"Find the solution of the following Differential Equations..." solved and explained

Tiffany Russell

Answered question

2021-12-31

Find the solution of the following Differential Equations by Exact

(2 y + x y) d x + 2 x d y = 0, y (3) = \sqrt{2}

Answer & Explanation

esfloravaou

Beginner2022-01-01Added 43 answers

Given:
$(2 x + x y) d x + 2 x d y = 0, y (3) = \sqrt{2}$
this equation is form $M (x, y) d x + N (x, y) d y = 0$
At this equation is exact if and only it
$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
$M = 2 y + x y \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} (2 y + x y) = 2 + x$
$N = 2 x \frac{\partial N}{\partial x} = 2$
Hence $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$
Now we find the integrating factor to made this equation exact.
$M y - N x = 2 + x - 2 = x$
$\frac{M y - N x}{N} = \frac{x}{2 x} = \frac{1}{2}$
Hence $I . F . = e^{\int \frac{1}{2} d x} = e^{\frac{x}{2}}$
$\Rightarrow e^{\frac{x}{2}} (2 y + x y) d x + 2 e^{\frac{x}{2}} x d y = 0$ (2)
$= 2 e^{2 / 2} y + x e^{x / 2} y$
$N = 2 e^{\frac{x}{2}} x$
$\frac{\partial M}{\partial y} = 2 e^{\frac{x}{2}} + x e^{\frac{x}{2}}$
$\frac{\partial N}{\partial x} = 2 e^{\frac{x}{2}} y + 2 x \cdot e^{\frac{x}{2}} \cdot \frac{1}{2} = 2 e^{\frac{x}{2}} + x e^{\frac{x}{2}}$
$\Rightarrow \frac{\partial M}{\partial y} = \frac{\partial M}{\partial x}$
equation (2) is exact equation:
Now solution of equation
$\int M d x + \int M d y = c$
$\int (2 y e^{\frac{x}{2}} + x y e^{\frac{x}{2}}) d x + \int d y = c$

eskalopit

Beginner2022-01-02Added 31 answers

(2 y + x y) d x + 2 x d y = 0

rearranging,

\frac{d y}{d x} = - \frac{2 y + x y}{2 x} = - \frac{2 + x}{2 x} . y = - g (x) . y

\frac{d y}{d x} + g (x) . y = 0

Integrating Factor

I F = \exp (\int g (x) d x)

\int g (x) d x = \int \frac{2 + x}{2 x} d x = \int {\frac{1}{x} + \frac{1}{2}} d x

\int g (x) d x = \ln (x) + \frac{x}{2}

Hence,

I F = \exp (\ln (x) + \frac{x}{2}) = x \cdot e^{\frac{x}{2}}

The DE now becomes,

d (I F \cdot y) = 0

d (x \cdot e^{\frac{x}{2}} . y) = 0

Integrating,

x \cdot e^{\frac{x}{2}} . y = c o n s t

Answer:

x y . e^{\frac{x}{2}} = c

karton

Expert2022-01-09Added 613 answers

\(\begin{array}{}(2y+xy)dx+2ydy=0 \\Rightarrow 2y+xy+2y \frac{dy}{dx}=0 \\text{Substitute}\ \frac{dy}{dx}\ \text{with y}

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