Question

Solve the initial-value problems. $$ (4 x+3 y+1) d x+(x+y+1) d y=0, \quad y(3)=-4. $$

Short answer

The short answer to the given initial-value problem is the implicit solution: \(U(x, y) = 2x^2 + 3xy + x + y^2 - 2xy + y + 51 = 0.\)

Step-by-step solution

1. Check if the equation is exact

In order to check for exactness, we first need to find the partial derivatives M_y and N_x and check if they are equal. If they are equal, then the given equation is exact and can be solved using exact differential equation methods. Given the equation \((4x + 3y + 1)dx + (x + y +1)dy = 0 \), we have \(M(x, y) = 4x + 3y + 1\) and \(N(x, y) = x + y + 1\). Now let's find the partial derivatives. - Partial derivative of M with respect to y, \(M_y = \frac{\partial M}{\partial y} = 3\) - Partial derivative of N with respect to x, \(N_x = \frac{\partial N}{\partial x} = 1\) Since \(M_y = N_x\), the given differential equation is exact.

2. Find the potential function U(x, y)

When the differential equation is exact, the potential function U(x, y) can be found by integrating M(x, y) with respect to x and N(x, y) with respect to y. First, integrate M(x, y) with respect to x: \[ U(x, y) = \int M(x, y) dx = \int (4x + 3y + 1) dx = 2x^2 + 3xy + x + c(y) \] Now, differentiate this function U with respect to y, and then equate the result with N(x, y) to find the function c(y). \[ \frac{\partial U}{\partial y} = N(x, y) \Rightarrow 3x + c'(y) = x + y + 1 \] Solve for c'(y): \[ c'(y) = x + y + 1 - 3x \] Now, integrate c'(y) with respect to y: \[ c(y) = \int (y - 2x + 1) dy = y^2 -2xy + y + C \] Thus, the potential function U(x, y) is: \[ U(x, y) = 2x^2 + 3xy + x + y^2 - 2xy + y + C \]

3. Use the initial condition to find the constant C

Now that we've found the potential function U(x, y), we can use the initial condition \(y(3) = -4\) to find the constant of integration C. Plug the initial conditions into U(x, y): \[ U(3, -4) = 2(3)^2 + 3(3)(-4) + 3 - 4^2 - 2(3)(-4) - 4 + C = 0 \] Solve for C: \[ C = 51 \] So, the particular solution of the exact differential equation is: \[ U(x, y) = 2x^2 + 3xy + x + y^2 - 2xy + y + 51 = 0 \] This solution yields the implicit solution for the given initial value problem.

Key concepts

Understanding Initial-Value Problems

When solving differential equations, 'initial-value problems' are those in which you're provided with the value of the unknown function at a specific point. This piece of information is crucial as it dictates one specific solution among the possible infinite solutions to the differential equation.

To demonstrate this concept, let's consider our given problem \[ (4x + 3y + 1)dx + (x + y + 1)dy = 0, \quad y(3)=-4. \] The part \(y(3)=-4\) is our initial condition. It specifies that when \(x = 3\), \(y = -4\). By applying this initial condition after finding the general solution, we determine the specific 'constant of integration,' which leads to the unique solution that satisfies both the differential equation and the initial condition.

The Role of Partial Derivatives in Exact Differential Equations

Understanding 'partial derivatives' is central to solving exact differential equations. Partial derivatives tell us how a multivariable function changes as one variable changes while keeping all other variables constant. To solve an exact differential equation, we must first check if it meets the criterion of exactness, which involves comparing the partial derivatives of two functions involved.

In our equation, \(M(x, y) = 4x + 3y + 1\) and \(N(x, y) = x + y + 1\), the checks \(M_y = 3\) and \(N_x = 1\) ensure that they are indeed equal, confirming that our equation is exact. By discerning this, we can assert that a function \(U(x, y)\), called a 'potential function', exists—such that \(dU = Mdx + Ndy\). With this piece of knowledge, we proceed to find \(U(x, y)\), which will lead us to the solution of our differential equation.

Finding the Potential Function in Exact Equations

A 'potential function' \(U(x, y)\) is a scalar function whose gradients are the components of our given differential equation. For exact equations like ours, finding this potential function is the key to solving the problem since it encapsulates all information of the system described by the differential equation.

We obtain \(U(x, y)\) by integrating \(M\) with respect to \(x\), and \(N\) with respect to \(y\). However, after the first integration, we include an arbitrary function (\(c(y)\) or \(g(x)\)) which accounts for the constant of integration—the function’s dependency on the other variable. By taking the derivative of our first integral with respect to the other variable, we can solve for this arbitrary function, providing us with our potential function \(U(x, y)\).

Integration of Differential Equations to Find Solutions

Finally, 'integration of differential equations' is the process we use to find the potential function that ultimately leads us to the solution of the given problem. It is through integration that we progress from the differential form to a function that represents the equations. In doing so, we systematically add the arbitrary functions that reflect the integration constant dependencies on the other variables.

In the case of exact equations, integration helps us construct the potential function \(U(x, y)\) that, when differentiated, gives back the components of the original differential equation. Consequently, the integral of \(M(x, y)dx\) and \(N(x, y)dy\), combined with applying the initial-value condition, grants us the explicit solution we seek, solving our initial-value problem.