Expert answers to "Find the solution of the following Differential Equations..."

reproacht3

Answered question

2021-12-31

Find the solution of the following Differential Equations

(2 x + e^{y}) d x + x e^{y} d y = 0

Answer & Explanation

nghodlokl

Beginner2022-01-01Added 33 answers

Step 1

(2 x + e^{y}) d x + x e^{y} d y = 0

\Rightarrow 2 x d x + e^{y} d x + x e^{y} d y = 0

\Rightarrow 2 x d x + d (x e^{y}) = 0

∴ d (f \cdot g) = f d (g) + g d (f)

Step 2

\Rightarrow \int 2 x d x + \int d (x e^{y}) = 0

\Rightarrow 2 \frac{x^{2}}{2} + x e^{y} = c

\Rightarrow x^{2} + x e^{y} = c

Pademagk71

Beginner2022-01-02Added 34 answers

Solution:

(2 x + e^{y}) d x + x e^{y} d y = 0, y (1) = 0

Compare above with

M d x + N d y = 0

M = 2 x + e^{y}, N = x e^{y}

\frac{\partial M}{\partial y} = e^{y}, \frac{\partial N}{\partial x} = e^{y}

Since

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

, therefore differential equation is exact

\frac{\partial f}{\partial x} = 2 x + e^{y}

(i)

\frac{\partial f}{\partial y} = x e^{y}

(ii)
integrate (i) wrt x, we get

f (x, y) = \int (2 x + e^{y}) d x

f (x, y) = x^{2} + x e^{y} + h (y)

Differentiate above wrt y, we get

\frac{\partial f}{\partial y} = x e^{y} + h^{'} (y)

compare above and (ii), we get

x e^{y} + h^{'} (y) = x e^{y}

h^{'} (y) = 0

integrate both sides wrt y, we get

h (y) = C

therefore, we get

f (x, y) = x^{2} + x e^{y} + C

or we can write as

x^{2} + x e^{y} + C = 0

karton

Expert2022-01-09Added 613 answers

Consider the differential equation
$(2 x + e^{y}) d x + x e^{y} d y = 0$
Compare it with $M (x, y) d x + N (x, y) d y = 0$ , get
$M (x, y) = 2 x + e^{y} and N (x, y) = x e^{y}$
$\frac{\partial M}{\partial y} = e^{y}, \frac{\partial N}{\partial x} = e^{y}$
Observe that, $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
So, the given differential equation is in exact form
$(2 x + e^{y}) d x + x e^{y} d y = 0$
$2 x d x + e^{y} d x + x e^{y} d y = 0$
$2 x d x + d (x e^{y}) = 0$
on taking integration get
$x^{2} + x e^{y} = C$
which is the required solution.

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