Expert answers to "Obtain the solution on the following differential equations:..."

Pam Stokes

Answered question

2021-12-18

Obtain the solution on the following differential equations:

{(1 - x y)}^{- 2} d x + [y^{2} + x^{2} {(1 - x y)}^{- 2}] d y = 0

Piosellisf

Beginner2021-12-19Added 40 answers

Step 1
The differential equation

M d x + N d y = 0

is said to be exact. If

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

.
So, the solution of the differential equation is:

\int M d x + \int (Term in "N" free from x) d x = C

Given:

{(1 - x y)}^{- 2} d x + [y^{2} + x^{2} {(1 - x y)}^{- 2}] d y = 0

So,

M = {(1 - x y)}^{- 2}, N = [y^{2} + x^{2} {(1 - x y)}^{- 2}] d y = 0

\frac{\partial M}{\partial y} = \frac{\partial}{\partial y} ({(1 - x y)}^{- 2}) = - \frac{2}{{(1 - x y)}^{3}} \frac{\partial a}{\partial y} (1 - x y) = \frac{2 x}{{(1 - x y)}^{3}}

\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} (y^{2} + x^{2} {(1 - x y)}^{- 2}) = \frac{\partial}{\partial x} (y^{2}) + \frac{\partial}{\partial x} (x^{2} {(1 - x y)}^{- 2}) = \frac{2 x}{{(- y x + 1)}^{3}}

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

... (Exact)
So, the given differential equation is exact.
Step 2
Now find the solution.

\int {(1 - x y)}^{- 2} d x + \int (y^{2}) d x = C

\int - \frac{1}{y u^{2}} d u + y^{2} \int d x = C \dots [L e t, u = 1 - y x; d x = - \frac{1}{y} d u]

- \frac{1}{y} \cdot \int \frac{1}{u^{2}} d u + y^{2} (x) = C

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eninsala06

Beginner2021-12-20Added 37 answers

Solve

\frac{1}{{(- x y (x) + 1)}^{2}} + \frac{d y (x)}{d x} (\frac{x^{2}}{{(- x y (x) + 1)}^{2}} + y {(x)}^{2}) = 0

:
Let

P (x, y) = \frac{1}{{(- x y + 1)}^{2}} and Q (x, y) = y^{2} + \frac{x^{2}}{{(- x y + 1)}^{2}}

.
This is an exact equation, because

\frac{d P (x, y)}{d y} = \frac{2 x}{{(- x y + 1)}^{3}} = \frac{d Q (x, y)}{d x}

.
Define f(x, y) such that

\frac{d f (x, y)}{d x} = P (x, y) and \frac{d f (x, y)}{d y} = Q (x, y)

.
Then, the solution will be given by

f (x, y) = c_{1}

, where

c_{1}

is an arbitrary constant.
Integrate

\frac{d f (x, y)}{d x}

with respect to x in order to find f(x, y):

f (x, y) = \int e g r a l \frac{1}{{(- y x + 1)}^{2}} d x = - \frac{1}{y (x y - 1)} + g (y)

where g(y) is an arbitrary function of y.
Differentiate f(x, y) with respect to y in order to find g(y):

\frac{d f (x, y)}{d y} = \frac{d}{d y} (- \frac{1}{y (y x - 1)} + g (y)) = \frac{x}{y {(x y - 1)}^{2}} + \frac{1}{y^{2} (x y - 1)} + \frac{d g (y)}{d y}

Substitute into

\frac{d f (x, y)}{d y} = Q (x, y)

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