Question

In Problems, 1-20 determine whether the given differential equation is exact. If it is exact, solve it.

$\left(5y-2x\right)\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The solution is $-2xy+\frac{5}{2}{y}^2=c$

Step-by-step solution

Given information

The given equation is $\left(5y-2x\right)y\text{'}-2y=0$.

The given equation can be written as,

$\begin{array}{c}\left(5y-2x\right)\frac{dy}{dx}-2y=0\\ \left(5y-2x\right)dy-2ydx=0\end{array}$

.

Compare the given equation with

$Mdx+Ndy=0$

$M(x,y)=-2y$

$N(x,y)=5y-2x$

Condition for exactness

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

Determining the exactness of the differential equation

Checking whether the differential equation is exact by finding $\frac{\partial M}{\partial y},\frac{\partial N}{\partial x}$.

$\begin{array}{l}\frac{\partial M}{\partial y}=-2\\ \frac{\partial N}{\partial y}=-2\end{array}$

The differential equation is exact, because $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.

Next, we have to find the solution of the given equation. Where$\frac{\partial f}{\partial y}=-2x+g\text{'}\left(y\right)$ .

Find the solution

Now, integrate $\frac{\partial f}{\partial y}=-2x+g\text{'}\left(y\right)$, for $g\text{'}\left(y\right)=5y$.

Where $N(x,y)=\frac{\partial f}{\partial y}$. Then

$-2xy+\frac{5}{2}{y}^2=c$

The solution is$-2xy+\frac{5}{2}{y}^2=c$