Question

In Problems 27 and 28 find the value of k so that the given differential equation is exact.

$\left(6x{y}^3+cosy\right)\mathit{d}\mathit{x}\mathbf{+}\left(2k{x}^2{y}^2-xsiny\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The value of k is$\frac{9}{2}$

Step-by-step solution

Condition for exactness

An equation of the form $Mdx+Ndy=0$, is said to be exact, if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

To Find the differential equation is exact

Using the labels from the question,

$$\begin{gathered} M(x,y)=6xy+cosy \\ N(x,y)=2k{x}^2{y}^2-xsiny \end{gathered}$$

Find$\frac{\partial M}{\partial y},\frac{\partial N}{\partial x}$

$$\begin{gathered} \frac{\partial M}{\partial y}=18x{y}^2-siny \\ \frac{\partial N}{\partial x}=4kx{y}^2-siny \end{gathered}$$

The given differential equation is exact when

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

Then we get

$18x{y}^2-\sin y=2k{x}^2{y}^2-x\sin y$

$\begin{array}{c}4k=18\\ k=\frac{18}{4}\\ =\frac{9}{2}\end{array}$

Hence, the required value is, .$k=\frac{9}{2}$