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. $(2x-5y+1)dx-(5y-2x)dy=0$

Laura Worden

Beginner2021-12-28Added 45 answers

We have:

$(2x-5y+1)dx-(5y-2x)dy=0$

Or,$(2x-5y+1)dx+(2x-5y)dy=0$

Clearly the differential equation is of parallel line.

Now,$\frac{dy}{dx}=-\frac{2x-5y+1}{2x-5y}$

Put$2x-5y=v$

Or,$2-5\frac{dy}{dx}=\frac{dv}{dx}$

$\frac{dy}{dx}=\frac{2}{5}-\frac{1}{5}\cdot \frac{dv}{dx}$

$-\frac{v+1}{v}=\frac{2}{5}-\frac{dv}{5dx}$

$\frac{1}{5}\cdot \frac{dv}{dx}=\frac{2}{5}+\frac{v+1}{v}$

$\frac{1}{5}\cdot \frac{dv}{dx}=\frac{2v+5v+5}{5v}$

$\frac{dv}{dx}=\frac{7v+5}{v}$

$\frac{v}{7v+5}dv=dx$

$\int \frac{v}{7v+5}dv=\int dx$

$\frac{1}{7}\int \frac{7v+5-5}{7v+5}dv=\int dx$

$\frac{1}{7}\int (1-\frac{5}{7v+5})dv=x+C$

$\frac{1}{7}(v-\frac{5}{7}\mathrm{ln}(7v+5))=x+C$

$div>$

Or,

Clearly the differential equation is of parallel line.

Now,

Put

Or,

0

Annie Levasseur

Beginner2021-12-29Added 30 answers

Step 1

The given differential equation is:

$(2x-5y+1)dx-(5y-2x)dy=0$

Step 2

Now compare the differential equation with:$Mdx+Ndy=0$

$M(x,y)=2x-5y+1\text{}\text{and}\text{}N(x,y)=-5x+2y$

Now:$\frac{\partial M}{\partial y}=-5\text{}\text{and}\text{}\frac{\partial N}{\partial x}=-5$

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

Therefore the given differential equation is Exact.

Now consider the solution of the differential equation is:$F(x,y)=0$

Therefore:$\frac{\partial F}{\partial x}=2x-5y+1\text{}\text{and}\text{}\frac{\partial F}{\partial y}=-5x+2y$

Consider:$\frac{\partial F}{\partial x}=2x-5y+1$

$F(x,y)=\int (2x-5y+1)dx+\varphi \left(y\right)$

$F(x,y)={x}^{2}-5xy+x+\varphi \left(y\right)$ (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:$F(x,y)={x}^{2}-5xy+x+{y}^{2}+C$

Hence the solution of the given differential equation is:

${x}^{2}-5xy+x+{y}^{2}+C=0$

The given differential equation is:

Step 2

Now compare the differential equation with:

Now:

Therefore the given differential equation is Exact.

Now consider the solution of the differential equation is:

Therefore:

Consider:

Therefore:

Hence the solution of the given differential equation is:

karton

Expert2022-01-09Added 613 answers

Given diff equation is exact

Its general solution is

C: arbitraty constant

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