Agohofidov6

2021-12-27

Classify the following differential equations as separable, homogeneous, parallel line, or exact. Explain briefly your answers. Then, solve each equation according to their classification. $\left(2x-5y+1\right)dx-\left(5y-2x\right)dy=0$

Laura Worden

We have:
$\left(2x-5y+1\right)dx-\left(5y-2x\right)dy=0$
Or, $\left(2x-5y+1\right)dx+\left(2x-5y\right)dy=0$
Clearly the differential equation is of parallel line.
Now, $\frac{dy}{dx}=-\frac{2x-5y+1}{2x-5y}$
Put $2x-5y=v$
Or, $2-5\frac{dy}{dx}=\frac{dv}{dx}$
$\frac{dy}{dx}=\frac{2}{5}-\frac{1}{5}\cdot \frac{dv}{dx}$
$-\frac{v+1}{v}=\frac{2}{5}-\frac{dv}{5dx}$
$\frac{1}{5}\cdot \frac{dv}{dx}=\frac{2}{5}+\frac{v+1}{v}$
$\frac{1}{5}\cdot \frac{dv}{dx}=\frac{2v+5v+5}{5v}$
$\frac{dv}{dx}=\frac{7v+5}{v}$
$\frac{v}{7v+5}dv=dx$
$\int \frac{v}{7v+5}dv=\int dx$
$\frac{1}{7}\int \frac{7v+5-5}{7v+5}dv=\int dx$
$\frac{1}{7}\int \left(1-\frac{5}{7v+5}\right)dv=x+C$
$\frac{1}{7}\left(v-\frac{5}{7}\mathrm{ln}\left(7v+5\right)\right)=x+C$

Annie Levasseur

Step 1
The given differential equation is:
$\left(2x-5y+1\right)dx-\left(5y-2x\right)dy=0$
Step 2
Now compare the differential equation with: $Mdx+Ndy=0$

Now:
$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.
Therefore the given differential equation is Exact.
Now consider the solution of the differential equation is: $F\left(x,y\right)=0$
Therefore:
Consider: $\frac{\partial F}{\partial x}=2x-5y+1$
$F\left(x,y\right)=\int \left(2x-5y+1\right)dx+\varphi \left(y\right)$
$F\left(x,y\right)={x}^{2}-5xy+x+\varphi \left(y\right)$ (1)
$\frac{\partial F}{\partial y}=-5x+{\varphi }^{\prime }\left(y\right)$
$-5x+2y=-5x+{\varphi }^{\prime }\left(y\right)$
${\varphi }^{\prime }\left(y\right)=2y$
$\varphi \left(y\right)={y}^{2}+C$ where C is a constant.
Therefore: $F\left(x,y\right)={x}^{2}-5xy+x+{y}^{2}+C$
Hence the solution of the given differential equation is:
${x}^{2}-5xy+x+{y}^{2}+C=0$

karton

$\begin{array}{}\left(2x-5y+1\right)dx-\left(5x-2y\right)dy=0\\ \left(2x-5y+1\right)dx+\left(-5x+2y\right)dy=0\\ Mdx+Ndy=0\\ M=2x-5y+1,N=-5x+2y\\ \frac{\partial M}{\partial y}=-5,\frac{\partial N}{\partial x}=-5\\ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\end{array}$
Given diff equation is exact
Its general solution is
$\begin{array}{}\\ \int Mdx+\int Ndy=C\\ \int \left(2x-5y+1\right)dx+\int 2ydy=c\\ \frac{2{x}^{2}}{2}-5xy+x+\frac{2{y}^{2}}{2}=C\\ {x}^{2}+{y}^{2}-5xy+x=c\end{array}$
C: arbitraty constant

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