Exact Differential Equations: Identifying & Solving via ψ(x,y)

First, verify exactness: M = 3x² + 2xy, N = x² + y², so ∂M/∂y = 2x and ∂N/∂x = 2x (equal, exact). Find ψ by integrating M with respect to x: ψ = x³ + x²y + h(y). Differentiate with respect to y: ∂ψ/∂y = x² + h'(y). Set equal to N: x² + h'(y) = x² + y², so h'(y) = y². Integrate: h(y) = (⅓)y³ + C. Thus, ψ = x³ + x²y + (⅓)y³ = C. Using the point (1,1): 1³ + 1²·1 + (⅓)·1³ = 1 + 1 + 1/3 = 7/3. Therefore, the solution curve is x³ + x²y + (⅓)y³ = 7/3.

To determine if a differential equation M dx + N dy = 0 is exact, we need to check if ∂M/∂y = ∂N/∂x. For option A, M = 2x + y and N = x + 2y. Computing partial derivatives gives ∂M/∂y = 1 and ∂N/∂x = 1. Since these are equal, the equation is exact. For option B, M = x² + y and N = xy + 1 gives ∂M/∂y = 1 and ∂N/∂x = y (not equal). For option C, M = 3x² + 2xy and N = 2x² + 3y gives ∂M/∂y = 2x and ∂N/∂x = 4x (not equal). For option D, M = y² + x and N = 3xy + 1 gives ∂M/∂y = 2y and ∂N/∂x = 3y (not equal). Therefore, only option A is exact.

To verify if this statement is true, we check that the partial derivatives of ψ match M and N. For ψ(x,y) = x² + xy + y², we have ∂ψ/∂x = 2x + y = M and ∂ψ/∂y = x + 2y = N. Since both conditions are satisfied, this is indeed the correct potential function. For option A, the equation is not exact because ∂M/∂y = 1 and ∂N/∂x = y (not equal). For option B, an integrating factor is only needed when ∂M/∂y ≠ ∂N/∂x. For option D, not all first-order differential equations can be made exact with an integrating factor. For option E, the condition for exactness is ∂M/∂y = ∂N/∂x, not ∂M/∂x = ∂N/∂y.

For an exact equation M dx + N dy = 0, we find the potential function ψ by integrating M with respect to x (treating y as constant) or N with respect to y (treating x as constant). Integrating M = 2x + y with respect to x gives ψ = x² + xy + h(y). Differentiating this with respect to y gives ∂ψ/∂y = x + h'(y). Setting this equal to N = x + 2y gives x + h'(y) = x + 2y, so h'(y) = 2y. Integrating h'(y) with respect to y gives h(y) = y² + C. Therefore, ψ(x,y) = x² + xy + y² + C. The constant C is omitted in the potential function, so ψ(x,y) = x² + xy + y².

For an exact equation M dx + N dy = 0, if ψ(x,y) is the potential function, then the general solution is ψ(x,y) = C, where C is a constant. From the previous question, we found ψ(x,y) = x² + xy + y². Therefore, the general solution is x² + xy + y² = C.

For an exact equation M dx + N dy = 0, the potential function ψ satisfies ∂ψ/∂x = M and ∂ψ/∂y = N. Here, M = sin(x) + 2y and N = 2x + cos(y). Therefore, ∂ψ/∂y = N = 2x + cos(y).

For any first-order differential equation in the form M dx + N dy = 0, the first step is to check if it is exact. We compute ∂M/∂y and ∂N/∂x. If they are equal, the equation is exact and can be solved by finding the potential function. If not equal, we might need to find an integrating factor or use another method. For this equation, M = x² + 2y and N = 2x - y, so ∂M/∂y = 2 and ∂N/∂x = 2, making it exact.

The solution to an exact differential equation is given by ψ(x,y) = C. Substituting the given point (1,1) into the equation gives ψ(1,1) = 1²·1 + 1·1² + 1³ = 1 + 1 + 1 = 3. Therefore, C = 3.

The solution ψ(x,y) = C represents the level curves (or contour lines) of the function ψ(x,y) = constant. These curves have the property that moving along them, the differential change dψ = (∂ψ/∂x)dx + (∂ψ/∂y)dy = M dx + N dy = 0, which means ψ is constant along these curves. Geometrically, the vector field (M,N) = (∂ψ/∂x, ∂ψ/∂y) is perpendicular to these level curves at every point, not parallel. The other options either misstate the relationship or confuse the interpretation.

The defining condition for an exact equation is that the mixed partial derivatives of the potential function ψ(x,y) are equal, which translates to ∂M/∂y = ∂N/∂x. This ensures that M dx + N dy is the total differential of some function ψ.

First, we verify the equation is exact. M = 2xy + y² and N = x² + 2xy, so ∂M/∂y = 2x + 2y and ∂N/∂x = 2x + 2y (equal, so exact). To find ψ, integrate M with respect to x: ψ = x²y + xy² + h(y). Differentiating with respect to y gives ∂ψ/∂y = x² + 2xy + h'(y). Setting this equal to N gives x² + 2xy + h'(y) = x² + 2xy, so h'(y) = 0 and h(y) = C. Thus, ψ = x²y + xy² = C. Using the initial condition y(1) = 2: 1² · 2 + 1 · 2² = 2 + 4 = 6. Therefore, the particular solution is x²y + xy² = 6.

While exact differential equations have solutions of the form ψ(x,y) = C, this does not guarantee a unique solution for every initial condition. If the initial condition (x₀, y₀) satisfies ψ(x₀, y₀) = C₀, then there exists a unique solution passing through that point only if certain conditions are met (such as the implicit function theorem conditions). However, it's possible to have initial conditions where multiple solutions exist or where no solution exists, particularly at points where the gradient of ψ is zero.

To find ∂M/∂y when M = x³ + xy², we differentiate M with respect to y while treating x as constant. The term x³ has derivative 0 with respect to y, and the term xy² has derivative 2xy with respect to y (treating x as constant). Therefore, ∂M/∂y = 2xy.

The general solution to an exact equation is ψ(x,y) = C, where ψ is the potential function. Given ψ(x,y) = xeˣ - eˣ + xy + eʸ + C, we omit the constant C (which is arbitrary) and write the solution as xeˣ - eˣ + xy + eʸ = C. The other options either miss terms or have incorrect coefficients.

After finding ψ = f(x,y) + h(y) by integrating M with respect to x, we need to determine h(y). We do this by differentiating ψ with respect to y to get ∂ψ/∂y = ∂f/∂y + h'(y). We then set this equal to N (from the original equation M dx + N dy = 0) to get ∂f/∂y + h'(y) = N. Solving for h'(y) gives h'(y) = N - ∂f/∂y, and then we integrate to find h(y).