Variables Separable

Problem 05 | Substitution Suggested by the Equation

Problem 05
$dy/dx = \sin (x + y)$
 

Solution 05
$dy/dx = \sin (x + y)$

$dy = \sin (x + y)~dx$
 

Let
$z = x + y$

$dz = dx + dy$

$dy = dz - dx$
 

$dz - dx = \sin z~dx$

$dz = \sin z~dx + dx$

$dz = (\sin z + 1)~dx$

$\dfrac{dz}{\sin z + 1} = dx$

$\dfrac{dz}{1 + \sin z} \cdot \dfrac{1 - \sin z}{1 - \sin z} = dx$

$\dfrac{(1 - \sin z)~dz}{1 - \sin^2 z} = dx$

$\dfrac{(1 - \sin z)~dz}{\cos^2 z} = dx$

$\dfrac{~dz}{\cos^2 z} - \dfrac{\sin z~dz}{\cos^2 z} = dx$

Problem 03 | Substitution Suggested by the Equation

Problem 03
$dy/dx = (9x + 4y + 1)^2$
 

Solution 03
$dy/dx = (9x + 4y + 1)^2$

$dy = (9x + 4y + 1)^2~dx$
 

Let
$z = 9x + 4y + 1$

$dz = 9~dx + 4~dy$

$dy = \frac{1}{4}(dz - 9~dx)$
 

$\frac{1}{4}(dz - 9~dx) = z^2~dx$

$dz - 9~dx = 4z^2~dx$

$dz = 4z^2~dx + 9~dx$

$dz = (4z^2 + 9)~dx$

$\dfrac{dz}{4z^2 + 9} = dx$

$\dfrac{dz}{(2z)^2 + 3^2} = dx$

$\displaystyle \dfrac{1}{2} \int \dfrac{2dz}{(2z)^2 + 3^2} = \int dx$

Problem 02 | Substitution Suggested by the Equation

Problem 02
$\sin y(x + \sin y)~dx + 2x^2 \cos y~dy = 0$
 

Solution 02
$\sin y(x + \sin y)~dx + 2x^2 \cos y~dy = 0$
 

Let
$z = \sin y$

$dz = \cos y~dy$
 

Hence,
$z(x + z)~dx + 2x^2~dz = 0$       → homogeneous equation
 

Let
$z = vx$

$dz = v~dx + x~dv$
 

$vx(x + vx)~dx + 2x^2(v~dx + x~dv) = 0$

$x^2(v + v^2)~dx + 2vx^2~dx + 2x^3~dv = 0$

Problem 01 | Substitution Suggested by the Equation

Problem 01
$(3x - 2y + 1)~dx + (3x - 2y + 3)~dy = 0$
 

Solution 01
$(3x - 2y + 1)~dx + (3x - 2y + 3)~dy = 0$
 

Let
$z = 3x - 2y$

$dz = 3~dx - 2~dy$

$dy = \frac{1}{2}(3~dx - dz)$
 

Thus,
$(z + 1)~dx + (z + 3)[ \, \frac{1}{2}(3~dx - dz) \, ] = 0$

$(z + 1)~dx + \frac{3}{2}(z + 3)~dx - \frac{1}{2}(z + 3)~dz = 0$

$[ \, (z + 1) + (\frac{3}{2}z + \frac{9}{2}) \, ]~dx - \frac{1}{2}(z + 3)~dz = 0$

