Solving Exact Differential Equations: (1 + e^(x/y))dx + e^(x/y)(1 - x/y)dy = 0: 19 MCQs with answers

$\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$

The exactness test for $M\,dx + N\,dy = 0$ is $\partial M/\partial y = \partial N/\partial x$. The other options mix up which variable each function is differentiated with respect to.

$1 + e^{x/y}$

In the form $M\,dx + N\,dy = 0$, $M$ is the coefficient of $dx$, which is $1 + e^{x/y}$.

$e^{x/y}(1 - x/y)$

$N$ is the coefficient of $dy$, which is $e^{x/y}(1 - x/y)$.

$-\dfrac{x}{y^2}e^{x/y}$

Differentiating $e^{x/y}$ with respect to $y$ gives $e^{x/y} \cdot (-x/y^2) = -\dfrac{x}{y^2}e^{x/y}$. The constant $1$ differentiates to $0$.

The equation is exact.

Because $\partial M/\partial y = \partial N/\partial x$, the exactness test passes and the equation is exact.

$\dfrac{\partial F}{\partial x} = M$ and $\dfrac{\partial F}{\partial y} = N$

By definition of an exact equation, $F$ satisfies $\partial F/\partial x = M$ and $\partial F/\partial y = N$.

$x + y\,e^{x/y} = C$

The potential function is $F = x + y\,e^{x/y}$ (with $f(y)$ absorbed into $C$), so the solution is $x + y\,e^{x/y} = C$.

True

A differential equation $M\,dx + N\,dy = 0$ is exact if and only if $\partial M/\partial y = \partial N/\partial x$, and its solution is $F(x,y) = C$ where $\partial F/\partial x = M$ and $\partial F/\partial y = N$.

$-\dfrac{x}{y^2}e^{x/y}$

Using the product rule: $e^{x/y}\cdot\frac{1}{y}(1 - x/y) + e^{x/y}\cdot(-\frac{1}{y}) = e^{x/y}[\frac{1}{y} - \frac{x}{y^2} - \frac{1}{y}] = -\dfrac{x}{y^2}e^{x/y}$. The $1/y$ terms cancel.

$x + y\,e^{x/y} + f(y)$

The antiderivative of $e^{x/y}$ with respect to $x$ is $y\,e^{x/y}$ because $\partial_x(y e^{x/y}) = e^{x/y}$. So the integral is $x + y\,e^{x/y} + f(y)$.

Differentiate $F$ with respect to $y$ and set it equal to $N$.

To find $f(y)$, use the second condition $\partial F/\partial y = N$. Differentiating $F$ with respect to $y$ and matching $N$ determines $f'(y)$.

$f'(y) = 0$

Since $N = e^{x/y}(1 - x/y)$, comparing $e^{x/y}(1 - x/y) + f'(y) = e^{x/y}(1 - x/y)$ gives $f'(y) = 0$.

It is fine; $f(y)$ is a constant that merges into $C$.

When $f'(y) = 0$, the unknown function $f(y)$ is simply a constant. That constant is absorbed into the final constant $C$, so no error has occurred.

Show $\partial F/\partial x = M$ and $\partial F/\partial y = N$.

Verification means differentiating $F$ back to recover the original coefficients: $\partial F/\partial x$ should equal $M$ and $\partial F/\partial y$ should equal $N$.

a vector field is conservative

The condition $\partial M/\partial y = \partial N/\partial x$ is exactly the criterion for a vector field $(M, N)$ to be conservative, i.e. to have a potential function.

State functions such as internal energy and entropy

Exact differentials describe state functions like internal energy and entropy, whose values depend only on the state, not the path taken. Heat and work are path-dependent and are not exact differentials.

Integrate $M$ with respect to $x$

The correct first step is to integrate $M$ with respect to $x$, which gives part of the potential function $F(x,y)$. This is essential for finding the solution to the exact differential equation.

Differentiate $F$ with respect to $y$ and set it equal to $N$

The next step is to differentiate $F$ with respect to $y$ and set it equal to $N$. This allows you to find the function $f(y)$ by matching the derivatives.

It means $f(y)$ is a constant function.

Finding $f'(y) = 0$ implies that $f(y)$ is a constant function. This is a normal outcome in the context of exact differential equations, indicating no additional $y$-dependent terms are present.