Answer to Question #181270 in Differential Equations for randz

Question

Give the general solution using determination of

Integrating Factors.

1. (x^2 + y^2 + 1)dx + x(x − 2y)dy = 0

2. y(4x + y)dx − 2((x^2 − y)dy = 0

3. y(y + 2x − 2)dx − 2(x + y)dy = 0

4. y(8x − 9y)dx − 2x(x − 3y)dy = 0

5. y(2x^2 − xy)dx − (x − y)dy = 0

6. 2(2y^2 + 5xy − 2y + 4)dx + x(2x + 2y − 1)dy = 0

7. (2x^2 + 3xy − 2y + 6x)dx + x(x + 2y − 1)dy = 0

Answer 1

1. $(x^2 + y^2 + 1)dx + x(x − 2y)dy = 0$

Integrating factor-

$I.F.=\dfrac{1}{Mdx+Ndy}=\dfrac{1}{\frac{x^3}{3}+xy^2+x}$

Given equation is of the form Mdx+Ndy, Integrate M w.r.t. x and integrate only those term which are

free from x-

The solution of the above equation is-

$=\dfrac{x^3}{3}+xy^2+x$

2. $y(4x + y)dx − 2(x^2 − y)dy = 0$

Integrating factor-

$I.F.=\dfrac{1}{Mdx+Ndy}=\dfrac{1}{2yx^2+xy^2+y^2}$

Given equation is of the form Mdx+Ndy, Integrate M w.r.t. x and integrate only those term which are

free from x-

The solution of the above equation is-

$=2yx^2+xy+y^2$

3. $y(y + 2x − 2)dx − 2(x + y)dy = 0$

Integrating factor-

$I.F.=\dfrac{1}{Mdx+Ndy}=\dfrac{1}{x^2+xy^2-2xy-y^2}$

Given equation is of the form Mdx+Ndy, Integrate M w.r.t. x and integrate only those term which are

free from x-

The solution of the above equation is-

$=x^2+xy^2-2xy-y^2$

4. $y(8x − 9y)dx − 2x(x − 3y)dy = 0$ 4

Integrating factor-

$I.F.=\dfrac{1}{Mdx+Ndy}=\dfrac{1}{4x^2y-9y^2x+3xy^2}$

Given equation is of the form Mdx+Ndy, Integrate M w.r.t. x and integrate only those term which are

free from x-

The solution of the above equation is-

$=4x^2y-9xy^2+3xy^2$

5. $y(2x^2 − xy)dx − (x − y)dy = 0$

Integrating factor-

$I.F.=\dfrac{1}{Mdx+Ndy}=\dfrac{1}{\frac{2x^3y}{3}-\frac{y^2x^2}{2}+\frac{y^2}{2}}$

Given equation is of the form Mdx+Ndy, Integrate M w.r.t. x and integrate only those term which are

free from x-

The solution of the above equation is-

$=\dfrac{2x^3y}{3}-\dfrac{y^2x^2}{2}+\dfrac{y^2}{2}$

6. $2(2y^2 + 5xy − 2y + 4)dx + x(2x + 2y − 1)dy = 0$

Integrating factor-

$I.F.=\dfrac{1}{Mdx+Ndy}=\dfrac{1}{4y^2x+5x^2y-4yx+8x}$

Given equation is of the form Mdx+Ndy, Integrate M w.r.t. x and integrate only those term which are

free from x-

The solution of the above equation is-

$=4y^2x+5x^2y-4yx+8x$

7. $(2x^2 + 3xy − 2y + 6x)dx + x(x + 2y − 1)dy = 0$

Integrating factor-

$I.F.=\dfrac{1}{Mdx+Ndy}=\dfrac{1}{\frac{2x^3}{3}+\frac{3x^2y}{2}-2yx}$

Given equation is of the form Mdx+Ndy, Integrate M w.r.t. x and integrate only those term which are

free from x-

The solution of the above equation is-

$=\dfrac{2x^3}{3}+\dfrac{3x^2y}{2}-2y$