Integrating Factor

Exercise 1.5

1. Find the general solution.

(a) $\ (x - y^{2})dx + 2xydy = 0$
(b) $\ (5xy + 4y^{2} + 1)dx + (x^{2} + 2xy)dy = 0$
(c) $\ (2xy^{2} + y)dx + (2y^{3} - x)dy = 0$
(d) $\ (2x + \tan{y})dx + (x - x^{2}\tan{y})dy = 0$

2. Solve the following initial value problem.

(a) $\ (2y^{2} - xy)dx + (2x^{2} - 3xy)dy = 0$
(b) $\ (y^{4} + 2x^{3}y)dx - (x^{4} + 2xy^{3})dy = 0$

Answer 1. (a) Note that $M_{y} = -2y, \ N_{x} = 2y$. Then it is not exact differential equation. So, we seek an integrating factor. Since $M_{y} - N_{x} = - 4y$, calculate $(1/N)[M_{y} - N_{x}]$. Then

$$\frac{1}{N}(M_{y} - N_{x}) = \frac{-4y}{2xy} = \frac{-2}{x}$$

This is a function of $x$ only. Thus, the integrating factor is given by

$$\mu = \exp(- \int \frac{2}{x} dx) = \exp(-2 \log{x}) = \frac{1}{x^{2}}$$

Multiply this integrating factor to the original equation. Then

$$(\frac{1}{x} - \frac{y^2}{x^2}) dx + \frac{2y}{x} dy = 0$$

This has to be exact. So, using the grouping method, we have

$$\frac{1}{x} dx + (- \frac{y^2}{x^2} dx + \frac{2y}{x} dy) = 0$$

Rewrite this using total differentials,

$$d(\log{\vert x\vert}) + d(\frac{y^2}{x}) = d(c)$$

Therefore,

$$\log{\vert x\vert} + \frac{y^2}{x} = c \ \ \$$

(b) Note that $M_{y} = 5x + 8y, \ N_{x} = 2x + 2y$. Thus it is not exact. So, we seek an integrating factor. Since $M_{y} - N_{x} = 3x + 6y$, calculate $(1/N)[M_{y} - N_{x}]$. Then we have

$$\frac{1}{N}(M_{y} - N_{x}) = \frac{3x + 6y}{x^2 + xy} = \frac{3(x+2y)}{x(x+2y)} = \frac{3}{x}$$

This is a function $x$ only. So, the integrating factor is given by

$$\mu = \exp(\int \frac{3}{x} dx) = \exp(3 \log{x}) = x^{3}$$

Multiply $\mu$ to the original euqation, w have

$$(5x^4 y + 4x^3 y^2 + x^3)dx + (x^5 + 2x^4 y ) dy = 0$$

This equation is exact. So, using the grouping method, we obtain

$$x^3 dx + [(5x^4 y + 4x^3 y^2) dx + (x^5 + 2x^4 y ) dy] = 0$$

Rewrite this using the total differentials, we have

$$d(\frac{x^4}{4}) + d(x^5 y + 2x^{4}y^{2}) = d(c)$$

Therefore, the general solution is

$$\frac{x^4}{4} + x^5 y + x^4 y^2 = c \ \ \$$

(c) Note that $M_{y} = 4xy + 1, N_{x} = -1$. Then it is not exact. So, we look for an integrating factor. Since $M_{y} - N_{x} = 4xy + 2$, we calculate $(1/M)[M_{y} - N_{x}]$. Then

$$\frac{1}{N}(M_{y} - N_{x}) = \frac{4xy + 2}{2xy^2 + y} = \frac{2(2xy + 1)}{y(2xy + 1)} = \frac{2}{y}$$

This is a function of $y$ only. Thus the integrating factor is given by

$$\mu = \exp(- \int \frac{2}{y} dy) = \exp(-2 \log{y}) = \frac{1}{y^2}$$

Multiply $\mu$ to the original equation.

$$(2x + \frac{1}{y})dx + (2y - \frac{x}{y^2})dy = 0$$

This equation is exact. So, using the grouping method, we have

$$2x dx + (\frac{1}{y}dx - \frac{x}{y^2}dy) + 2y dy = 0$$

Rewrite this using the total differentials,

$$d(x^2) + d(\frac{x}{y}) + d(y^2) = d(c)$$

Therefore, the general solution is given by

$$x^2 + \frac{x}{y} + y^2 = c \ \ \$$

(d) Note that $M_{y} = \sec^{2}{y}, \ N_{x} = 1 - 2x \tan{y}$. Then it is not exact. So, we look for an integrating factor.

