Question

Find the general solution to the differential equation.\(2 x \frac{d y}{d x}=y^{2}\)

Short answer

The general solution is \(y = \frac{1}{-\frac{1}{2}\ln|x| - C}\).

Step-by-step solution

Rewrite the Differential Equation

Start by rewriting the given differential equation in a form that helps us identify the separation of variables technique. The original differential equation is \(2x \frac{dy}{dx} = y^2\). Divide both sides by \(y^2\) and multiply by \(dx\) to get: \(\frac{1}{y^2}dy = \frac{1}{2x}dx\).

Separate Variables

The equation from Step 1 is now arranged to facilitate separation of variables. We have: \(\frac{1}{y^2}dy = \frac{1}{2x}dx\). Now, integrate both sides separately. The left side with respect to \(y\) and the right side with respect to \(x\).

Integrate Both Sides

Integrate both sides of the equation. The integral of \(\frac{1}{y^2}dy\) is \(-\frac{1}{y}\), and the integral of \(\frac{1}{2x}dx\) is \(\frac{1}{2}\ln|x|\). Hence, we obtain: \(-\frac{1}{y} = \frac{1}{2}\ln|x| + C\), where \(C\) is the constant of integration.

Solve for y

To express \(y\) in terms of \(x\), we rearrange the equation from Step 3. Multiply through by \(-1\) to get:\(\frac{1}{y} = -\frac{1}{2}\ln|x| - C\).Taking reciprocals: \(y = \frac{1}{-\frac{1}{2}\ln|x| - C}\). This gives us the general solution.

Key concepts

Separation of Variables

The technique of separation of variables is a fundamental approach for solving differential equations. It involves rearranging the equation so that all terms containing one variable are on one side and all terms containing the other variable are on the opposite side.
This allows us to "separate" the variables, enabling integration on each side regarding their corresponding variables. In the original exercise, we began with the differential equation:- \(2x \frac{dy}{dx} = y^2\)The goal was to isolate each variable. By dividing both sides by \(y^2\) and multiplying by \(dx\), we obtained:- \(\frac{1}{y^2}dy = \frac{1}{2x}dx\)This transformed equation is now set up for separation of variables, a method that works particularly well when each side of the equation involves only one variable (plus potentially constants). By setting up our equation this way, we create a clear path toward integrating each side separately, which is crucial in further solving the differential equation.

Integration Techniques

Integration is a key step in solving differential equations, especially once the variables have been separated. The task is to find the integral of each side of the equation corresponding to its variable.After separating the variables in our example to get:- \(\frac{1}{y^2}dy = \frac{1}{2x}dx\)We proceed with integration:- The left side, \(\frac{1}{y^2}dy\), integrates to \(-\frac{1}{y}\)- The right side, \(\frac{1}{2x}dx\), integrates to \(\frac{1}{2}\ln|x|\)These integrals are derived using basic techniques:- For \(\frac{1}{y^2}dy\), recognize it as \(y^{-2}\), which integrates to \(-y^{-1}\) or \(-\frac{1}{y}\), resulting from the power rule \(\int y^n dy = \frac{y^{n+1}}{n+1} + C\) when \(n=-2\).- For \(\frac{1}{2x}dx\), the integration is straightforward: since \(\int \frac{1}{x} dx = \ln|x|\), scaling by \(\frac{1}{2}\) gives \(\frac{1}{2}\ln|x|\).Thus, integration techniques allow us to express each side of the separated equation in terms of integrals, revealing relationships between the variables.

General Solution

To find the general solution of a differential equation, we seek an expression that encompasses all possible solutions. This involves integrating and including an arbitrary constant.Once we have integrated both sides of the equation, as seen in:- \(-\frac{1}{y} = \frac{1}{2}\ln|x| + C\)The general solution is obtained by solving for \(y\). By manipulating the equation, we took the reciprocal of both sides to isolate \(y\):- \(y = \frac{1}{-\frac{1}{2}\ln|x| - C}\)This represents the general solution, where \(C\) is the arbitrary constant reflecting the integration process. - The constant \(C\) implies that there are infinitely many solutions, each corresponding to different initial conditions or parameter values.- \(y\) is expressed in terms of \(x\), satisfying the original differential equation for any choice of \(C\).Solving for \(y\) provides a useful function relating \(y\) to \(x\), adaptable to specific conditions or constraints. The notion of a general solution is vital, as it forms the foundation for finding particular solutions under given circumstances.