Question

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

$\left(y^3+kx{y}^4-2x\right)\mathit{d}\mathit{x}\mathbf{+}\left(3x{y}^2+20x^2{y}^3\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The value of $k$ is 10.

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

$$\begin{gathered} M(x,y)=y^3+kx{y}^4-2x \\ N(x,y)=3x{y}^2+20x^2{y}^3 \end{gathered}$$

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

$\frac{\partial M}{\partial y}=3y^2+4kx{y}^{3}and\frac{\partial N}{\partial x}=3y^2+40x{y}^3$

The given differential equation is exact when

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

Then, we get

$3y^2+4kx{y}^3=3y^2+40x{y}^3$

From the obtained equation we obtain the solution $k=10$.