91Ó°ÊÓ

Separation of variables is a powerful method used to solve ordinary differential equations (ODEs). This technique simplifies an ODE by separating it into two distinct integrals, each involving only one variable. To do this, we rearrange the given equation so that all terms involving one variable are on one side, and the terms with the other variable are on the opposite side.

For the equation \(\frac{d y}{d x}=-y+2\), we start by manipulating it to isolate terms. We move \(-y+2\) to the denominator of the left side and rearrange to get:

• \(\frac{dy}{-y+2}=dx\)

This step accomplishes the separation of variables where each side of the equation is now in terms of a single variable. This allows us to integrate both sides independently, setting the stage for finding the general solution.

Integration is the mathematical process of finding the function, or antiderivative, that describes the accumulation of quantity. In solving the separated equation \(\frac{dy}{-y+2}=dx\), integration is used to find the relationship between \(y\) and \(x\).

We integrate both sides with respect to their variables, leading to:

• Left side: \(\int \frac{dy}{-y+2}\)
• Right side: \(\int dx\)

Using substitution on the left side, let \(u = -y+2\), then \(du = -dy\), simplifies the integral to:

• \(-\int \frac{du}{u} = \int dx\)

The left integral evaluates to \(-\ln|u|\), and the right to \(x + C\), resulting in the expression:

• \(-\ln|u| = x + C\)

This integral form is crucial as it will be used to express the relationship in terms of \(y\) by swapping back \(u\) for \(-y+2\).

Exponential functions are essential in expressing differential equation solutions. They describe growth or decay processes and have the form \(e^x\), where \(e\) is Euler's number, approximately 2.718.

After integrating the equation and substituting back for \(u = -y+2\), you obtain \(-\ln|-y+2| = C - x\). To solve for \(y\), we exponentiate both sides to clear the logarithm:

• \( |-y+2| = e^{C-x} \)

Exponentiating both sides helps revert the logarithmic transformation, leading to an expression for \(y\). Rearranging gives:

• \( y - 2 = - e^{C-x} \)

Finally, expressing \(y\) in terms of \(x\):

• \( y(x) = 2 - e^{C-x} \)

This form is common in solutions of differential equations where initial conditions are used to determine the constant \(C\). The key takeaway with exponential functions is how they transform the problem into a format that describes the behavior of the solution clearly and effectively.