 hvacwk

2021-12-29

Determine whether the following differential equations are exact or not exact. if exact, solve for the general solution
$\left(x+2y\right)dx-\left(2x+y\right)dy=0$ turtletalk75

Expert

Step 1
$\left(x+2y\right)dx-\left(2x+y\right)dy=0$
The differential equation $Mdx+Ndy=0$ is a exact if
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$
Step 2
From the given differential equation
$M=x+2y$
$N=-\left(2x+y\right)$
$N=-\left(2x+y\right)$
$\frac{\partial M}{\partial y}=2$
$\frac{\partial N}{\partial x}=-2$
Since, $\frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}$
The given differential equation is not exact differential equation. Buck Henry

Expert

Simplifying
$\left(x+2y\right)\cdot dx+-1\left(2x+-1y\right)\cdot dy=0$
Reorder the terms for easier multiplication:
$dx\left(x+2y\right)+-1\left(2x+-1y\right)\cdot dy=0$
$\left(x\cdot dx+2y\cdot dx\right)+-1\left(2x+-1y\right)\cdot dy=0$
Reorder the terms:
$\left(2dxy+{dx}^{2}\right)+-1\left(2x+-1y\right)\cdot dy=0$
$\left(2dxy+{dx}^{2}\right)+-1\left(2x+-1y\right)\cdot dy=0$
Reorder the terms for easier multiplication:
$2dxy+{dx}^{2}+-1dy\left(2x+-1y\right)=0$
$2dxy+{dx}^{2}+\left(2x\cdot -1dy+-1y\cdot -1dy\right)=0$
$2dxy+{dx}^{2}+\left(-2dxy+1{dy}^{2}\right)=0$
Reorder the terms:
$2dxy+-2dxy+{dx}^{2}+1{dy}^{2}=0$
Combine like terms: $2dxy+-2dxy=0$
$0+{dx}^{2}+1{dy}^{2}=0$
${dx}^{2}+1{dy}^{2}=0$
Solving
${dx}^{2}+1{dy}^{2}=0$
Solving for variable d.
Move all terms containing d to the left, all other terms to the right.
Factor out the Greatest Common Factor (GCF), d.
$d\left({x}^{2}+{y}^{2}\right)=0$ karton

Expert