Question

Suppose the differential equation \(N(t, y) y^{\prime}+\) \(M(t, y)=0\) is not exact; that is, \(N_{t}(t, y) \neq M_{y}(t, y) .\) Is it possible to multiply the equation by a function, call it \(\mu(t, y)\), so that the resulting equation is exact? (a) Show that if \(\mu(t, y) N(t, y) y^{\prime}+\mu(t, y) M(t, y)=0\) is exact, the function \(\mu\) must be a solution of the partial differential equation $$ N(t, y) \mu_{t}-M(t, y) \mu_{y}=\left[M_{y}(t, y)-N_{t}(t, y)\right] \mu . $$ Parts (b) and (c) of this exercise discuss special cases where the function \(\mu\) can be chosen to be a function of a single variable. In these special cases, the partial differential equation in part (a) reduces to a first order linear ordinary differential equation and can be solved using the techniques of Section 2.2. (b) Suppose the quotient \(\left[M_{y}(t, y)-N_{t}(t, y)\right] / N(t, y)\) is just a function of \(t\), call it \(p(t)\). Let \(P(t)\) be an antiderivative of \(p(t)\). Show that \(\mu\) can be chosen as \(\mu(t)=e^{P(t)}\). (c) Suppose the quotient \(\left[N_{t}(t, y)-M_{y}(t, y)\right] / M(t, y)\) is just a function of \(y\), call it \(q(y)\). Let \(Q(y)\) be an antiderivative of \(q(y)\). Show that \(\mu\) can be chosen as \(\mu(y)=e^{Q(y)}\).

Short answer

Answer: The inexact differential equation can be made exact by multiplying with \(\mu(t, y)\) in two special cases: 1. When \(\frac{M_{y}(t, y)-N_{t}(t, y)}{N(t, y)}\) is a function of \(t\) only, denoted by \(p(t)\). In this case, \(\mu(t)\) depends only on \(t\) and has the form \(\mu(t) = e^{P(t)}\), where \(P'(t) = p(t)\). 2. When \(\frac{N_{t}(t, y)-M_{y}(t, y)}{M(t, y)}\) is a function of \(y\) only, denoted by \(q(y)\). In this case, \(\mu(y)\) depends only on \(y\) and has the form \(\mu(y) = e^{Q(y)}\), where \(Q'(y) = q(y)\).

Step-by-step solution

Derive the partial differential equation for \(\mu\)

First, we are given the inexact equation: $$ N(t, y) y^{\prime}+M(t, y)=0 $$ Now, suppose that multiplying the equation by \(\mu(t, y)\) turns it into an exact equation: $$ \mu(t, y) N(t, y) y^{\prime}+\mu(t, y) M(t, y)=0 $$ For this equation to be exact, we must have: $$ (\mu N)_{y} = (\mu M)_{t} $$ Using the product rule for differentiation, we can write the above condition as: $$ N \mu_{y}+\mu N_{y}=M \mu_{t}+\mu M_{t} $$ Now, rearrange to isolate the partial derivatives of \(\mu\): $$ N(t, y) \mu_{t}-M(t, y) \mu_{y}=\left[M_{y}(t, y)-N_{t}(t, y)\right] \mu $$ So, \(\mu\) must be a solution of the partial differential equation: $$ N(t, y) \mu_{t}-M(t, y) \mu_{y}=\left[M_{y}(t, y)-N_{t}(t, y)\right] \mu . $$ In parts (b) and (c), we will focus on special cases where \(\mu\) depends on a single variable.

Case when \(\mu\) depends only on \(t\)

In this case, we are given that the quotient \(\frac{M_{y}(t, y)-N_{t}(t, y)}{N(t, y)}\) is just a function of \(t\), which we denote as \(p(t)\). We can write: $$ \frac{M_{y}(t, y)-N_{t}(t, y)}{N(t, y)} = p(t) $$ Let \(P(t)\) be an antiderivative of \(p(t)\), so: $$ P'(t) = p(t) $$ Now, we consider a function \(\mu(t)\) dependent only on \(t\): $$ \mu(t)=e^{P(t)} $$ We will now show that this function indeed satisfies the partial differential equation derived in Step 1. To this end, we first note that: $$ \mu_{t}(t,y) = P'(t) e^{P(t)} = p(t) \mu(t)$$ $$ \mu_{y}(t,y) = 0 $$ Plugging these into the partial differential equation in Step 1, we obtain: $$ N(t, y) p(t) \mu(t) - M(t, y) (0) = \left[M_{y}(t, y)-N_{t}(t, y)\right] \mu(t) $$ This proves that \(\mu(t) = e^{P(t)}\) and satisfies the partial differential equation in Step 1 when \(\mu\) is a function of \(t\) only.

