91Ó°ÊÓ

When dealing with first-order linear differential equations, just like the one given in the problem \(y' - e^x y = 0\), it is often helpful to transform it into an easier format for solving.The standard form looks like this: \(y' + P(x) y = Q(x)\).In this format, \(P(x) = -e^x\) and \(Q(x) = 0\).The integrating factor is a powerful tool to simplify such equations.To find the integrating factor, we use the formula \(\mu(x) = e^{\int P(x) \, dx}\).It is a multiplier that we apply to the entire differential equation.By doing so, it allows us to express the left side of the equation as the derivative of a product.In our example, we calculated:

• \(\int - e^x \, dx = - e^x + C\)
• \(\mu(x) = e^{-e^x}\)

Using this integrating factor, we simplify the original equation, making it ready for further operations.

Finding the general solution of a differential equation involves solving it in a way that includes all possible solutions.In our exercise, once transformed using the integrating factor, the differential equation becomes simpler to solve.After multiplying \(y' - e^x y = 0\) by \(e^{-e^x}\), we get:\(e^{-e^x} y' - y = 0\).Recognizing this as a derivative of a product, we rewrite:\(\frac{d}{dx}(y(x)e^{-e^x}) = 0\).Integrating both sides, we obtain:\(y(x)e^{-e^x} = C\),where \(C\) is a constant of integration.This integration captures every possible solution by including this arbitrary constant.Finally, isolating \(y(x)\), we find:\(y(x) = Ce^{e^x}\).This is the general solution of the differential equation, encompassing all potential shifts via \(C\).

The product rule for derivatives is an essential concept, particularly when solving differential equations like the one provided.This rule states that if you have two functions, \(u(x)\) and \(v(x)\), the derivative of their product is:\((uv)' = u'v + uv'\).In the context of our problem:

• The product is \(y(x)e^{-e^x}\).
• The derivative became the expression \(e^{-e^x} y' - y\).

The beauty here is that we adjusted the equation into a perfect derivative using the integrating factor, which disguises a part of the expression as a product derivative.By recognizing the structure of \(e^{-e^x} y' - y\) as a derivative,we connected it back to its integral form: \(\frac{d}{dx}(y(x)e^{-e^x}) = 0\).Understanding this principle further facilitates integrating separately and simplifying to reach the general solution.