"(y−xy2)dx+(x+x2y2)dy=0;when y(1)=1" was answered by our experts

Name: (y−xy2)dx+(x+x2y2)dy=0;when y(1)=1
Uploaded: 2022-02-23T17:22:57+00:00
Description: (y−xy2)dx+(x+x2y2)dy=0;when y(1)=1

Priscilla Johnston

Answered question

2022-01-06

(y - x y^{2}) d x + (x + x^{2} y^{2}) d y = 0; when y (1) = 1

movingsupplyw1

Beginner2022-01-07Added 30 answers

Solution:

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

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

d (x y) - {(x y)}^{2} [\frac{d x}{x} - d y] = 0

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

\Rightarrow - \frac{1}{x \cdot y} - \ln x + y = C

put

4 (1) = 1

- \frac{1}{1 \cdot 1} - \ln 1 + 1 = C \Rightarrow - 1 - 0 + 1 = C

c = 0

Hence,

- \frac{1}{x \cdot y} \cdot \ln x + y = 0

SlabydouluS62

Skilled2022-01-08Added 52 answers

Consider the differential equation

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

y d x - x y^{2} d x + x d y + x^{2} y^{2} d y = 0

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

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

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

Dividing by

x^{2} y^{2}

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

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

Integrating the equation above we get

\int [d \frac{x y}{{(x y)}^{2}}] + \int d y - \int \frac{d x}{x} = C

…..(1)
Where C is constant of integration
Now say

\int \frac{d z}{z^{2}} = - \frac{1}{z}

\int x^{n} d x = x^{n} + \frac{1}{n + 1} + c for n \neq - 1

\int \frac{d x}{x} = \ln x + c

Using these identities if

z = x y

then the above eqn.(1) becomes

- \frac{1}{x y} + y - \ln x = C

Multiply throughout by xy in the above equation we get

- 1 + x y^{2} - x y \ln x = C x y

Therefore the solution of the equation is

x y^{2} - x y \ln x - 1 = C x y

.
Where C is any arbitrary constant.

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