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 04 | Bernoulli's Equation

Problem 04
$y' = y - xy^3e^{-2x}$
 

Solution 04
$y' = y - xy^3e^{-2x}$

$\dfrac{dy}{dx} - y = -xe^{-2x}y^3$

$dy - y~dx = -xe^{-2x}y^3~dx$       → Bernoulli's equation

$dy + Py~dx = Qy^n~dx$
 

From which
$P = -1$

$Q = -xe^{-2x}$

$n = 3$

$(1 - n) = -2$

$z = y^{1 - n} = y^{-2}$
 

Integrating factor,
$u = e^{(1 - n)\int P~dx} = e^{-2\int (-1)~dx}$

$u = e^{2\int dx} = e^{2x}$
 

Thus,
$\displaystyle zu = (1 - n)\int Qu~dx + C$

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 04 | Determination of Integrating Factor

Problem 04
$y(4x + y)~dx - 2(x^2 - y)~dy = 0$
 

Solution 04
$M~dx + N~dy = 0$

$y(4x + y)~dx - 2(x^2 - y)~dy = 0$
 

$M = y(4x + y) = 4xy + y^2$

$N = -2(x^2 - y) = -2x^2 + 2y$
 

$\dfrac{\partial M}{\partial y} = 4x + 2y$

$\dfrac{\partial N}{\partial x} = -4x$
 

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = (4x + 2y) - (-4x)$

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = 8x + 2y$
 

Problem 03 | Determination of Integrating Factor

Problem 03
$y(2x - y + 1)~dx + x(3x - 4y + 3)~dy = 0$
 

Solution 03
$M~dx + N~dy = 0$

$y(2x - y + 1)~dx + x(3x - 4y + 3)~dy = 0$
 

$M = y(2x - y + 1) = 2xy - y^2 + y$

$N = x(3x - 4y + 3) = 3x^2 - 4xy + 3x$
 

$\dfrac{\partial M}{\partial y} = 2x - 2y + 1$

$\dfrac{\partial N}{\partial x} = 6x - 4y + 3$
 

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = (2x - 2y + 1) - (6x - 4y + 3)$

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = -4x + 2y - 2$

Problem 02 | Determination of Integrating Factor

Problem 02
$2y(x^2 - y + x)~dx + (x^2 - 2y)~dy = 0$
 

Solution 02
$M~dx + N~dy = 0$

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

$M = 2y(x^2 - y + x) = 2x^2y - 2y^2 + 2xy$

$N = x^2 - 2y$
 

$\dfrac{\partial M}{\partial y} = 2x^2 - 4y + 2x$

$\dfrac{\partial N}{\partial x} = 2x$
 

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = (2x^2 - 4y + 2x) - 2x$

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = 2x^2 - 4y = 2(x^2 - 2y)$
 

Problem 01 | Determination of Integrating Factor

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

Solution 01
$M~dx + N~dy = 0$

$(x^2 + y^2 + 1)~dx + x(x - 2y)~dy = 0$
 

$M = x^2 + y^2 + 1$

$N = x(x - 2y) = x^2 - 2xy$
 

$\dfrac{\partial M}{\partial y} = 2y$

$\dfrac{\partial N}{\partial x} = 2x - 2y$
 

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = 2y - (2x - 2y)$

$\dfrac{\partial M}{\partial y} - \dfrac{\partial N}{\partial x} = -2x + 4y$

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