Question

In problems, determine whether the given differential equation is exact. If it is exact, solve it.

$(2x-1)dx+(3y+7)dy=0$

Short answer

The given differential equation is exact and the solution is $x^2-x+\frac{3}{2}{y}^2+7y=C$.

Step-by-step solution

Given Information.

The given equation is$(2x-1)dx+(3y+7)dy=0$. An equation of the form$Mdx+Ndy=f\left(\text{x}\right)$ is exact only if it satisfies$\frac{\partial \text{M}}{\partial \text{x}}=\frac{\partial \text{N}}{\partial \text{y}}$ .

Determining the exactness of the differential equation

Find M and N using the given equation.

$$\begin{gathered} M(x,y)=2x-1 \\ N(x,y)=3y+7 \end{gathered}$$

Find $\frac{\partial \text{M}}{\partial \text{x}}$and $\frac{\partial \text{N}}{\partial \text{y}}$

$$\begin{gathered} \frac{\partial M}{\partial y}=0 \\ \frac{\partial N}{\partial x}=0 \end{gathered}$$

As$\frac{\partial \text{M}}{\partial \text{x}}=\frac{\partial \text{N}}{\partial \text{y}}$, so, the equation is exact.

Find the solution of differential equation

The solution of the differential equation is a function$f\left(x,y\right)$ such that $\frac{\partial f}{\partial x}=\text{M}$and$\frac{\partial f}{\partial \text{y}}=\text{N}$ . Here data-custom-editor="chemistry" $\frac{\partial f}{\partial x}=2x-1$and$\frac{\partial f}{\partial y}=3x+7$ .After integrating the first of these equations, we get$f(x,y)=x^2-x+g\left(y\right)$

$N(x,y)$is obtained by taking the partial derivative of the previous expression with respect to y and setting the result to$N(x,y)$ ,

$$\begin{gathered} \frac{\partial f}{\partial y}=g\text{'}\left(y\right) \\ =3y+7 \end{gathered}$$

On comparing these two equations, we get

$$\begin{gathered} g\text{'}\left(y\right)=3y+7 \\ g\left(y\right)=\frac{3}{2}{y}^2+7y \end{gathered}$$

Therefore, function $f(x,y)$becomes$f(x,y)=x^2-x+\frac{3}{2}{y}^2+7y$ , As a result, the implicit solution of the problem is$x^2-x+\frac{3}{2}{y}^2+7y=C$