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Solve the differential equation. $$ y^{\prime}=\sqrt{x} y $$

Short Answer

Expert verified
The solution of the differential equation \(y'=\sqrt{x} y\) is \(y(x) = e^{\frac{2}{3} x^{3 / 2}+C}\).

Step by step solution

01

Recognizing the differential equation

The original equation \(y'=\sqrt{x} y\) is a first-order linear differential equation (variable separable form).
02

Perform Separation of Variables

Rearrange the equation to separate variables. This gives \(\frac{y'}{y}=\sqrt{x}\).
03

Integrate Both Sides

After separating the variables, the equation will be integrated on both sides with respect to x. This leaves us \(\int \frac{y'}{y} dx = \int \sqrt{x} dx\).
04

Simplify the Equation

After integrating we get \(\ln y = \frac{2}{3} x^{3 / 2} + C\), where C is the constant of integration.
05

Final Solution

Finally to get y, we make use of the exponent to clear the natural logarithm giving us \(y(x) = e^{\frac{2}{3} x^{3 / 2}+C}\), which is the general form of solutions.

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Key Concepts

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

Separation of Variables
The method of Separation of Variables is a fundamental technique used to solve first-order linear differential equations, where the equation can be manipulated so that each variable appears on opposite sides of the equation.

In our exercise, the differential equation given was \(y' = \sqrt{x} y\). To implement separation of variables, we want to isolate \(y\) and its derivative \(y'\) on one side, and \(x\) with its derivatives on the other. This step involves algebraic manipulation, resulting in \(\frac{y'}{y} = \sqrt{x}\). Now, all instances of \(y\) are on the left and all instances of \(x\) are on the right, ready for integration.
Integration of Functions
The process of Integration of Functions allows us to find the original functions from their derivatives, which is the reverse operation of differentiation.

In the given problem, after separating variables, we are led to integrate both sides: \(\int \frac{y'}{y} dx = \int \sqrt{x} dx\). Integrating \(\frac{y'}{y}\) with respect to \(x\) yields \(\ln|y|\), the natural logarithm of the absolute value of \(y\), since the derivative of \(\ln|y|\) with respect to \(y\) is \(\frac{1}{y}\). On the right side, integrating \(\sqrt{x}\) with respect to \(x\) gives us \(\frac{2}{3}x^{3/2}\), as the integral of \(x^{n}\) is generally \(\frac{x^{n+1}}{n+1}\).

Integration is a powerful tool in solving differential equations as it helps to reconstruct the unknown function from its rate of change.
Exponential Functions
The concept of Exponential Functions plays a crucial role in solving differential equations, particularly after taking the integral of each side.

Once we obtain the form \(\ln|y| = \frac{2}{3}x^{3/2} + C\), where \(C\) is the constant of integration, we use the property of logarithms to solve for \(y\). Recognizing that \(e^{\ln|y|} = |y|\), we can convert the logarithmic form into its exponential counterpart, thereby getting rid of the logarithm. This leads to \(y = e^{\frac{2}{3}x^{3/2} + C}\) or \(y = e^{C}e^{\frac{2}{3}x^{3/2}}\), showcasing the exponential function's flexibility in simplifying expressions and solving equations.
Constant of Integration
When integrating, the Constant of Integration represents an arbitrary constant added to the function to account for the fact that indefinite integration is not unique.

In the equation found through integrating \(\ln|y| = \frac{2}{3}x^{3/2} + C\), \(C\) symbolizes all the possible constants that could be added to the integral to make it a family of functions, rather than a single solution. It is important in the context of initial value problems where a specific solution is required, and the value of \(C\) can be determined when a condition (\(y(x_0) = y_0\)) is given. Without an initial condition, we express the general solution to our differential equation involving \(C\), illustrating that there are infinitely many solutions, each differing by a constant.

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Most popular questions from this chapter

On the Richter scale, the magnitude \(R\) of an earthquake of intensity \(I\) is \(R=\frac{\ln I-\ln I_{0}}{\ln 10}\) where \(I_{0}\) is the minimum intensity used for comparison. Assume that \(I_{0}=1\). (a) Find the intensity of the 1906 San Francisco earthquake \((R=8.3)\). (b) Find the factor by which the intensity is increased if the Richter scale measurement is doubled. (c) Find \(d R / d I\).

(a) use Euler's Method with a step size of \(h=0.1\) to approximate the particular solution of the initial value problem at the given \(x\) -value, (b) find the exact solution of the differential equation analytically, and (c) compare the solutions at the given \(x\) -value. $$ \begin{array}{lll} \text { Differential Equation } & \text { Initial Condition } & x \text { -value } \\ \frac{d y}{d x}+6 x y^{2}=0 & (0,3) & x=1 \end{array} $$

Consider a tank that at time \(t=0\) contains \(v_{0}\) gallons of a solution of which, by weight, \(q_{0}\) pounds is soluble concentrate. Another solution containing \(q_{1}\) pounds of the concentrate per gallon is running into the tank at the rate of \(r_{1}\) gallons per minute. The solution in the tank is kept well stirred and is withdrawn at the rate of \(r_{2}\) gallons per minute. A 200 -gallon tank is full of a solution containing 25 pounds of concentrate. Starting at time \(t=0\), distilled water is admitted to the tank at a rate of 10 gallons per minute, and the well-stirred solution is withdrawn at the same rate. (a) Find the amount of concentrate \(Q\) in the solution as a function of \(t\). (b) Find the time at which the amount of concentrate in the tank reaches 15 pounds. (c) Find the quantity of the concentrate in the solution as \(t \rightarrow \infty\).

Find the orthogonal trajectories of the family. Use a graphing utility to graph several members of each family. \(x^{2}+y^{2}=C\)

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(y=f(x)\) is a solution of a first-order differential equation, then \(y=f(x)+C\) is also a solution.

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