Question

Show that the given nonlinear differential equation is exact. (Some algebraic manipulation may be required. Also, recall the remark that follows Example 1.) Find an implicit solution of the initial value problem and (where possible) an explicit solution. \((2 y-t) y^{\prime}-y+2 t=0, \quad y(1)=0\)

Short answer

Question: Verify if the given differential equation \((2y - t)y' - y + 2t = 0\) is exact and solve it when \(y(1)=0\). Answer: The differential equation is exact, and the implicit solution for the given initial value \(y(1)=0\) is given by \(-yt + t^2 - y^2 -1= 0\).

Step-by-step solution

Rewrite the given equation in the form \(M(t, y) + N(t, y)y' = 0\)

The equation \((2y - t)y' - y + 2t = 0\) can be rewritten as: $$(-y+2t) + (2y - t)y' = 0,$$ where \(M(t, y) = -y + 2t\) and \(N(t, y) = 2y - t\).

Check for exactness

To check if the given equation is exact, we need to verify if $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}.$$ Find the partial derivatives: $$\frac{\partial M}{\partial y} = \frac{\partial (-y + 2t)}{\partial y} = -1$$ $$\frac{\partial N}{\partial t} = \frac{\partial (2y - t)}{\partial t} = -1$$ The exactness condition is satisfied. Thus, the given differential equation is exact.

Find an implicit solution

Since the equation is exact, there exists a function \(\psi(t, y)\) such that $$\frac{\partial \psi}{\partial t} = M(t, y) \quad \text{and} \quad \frac{\partial \psi}{\partial y} = N(t, y).$$ We first integrate \(M(t, y)\) with respect to \(t\): $$\psi(t, y) = \int M(t, y) \,\mathrm{d}t = \int (-y + 2t) \,\mathrm{d}t = -yt + t^2 + f(y),$$ where \(f(y)\) is an arbitrary function of \(y\). Next, differentiate \(\psi\) with respect to \(y\) and match it with \(N(t, y)\): $$\frac{\partial \psi}{\partial y} = -t + f'(y) = N(t, y) = 2y - t.$$ Now, integrate with respect to \(y\): $$f(y) = \int -2y\,\mathrm{d}y = -y^2 + C,$$ where C is a constant. The implicit solution is: $$\psi(t, y) = -yt + t^2 - y^2 + C = 0.$$

Apply the initial conditions and find the explicit solution

Apply the initial condition \(y(1) = 0\): $$\psi(1, 0) = -(1)(0) + (1)^2 - (0)^2 + C = 0,$$ which results in $$C = -1.$$ Now, we have the implicit solution: $$-yt + t^2 - y^2 -1= 0.$$ However, we cannot solve for \(y\) explicitly as a function of \(t\) in this case. Therefore, the implicit solution is the final answer: $$-yt + t^2 - y^2 -1= 0.$$

Key concepts

Exact Differential Equations

Nonlinear differential equations can often seem complex, but they can be simplified when they are "exact." A differential equation is said to be exact if it can be expressed in a form such that there exists a function \(\psi(t, y)\), whose total differential matches the given differential equation. To identify this, the equation needs to be written in the form \(M(t, y) + N(t, y)y' = 0\). Here, \(M\) and \(N\) are functions of \(t\) and \(y\).

To verify exactness, you check if the cross-partial derivatives are equal: \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\). In our exercise, by calculating these partial derivatives, we observed that indeed \(-1 = -1\), confirming that the differential equation is exact. This step essentially tells us that the equation has a well-defined path to a solution, making it equivalent to finding the exact derivative of some potential function \(\psi(t, y)\).

Implicit Solutions

Implicit solutions arise naturally from exact differential equations. Once the exactness condition is satisfied, the goal is to find a function \(\psi(t, y)\) such that \(\frac{\partial \psi}{\partial t} = M(t, y)\) and \(\frac{\partial \psi}{\partial y} = N(t, y)\). This process involves integrating \(M\) with respect to \(t\) and \(N\) with respect to \(y\).

In our example, integrating \(-y + 2t\) with respect to \(t\), we obtained \(-yt + t^2 + f(y)\). Similarly, adjusting \(f(y)\) through integration, led us to the implicit solution \(\psi(t, y) = -yt + t^2 - y^2 + C\). Implicit solutions highlight the nature of the solution in its entirety, encapsulating all relationships between \(t\) and \(y\), even if they don’t plainly say \(y = \text{function of } t\).

Initial Value Problem

Solving an initial value problem (IVP) in differential equations involves not only finding the solution of the differential equation but ensuring this solution satisfies given initial conditions. These conditions help pinpoint exactly which solution to the differential equation applies to a given scenario.

In our case, the condition \(y(1) = 0\) was applied to the implicit solution \(-yt + t^2 - y^2 + C = 0\). By substituting \(t = 1\) and \(y = 0\) into the equation, we determined that the constant \(C\) is \(-1\). This step helps tailor the general solution to the specific problem. However, sometimes, as in this case, the solution remains implicit. While not always readily convertable to \(y\) in terms of \(t\), this implicit form still holds the complete answer as the relationship is fully defined.