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. \(\left(2 t y+\frac{1}{y}\right) y^{\prime}+y^{2}=1, \quad y(1)=1\)

Short answer

Answer: No, it's not possible to find an explicit solution for \(y\). However, we can find an implicit solution \(2y^2+y=y(1+L)\), where \(L\) is a constant.

Step-by-step solution

Verify the exactness of the differential equation

We are given the nonlinear differential equation: \((2ty+\frac{1}{y})y'+y^2=1\) Here, \(M(t, y) = (2ty+\frac{1}{y})\) and \(N(t, y) = y^2\). We now compute the partial derivatives: \(\frac{\partial M}{\partial y} = 2t-\frac{1}{y^2}\) \(\frac{\partial N}{\partial t} = 0\) Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial t}\), the given equation is not exact. Now, we can try multiplying the equation by an integrating factor to make it exact.

Find the integrating factor

Let's try finding an integrating factor (\(\mu(t)\)) that depends on \(t\) only. To find the integrating factor we can use this formula: \(\frac{d \mu(t)}{dt} = (\frac{\partial M}{\partial y} - \frac{\partial N}{\partial t}) \mu(t)\) Here we have, \(\frac{d \mu(t)}{dt} = (2t-\frac{1}{y^2}) \mu(t)\) Since the right-hand side of the equation only contains \(t\), and we're looking for an integrating factor that depends on \(t\) only, we can divide both sides by \(\mu(t)\): \(\frac{d \mu(t)}{dt} \frac{1}{\mu(t)} = 2t-\frac{1}{y^2}\) Integrating both sides with respect to \(t\), we get: \(ln(\mu)=t^2+C \Rightarrow \mu=e^{t^2+C}=e^{t^2}\) (since \(e^C\) is a constant)

Multiply the original equation by the integrating factor

Now, multiply the given equation by the integrating factor \(e^{t^2}\): \(e^{t^2}(2ty+\frac{1}{y})y'+e^{t^2}y^2=e^{t^2}\)

Verify exactness again and find the potential function F(t,y)

Calculate the partial derivatives again for the new equation: \(\frac{\partial M}{\partial y} = e^{t^2}(2t-\frac{1}{y^2})\) \(\frac{\partial N}{\partial t} = 2ty^2e^{t^2}\) We can now see that \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\), so the new equation is exact. We can now find the potential function (\(F(t, y)\)) from the equation. We have: \(\frac{\partial F}{\partial t} = e^{t^2}(2ty+\frac{1}{y})\) and \(\frac{\partial F}{\partial y} = y^2e^{t^2}\) Integrating the first equation with respect to \(t\), we get: \(F(t, y) = \int e^{t^2}(2ty+\frac{1}{y}) dt\) Now, use substitution \(u=t^2 \Rightarrow \frac{1}{2}du=tdt\). So we have: \(F(t, y) = 2\int (yu+\frac{1}{2y}) e^u du\) Now integrate with respect to 'u' to find the potential function: \(F(t, y) = 2ye^u+\frac{1}{y}(e^u+C(y))\) Substituting \(u=t^2\) back and using the second condition: \(\frac{\partial}{\partial y} (2ye^{t^2}+\frac{1}{y}(e^{t^2}+C(y))) = y^2e^{t^2}\) Comparing this with the exact equation \(e^{t^2}y^2=e^{t^2}\), we get: \(C^{\prime}(y)=0 \Rightarrow C(y)=K\), where 'K' is a constant. Now, we have the potential function: \(F(t, y) = 2ye^{t^2}+\frac{1}{y}(e^{t^2}+K)\) To find the implicit solution, consider \(F(t, y)=C\), and replace 'C' by a constant 'c': \(2ye^{t^2}+\frac{1}{y}(e^{t^2}+K)=c\)

Solve for y and apply the initial condition

Now apply the initial condition \(y(1)=1\): \(2(1)e+(1)e^{1}+K=c\) To simplify the equation, let's use the substitution \(K+c=L\), where \(L\) is a constant: \(2ye^{t^2}+e^{t^2}=\frac{1}{y}e^{t^2}(1+L)\) Multiply both sides by \(y\): \(2ye^{t^2}y+e^{t^2}y=e^{t^2}(1+L)\) Divide both sides by \(e^{t^2}\): \(2y^2+y=y(1+L)\) Now, we can't solve for \(y\) explicitly, but we have an implicit solution: \(2y^2+y=y(1+L)\)

Key concepts

Exact Differential Equation

In the realm of differential equations, an exact differential equation is a goldmine for students looking for a clear-cut method to finding solutions. Imagine a situation where you're given a formula, and rather than tackling it through trial and error, there's a direct path to the answer. That's what you get with exact equations.

An exact differential equation looks something like this: \(M(t, y) + N(t, y)y' = 0\), where \(M\) and \(N\) are functions of \(t\) and \(y\), respectively. The clincher that makes an equation exact is this nifty condition: \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\). This tells us that both partial derivatives are mirror images of each other, and the solution involves a potential function \(F(t, y)\) such that \(M\) is the partial derivative of \(F\) with respect to \(y\), and \(N\) the derivative with respect to \(t\).

It's like discovering two puzzle pieces that fit perfectly together, revealing the bigger picture - the solution to our differential equation puzzle.

Integrating Factor

Sometimes our differential equations aren't as cooperative as we'd like, refusing to fit the mold of an exact equation. In these moments of mathematical stubbornness, we call upon the superhero of the equation world: the integrating factor.

The integrating factor is like a mathematical charm, transfiguring a non-exact equation into an exact one. It’s chosen carefully to ensure that once it’s multiplied by the original equation, the magical condition for exactness is satisfied.

The trick lies in finding an integrating factor \(\mu(t)\) that depends solely on \(t\), or \(\mu(y)\) that's a function of \(y\). With the right integrating factor, we breathe new life into the original equation, multiplying through to set the stage for exactness. The end goal? To reach a point where we can finally unravel the potential function \(F(t, y)\) which holds the keys to our elusive solution.

Initial Value Problem

Transitioning from the abstract to the concrete, we encounter the initial value problem. This isn't just any run-of-the-mill puzzle; it's one with a starting point, a 'known' in a sea of unknowns. When you have a differential equation paired with an initial condition, like \(y(1) = 1\), you're looking at an initial value problem.

This initial condition serves as an anchor, locking in the value of \(y\) at a specific instant \(t\), and guiding us to a unique solution. It's like being handed a map with a marked spot as you start a treasure hunt, except here, the treasure is the particular solution that fits that initial stipulation.

Implicit Solution

As we dive deeper into the world of differential equations, we may encounter thickets where an explicit path isn't evident, and wandering off-road is our only option. Enter the implicit solution.

An implicit solution is a solution still entangled in the relationship between \(t\) and \(y\), without isolating \(y\) on one side by itself. It represents a valid solution, more often seen as an equation involving both variables, like a secret hand-in-hand agreement between them. Sometimes, these solutions are best left in this form, as the effort to separate them might lead to complications or impossibilities.

Explicit Solution

After grappling with implicitness, students often breathe a sigh of relief when they come upon the explicit solution. It's the moment when \(y\) finally stands independent, with every other variable on the opposite side of the equals sign waving goodbye.

An explicit solution puts \(y\) in the spotlight - clear, untangled, and solo. It's the preferred form when practicality beckons, as it directly expresses \(y\) as a function of \(t\). This is the form that allows for straightforward plotting on graphs or running through calculators without a hitch, a beacon of clarity in the often murky waters of differential equations.