Question

The given equation is an implicit solution of \(N(t, y) y^{\prime}+M(t, y)=0\), satisfying the given initial condition. Assuming the equation \(N(t, y) y^{\prime}+M(t, y)=0\) is exact, determine the functions \(M(t, y)\) and \(N(t, y)\), as well as the possible value(s) of \(y_{0}\). \(y^{3}+4 t y+t^{4}+1=0, \quad y(0)=y_{0}\)

Short answer

Answer: The functions M(t, y) and N(t, y) are \(M(t, y) = 4y + 4t^3\) and \(N(t, y) = 3y^2 + 4t\), and the possible value of \(y_0\) is \(-1\).

Step-by-step solution

Differentiate the implicit solution with respect to t

Starting with the given implicit solution equation, differentiate it with respect to \(t\). Apply the chain rule when differentiating terms containing \(y(t)\): \[\frac{d}{dt}(y^3 + 4ty + t^4 + 1) = \frac{d}{dt}(0)\] Now, differentiate each term on the left-hand side: \[\frac{d}{dt}(y^3) + \frac{d}{dt}(4ty) + \frac{d}{dt}(t^4) + \frac{d}{dt}(1) = 0\]

Find y' term in the derivative

Using the chain rule and product rule, we find the terms involving \(y'\): \[3y^2y' + 4y + 4ty' + 4t^3 = 0\] Now, we can rewrite this equation in the form of \(N(t, y) y^{\prime}+M(t, y)=0\): \[(3y^2 + 4t)y' + (4y + 4t^3) = 0\] Thus, we found the functions \(N(t, y) = 3y^2 + 4t\) and \(M(t, y) = 4y + 4t^3\).

Find possible value(s) of \(y_{0}\)

Now let's find the value(s) of \(y_{0}\) for the given initial condition, \(y(0)=y_{0}\). Substitute \(y=y_0\) and \(t=0\) into the given implicit equation: \[y_0^3 + 4(0)(y_0) + 0^4 + 1 = 0\] Solve for \(y_0\): \[y_0^3 + 1 = 0\] \[y_0^3 = -1\] This implies that \(y_0 = \sqrt[3]{-1} = -1\). So, the functions \(M(t, y) = 4y + 4t^3\) and \(N(t, y) = 3y^2 + 4t\), and the possible value of \(y_0\) is \(-1\).

Key concepts

Implicit Differentiation

Implicit differentiation is a technique used when dealing with equations where the variables cannot be easily separated, which is common in many calculus problems. It involves differentiating both sides of the equation with respect to a single variable, even if the other variable appears on both sides of the equation.

In our case, the equation involves both variables, t, which is the independent variable, and y as the dependent variable. To differentiate such an implicit equation, we apply the derivative operator to each term as demonstrated in the solution. We treat y as a function of t, which implies whenever we differentiate a term with y in it, we must also incorporate y', the derivative of y with respect to t, using the chain rule.

This method reveals the underlying structure of the differential equation by clearly differentiating between the parts that vary with t and those that vary with y. It’s critical in setting up differential equations in a form that can be solved using various methods.

Initial Value Problem

An initial value problem is a type of differential equation accompanied by a specific condition that gives the value of the unknown function at a particular point. The condition, often in the form y(a) = b, is called the initial condition because it specifies where the solution begins.

Solving an initial value problem involves finding a function that not only satisfies the differential equation but also meets the stated initial condition. In the example given, we are provided with the initial condition y(0) = y_0. By substituting the initial condition into the differential equation, we can solve for y_0, thereby providing the specific solution to the initial value problem. The value y_0 is crucial because it anchors the solution curve at a known point, allowing a unique solution curve to be drawn from a family of possible solutions.

Chain Rule

The chain rule is a fundamental theorem in calculus that is used when finding the derivative of a composite function. In essence, it provides a way to differentiate a function that is 'nested' inside another. If you have a function u(g(t)), where u is a function of g, and g is a function of t, the derivative of u with respect to t is the product of the derivative of u with respect to g and the derivative of g with respect to t.

For the problem at hand, we apply the chain rule to differentiate terms like y^3, because y is a function of t. We multiply the outer derivative \(3y^2\) by the derivative of y, which is y'. This is an example of the chain rule in action, critical for handling derivatives of composite functions in differential equations.

Product Rule

The product rule is another essential concept in calculus when we need to differentiate products of two functions. According to the product rule, the derivative of a product of two functions f(t) and g(t) is given by f'(t)g(t) + f(t)g'(t). This means we take the derivative of the first function and multiply it by the second function as is, then we add the product of the first function as is with the derivative of the second function.

In our exercise, the term 4ty required the use of the product rule since it is the product of t and y. We differentiate t with respect to itself to get 1 (leaving y unchanged), and then multiply this by the derivative of y with respect to t (which is y'), and sum the two results. Correct application of the product rule helps us split the derivatives into clearly distinguishable parts, enabling the identification of functions M(t, y) and N(t, y) in exact differential equations.