$$M_{y} - N_{x} = \sec^{2}{y} - 1 + 2x \tan{y}$$

$$= \tan^{2}{y} + 2x \tan{y} = \tan{y}(\tan{y} + 2x)$$

Then calculate $(1/M)[M_{y} - N_{x}]$, we have

$$\frac{1}{M}(M_{y} - N_{x}) = \frac{\tan{y}(\tan{y} + 2x)}{2x + \tan{y}} = \tan{y}$$

This is a function of $y$ only. Thus the integrating factor is given by

$$\mu = \exp(- \int \tan{y} dy) = \exp(- \int \frac{\sin{y}}{\cos{y}} dy) = \exp( \log{\vert\cos{y}\vert}) = \cos{y}$$

Multiply $\mu$ to the original equation.

$$(2x \cos{y} + \sin{y})dx + (x \cos{x} - x^2 \sin{y}) dy = 0$$

This equation is exact. So, using the total differentials, we can write

$$d(x^2 \cos{y} + x \sin{y}) = d(c)$$

Therefore, the general solution is given by

$$x^2 \cos{y} + x \sin{y} = c \ \ \$$

2. (a) Note that $M_{y} = 4y - x, \ N_{x} = 4x - 3y$. Then this is not exact. So, we look for an integrating factor.

$$M_{y} - N_{x} = -5x + 7y$$

Then $(1/N)[M_{y} - N_{x}], (1/M)[M_{y} - N_{x}]$ are not function of $x, y$ only. So, we let the integrating factor $\mu = x^{m}y^{n}$. Then we find $m,n$.

$$(2x^{m}y^{n+2} - x^{m+1}y^{n+1})dx + (2x^{m+2}y^{n} - 3x^{m+1}y^{n+1})dy = 0$$

This equation is exact if and only if

$$(\mu M)_{y} = 2(n+2)x^{m}y^{n+1} - (n+1)x^{m+1}y^{n}$$

$$(\mu N)_{x} = 2(m+2)x^{m+1}y^{n} - 3(m+1)x^{m}y^{n+1}$$

Thus,

$$2(n+2) = -3(m+1), \ 2(m+2) = - (n + 1)$$

or

$$\begin{array}{ll} 3m + 2n &= -7\\ 2m + n &= - 5 \end{array}$$

Solv this using Cramer's rule, we have

$$m = \frac{\left \vert \begin{array}{cc} -7 & 2\\ -5 & 1 \end{... ...gin{array}{cc} 3 & 2\\ 2 & 1 \end{array} \right \vert} = \frac{3}{-1} = -3$$

Also, $n = 1$. Thus, $\mu = x^{-3}y$. Now multiply $\mu$ to the original equation, we have

$$(2x^{-3}y^{3} - x^{-2}y^{2})dx + (2x^{-1}y - 3x^{-2}y^{2})dy = 0$$

This equation is exact. Thus by using total differentials, we can write as follows:

$$d(x^{-1}y^{2} - x^{-2}y^{3}) = d(c)$$

Therefore, the general solution is

$$x^{-1}y^{2} - x^{-2}y^{3} = c \ \ \$$

(b) Note that $M_{y} = 4y^3 + 2x^3, \ N_{x} = -4x^3 - 2y^{3}$. Thus it is not exact. So, we look for an integrating factor. By the form of $M_{y},N_{x}$, we assume that $\mu = x^{m}y^{n}$. Then we find $m,n$.

$$(x^{m}y^{n+4} + 2x^{m+3}y^{n+1})dx - (x^{m+4}y^{n} + 2x^{m+1}y^{n+3})dy = 0$$

This equation is exact if and only if

$$(\mu M)_{y} = (n+4)x^{m}y^{n+3} + 2(n+1)x^{m+3}y^{n}$$

$$(\mu N)_{x} = -(m+4)x^{m+3}y^{n} - 2(m+1)x^{m}y^{n+3}$$

Thus,

$$(n+4) = -2(m+1), \ -(m+4) = 2(n + 1)$$

writing in the system of linear equations,

$$\begin{array}{ll} 2m + n &= -6\\ -m - 2n &= 6 \end{array}$$

Solve this system using Cramer's rule, we have

$$m = \frac{\left \vert \begin{array}{cc} -6 & 1\\ 6 & -2 \end{... ...n{array}{cc} 2 & 1\\ -1 & -2 \end{array} \right \vert} = \frac{6}{-3} = -2$$

and $n = -2$. Thus, $\mu = x^{-2}y^{-2}$. Multiply $\mu$ to the original equation. Then we have

$$(x^{-2}y^{2} + 2xy^{-1})dx - (x^{2}y^{-2} + 2x^{-1}y)dy = 0$$

This equation is exact. So, rewrite this using the total differentials, we have

$$d(-x^{-1}y^{2} + x^{2}y^{-1}) = d(c)$$

Therefore, the general solution is

$$-x^{-1}y^{2} + x^{2}y^{-1} = c \ \ \$$