substitution

02 - Solution to Radical Equations

Determine the value of $x$ from the following equations:

  1. $\sqrt{(4 - x^2)^3} + 3x^2\sqrt{4 - x^2} = 0$
     
  2. $\dfrac{1}{3x - 2} - \dfrac{8}{\sqrt{3x - 2}} = 9$
     

01 - Solution to Radical Equations

Solve for $x$ from the following equations

  1. $\sqrt{3 - x} + \sqrt{4 - 2x} = \sqrt{3 - 3x}$
     
  2. $\sqrt{\dfrac{2x + 4}{x - 5}} + 8\sqrt{\dfrac{x - 5}{2x + 4}} = 6$

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

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

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

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

Substitution Suggested by the Equation | Bernoulli's Equation

Substitution Suggested by the Equation
Example 1

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

The quantity (2x - y) appears twice in the equation. Let
$z = 2x - y$

$dz = 2~dx - dy$

$dy = 2~dx - dz$
 

Substitute,
$(z + 1)~dx - 3z(2~dx - dz) = 0$

then continue solving.

 

Problem 04 | Equations with Homogeneous Coefficients

Problem 04
$xy \, dx - (x^2 + 3y^2) \, dy = 0$
 

Solution 04
$xy \, dx - (x^2 + 3y^2) \, dy = 0$
 

Let
$x = vy$

$dx = v \, dy + y \, dv$
 

$vy^2(v \, dy + y \, dv) - (v^2y^2 + 3y^2) \, dy = 0$

$vy^2(v \, dy + y \, dv) - y^2(v^2 + 3) \, dy = 0$

$v(v \, dy + y \, dv) - (v^2 + 3) \, dy = 0$

$v^2 \, dy + vy \, dv - v^2 \, dy - 3 \, dy = 0$

$vy \, dv - 3 \, dy = 0$

$v \, dv - \dfrac{3 \, dy}{y} = 0$

$\displaystyle \int v \, dv - 3 \int \dfrac{dy}{y} = 0$

Problem 03 | Equations with Homogeneous Coefficients

Problem 03
$2(2x^2 + y^2) \, dx - xy \, dy = 0$
 

Solution 03
$2(2x^2 + y^2) \, dx - xy \, dy = 0$
 

Let
$y = vx$

$dy = v \, dx + x \, dv$
 

$2(2x^2 + v^2x^2) \, dx - vx^2(v \, dx + x \, dv) = 0$

$4x^2 \, dx + 2v^2x^2 \, dx - v^2x^2 \, dx - vx^3 \, dv = 0$

$4x^2 \, dx + v^2x^2 \, dx - vx^3 \, dv = 0$

$x^2(4 + v^2) \, dx - vx^3 \, dv = 0$

$\dfrac{x^2(4 + v^2) \, dx}{x^3(4 + v^2)} - \dfrac{vx^3 \, dv}{x^3(4 + v^2)} = 0$

$\dfrac{dx}{x} - \dfrac{v \, dv}{4 + v^2} = 0$

Problem 02 | Equations with Homogeneous Coefficients

Problem 02
$(x - 2y) \, dx + (2x + y) \, dy = 0$
 

Solution 02
$(x - 2y) \, dx + (2x + y) \, dy = 0$
 

Let
$y = vx$

$dy = v \, dx + x \, dv$
 

Substitute,
$(x - 2vx) \, dx + (2x + vx)(v \, dx + x \, dv) = 0$

$x \, dx - 2vx \, dx + 2vx \, dx + 2x^2 \, dv + v^2x \, dx + vx^2 \, dv = 0$

$x \, dx + 2x^2 \, dv + v^2x \, dx + vx^2 \, dv = 0$

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

$x(1 + v^2) \, dx + x^2(2 + v) \, dv = 0$

Pages

 
 
Subscribe to RSS - substitution