$(\frac{5}{2}z + \frac{11}{2})~dx - \frac{1}{2}(z + 3)~dz = 0$

Problem 06 - 07 | Integrating Factors Found by Inspection

Problem 06
$y(y^2 + 1) \, dx + x(y^2 - 1) \, dy = 0$

Solution 06
$y(y^2 + 1) \, dx + x(y^2 - 1) \, dy = 0$

$y^3 \, dx + y \, dx + xy^2 \, dy - x \, dy = 0$

$(xy^2 \, dy + y^3 \, dx) + (y \, dx - x \, dy) = 0$

$y^2(x \, dy + y \, dx) + (y \, dx - x \, dy) = 0$

$(x \, dy + y \, dx) + \left( \dfrac{y \, dx - x \, dy}{y^2} \right) = 0$

$d(xy) + d\left( \dfrac{x}{y} \right) = 0$

$\displaystyle \int d(xy) + \int d\left( \dfrac{x}{y} \right) = 0$

$xy + \dfrac{x}{y} = c$

$xy^2 + x = cy$

Problem 23 | Separation of Variables

Problem 23
$x \cos^2 y \, dx + \tan y \, dy = 0$
 

Solution 23

Problem 22 | Separation of Variables

Problem 22
$(xy + x) \, dx = (x^2y^2 + x^2 + y^2 + 1) \, dy = 0$
 

Solution 22
$(xy + x) \, dx = (x^2y^2 + x^2 + y^2 + 1) \, dy = 0$

$x(y + 1) \, dx = (x^2 + 1)(y^2 + 1) \, dy = 0$

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

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

By long division
$\dfrac{y^2 + 1}{y + 1} = y - 1 + \dfrac{2}{y + 1}$
 

Thus,
$\dfrac{x \, dx}{x^2 + 1} = \left( y - 1 + \dfrac{2}{y + 1} \right) \, dy = 0$

Problem 21 | Separation of Variables

Problem 21
$x^2 \, dx + y(x - 1) \, dy = 0$
 

Solution 21
$x^2 \, dx + y(x - 1) \, dy = 0$

$\dfrac{x^2 \, dx}{x - 1} + y \, dy = 0$
 

By long division
$\dfrac{x^2 \, dx}{x - 1} = x + 1 + \dfrac{1}{x - 1}$
 

Thus,
$\left[ (x + 1) + \dfrac{1}{x - 1} \right] \, dx + y \, dy = 0$

$\displaystyle \int (x + 1) \, dx + \int \dfrac{dx}{x - 1} + \int y \, dy = 0$

$\frac{1}{2}(x + 1)^2 + \ln |x - 1| + \frac{1}{2}y^2 + \ln c = 0$

$(x + 1)^2 + 2\ln |x - 1| + y^2 + 2\ln c = 0$

Problem 20 | Separation of Variables

Problem 20
$xy \, dx - (x + 2) \, dy = 0$
 

Solution 20
$xy \, dx - (x + 2) \, dy = 0$

$\dfrac{xy \, dx}{y(x + 2)} - \dfrac{(x + 2) \, dy}{y(x + 2)} = 0$

$\dfrac{x \, dx}{x + 2} - \dfrac{dy}{y} = 0$

$\left( 1 - \dfrac{2}{x + 2} \right) \, dx - \dfrac{dy}{y} = 0$

$\displaystyle \int dx - 2 \int \dfrac{dx}{x + 2} - \int \dfrac{dy}{y} = 0$

$x - 2 \ln (x + 2) - \ln y = \ln c$

$x = \ln c + \ln y + 2 \ln (x + 2)$

$x = \ln c + \ln y + \ln (x + 2)^2$

$x = \ln cy(x + 2)^2$

$\ln e^x = \ln cy(x + 2)^2$

Problem 19 | Separation of Variables

Problem 19
$dr = b(\cos \theta \, dr + r \sin \theta \, d\theta)$
 

Solution 19
$dr = b(\cos \theta \, dr + r \sin \theta \, d\theta)$

$dr = b\cos \theta \, dr + br \sin \theta \, d\theta$

$dr - b\cos \theta \, dr = br \sin \theta \, d\theta$

$(1 - b\cos \theta) \, dr = br \sin \theta \, d\theta$

$\dfrac{(1 - b\cos \theta) \, dr}{r(1 - b\cos \theta)} = \dfrac{br \sin \theta \, d\theta}{r(1 - b\cos \theta)}$

$\dfrac{dr}{r} = \dfrac{b \sin \theta \, d\theta}{1 - b\cos \theta}$

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