(\sinh (x / a)\). Then integrate to get the shape function $$y(x)=a \cosh \left(\frac{x}{a}\right)+C$$ of the hanging cable. This curve is called a catenary, from the Latin word for chain.

$$ x(x+y) y^{\prime}+y(3 x+y)=0 $$

Solve the differential equations in Problems by regarding \(y\) as the independent variable rather than \(x\). \((1+2 x y) \frac{d y}{d x}=1+y^{2}\)

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