Case when \(\mu\) depends only on \(y\)

In this case, we are given that the quotient \(\frac{N_{t}(t, y)-M_{y}(t, y)}{M(t, y)}\) is just a function of \(y\), which we denote as \(q(y)\). We can write: $$ \frac{N_{t}(t, y)-M_{y}(t, y)}{M(t, y)} = q(y) $$ Let \(Q(y)\) be an antiderivative of \(q(y)\), so: $$ Q'(y) = q(y) $$ Now, we consider a function \(\mu(y)\) dependent only on \(y\): $$ \mu(y)=e^{Q(y)} $$ We will now show that this function indeed satisfies the partial differential equation derived in Step 1. To this end, we first note that: $$ \mu_{t}(t,y) = 0 $$ $$ \mu_{y}(t,y) = Q'(y) e^{Q(y)} = q(y) \mu(y) $$ Plugging these into the partial differential equation in Step 1, we obtain: $$ N(t, y) (0) - M(t, y) q(y) \mu(y) = \left[M_{y}(t, y)-N_{t}(t, y)\right] \mu(y) $$ This proves that \(\mu(y) = e^{Q(y)}\) satisfies the partial differential equation in Step 1 when \(\mu\) is a function of \(y\) only.

Key concepts

Partial Differential Equations

Partial Differential Equations (PDEs) are an important concept in mathematics where functions have more than one independent variable. They involve partial derivatives, which describe how a function changes with respect to one variable while keeping others constant. In this exercise, we see PDEs in action while finding conditions for a differential equation to become exact with a multiplicative function \( \mu(t, y) \). This essentially means adjusting the equation so its solutions can be more easily determined. Understanding PDEs helps us solve complex problems involving multiple variables, such as fluid dynamics or financial modeling. In this case, identifying the partial derivatives \( \mu_t \) and \( \mu_y \) is crucial to forming the condition required for exactness.

Integrating Factors

Integrating Factors serve as a bridge to make non-exact differential equations exact. A non-exact equation is one that doesn't directly satisfy the conditions required to find a solution easily. The function \( \mu(t, y) \) in our exercise is an integrating factor meant to multiply with the original equation to achieve exactness. It's akin to finding the missing piece that makes all the parts fit perfectly. The process involves finding \( \mu \) such that it converts the differential equation into one that can be solved using integration. The form of \( \mu \) can sometimes be a function of a single variable, simplifying the task significantly as shown in the step-by-step solutions. By converting these equations with integrating factors, they can be greatly simplified and solved much easier.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) involve functions of a single independent variable, unlike PDEs. They are equations containing the derivatives of this function. The exercise transforms a PDE into an ODE in special cases, where \( \mu \) depends on just one variable, either \( t \) or \( y \). This simplification means it can be solved with tools we apply to ODEs. When \( \mu \) is only a function of \( t \) or \( y \), the problem reduces to this more straightforward form, making it easier to find an explicit solution. If the derivative terms align in a special way to create such a function, then the originally complex math becomes a simple process of integrating in terms of the single variable.

Exactness Condition

The concept of the Exactness Condition is crucial in whether we can solve a differential equation straightforwardly. For a differential equation to be exact, it must satisfy the condition involving its partial derivatives. In the exercise, the exactness condition is that \((\mu N)_y = (\mu M)_t\), which means if we find the right \( \mu \), this relationship holds. Exact equations are generally easier to solve because they are balanced in a way that aligns with fundamental mathematical properties. Making a non-exact equation exact typically involves calculating the right integrating factor \( \mu \) that matches this condition. This transforms the problem, making it much more elegant and opens the path for direct integration to find solutions.