In Exercises 129–132 solve the initial value problem.
131. x dy - (y + √y)dx = 0, y(1) = 1

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Rewrite the given differential equation in the form \(M(x,y)\,dx + N(x,y)\,dy = 0\). Here, the equation is \(x\,dy - (y + \sqrt{y})\,dx = 0\), so we can write it as \(-(y + \sqrt{y})\,dx + x\,dy = 0\).

Check if the differential equation is exact by computing the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\), where \(M = -(y + \sqrt{y})\) and \(N = x\).

If the equation is not exact, look for an integrating factor that depends on either \(x\) or \(y\) to make it exact. Consider the form of \(M\) and \(N\) to decide which variable the integrating factor might depend on.

Once the equation is exact, find the potential function \(\Psi(x,y)\) such that \(\frac{\partial \Psi}{\partial x} = M\) and \(\frac{\partial \Psi}{\partial y} = N\). Integrate \(M\) with respect to \(x\) and then determine the function of \(y\) by comparing with \(N\).

Use the initial condition \(y(1) = 1\) to solve for the constant of integration after finding the implicit solution \(\Psi(x,y) = C\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Initial Value Problem (IVP)

An initial value problem involves solving a differential equation with a given condition specifying the value of the unknown function at a particular point. This condition allows us to find a unique solution that fits the problem's context.

Separable Differential Equations

A separable differential equation can be rewritten so that all terms involving one variable and its differential are on one side, and all terms involving the other variable are on the opposite side. This allows integration of both sides separately to find the solution.

Substitution Method for Nonlinear Terms

When a differential equation contains nonlinear terms like √y, substitution (e.g., setting v = √y) can simplify the equation into a more manageable form. This technique helps transform complex expressions into linear or separable forms for easier solving.

In Exercises 129–132 solve the initial value problem.131. x dy - (y + √y)dx = 0, y(1) = 1