linear differential equation

 
 

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 | Linear Equations

Problem 04
$(y + 1) \, dx + (4x - y) \, dy = 0$
 

Solution 04
$(y + 1) \, dx + (4x - y) \, dy = 0$

$\dfrac{(y + 1) \, dx}{(y + 1) \, dy} + \dfrac{(4x - y) \, dy}{(y + 1) \, dy} = 0$

$\dfrac{dx}{dy} + \dfrac{4x - y}{y + 1} = 0$

$\dfrac{dx}{dy} + \dfrac{4x}{y + 1} - \dfrac{y}{y + 1} = 0$

$\dfrac{dx}{dy} + \left( \dfrac{4}{y + 1} \right)x = \dfrac{y}{y + 1}$       → linear in x
 

Hence,
$P = \dfrac{4}{y + 1}$

$Q = \dfrac{y}{y + 1}$
 

Integrating factor:

Problem 03 | Linear Equations

Problem 03
$y' = x - 2y$
 

Solution 03
$y' = x - 2y$

$\dfrac{dy}{dx} + 2y = x$       → linear in y
 

Hence,
$P = 2$

$Q = x$
 

Integrating factor:
$\displaystyle e^{\int P \, dx} = e^{\int 2 \, dx}$

$\displaystyle e^{\int P \, dx} = e^{2x}$
 

Thus,
$\displaystyle ye^{\int P\,dx} = \int Qe^{\int P\,dx} \, dx + C$

$\displaystyle ye^{2x} = \int xe^{2x} \, dx + c$
 

Using integration by parts
$u = x$,   $du = dx$

Problem 02 | Linear Equations

Problem 02
$2(2xy + 4y - 3) \, dx + (x + 2)^2 \, dy = 0$
 

Solution 02
$2(2xy + 4y - 3) \, dx + (x + 2)^2 \, dy = 0$

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

$\dfrac{2[ \, 2y(x + 2) - 3 \, ] \, dx}{(x + 2)^2 \, dx} + \dfrac{(x + 2)^2 \, dy}{(x + 2)^2 \, dx} = 0$

$\dfrac{4y(x + 2) - 6}{(x + 2)^2} + \dfrac{dy}{dx} = 0$

$\dfrac{4y(x + 2)}{(x + 2)^2} - \dfrac{6}{(x + 2)^2} + \dfrac{dy}{dx} = 0$

$\left( \dfrac{4}{x + 2} \right) y - \dfrac{6}{(x + 2)^2} + \dfrac{dy}{dx} = 0$

Problem 01 | Linear Equations

Problem 01
$(x^5 + 3y) \, dx - x \, dy = 0$
 

Solution 01
$(x^5 + 3y) \, dx - x \, dy = 0$

$\dfrac{(x^5 + 3y) \, dx}{x \, dx} - \dfrac{x \, dy}{x \, dx} = 0$

$\dfrac{x^5 + 3y}{x} - \dfrac{dy}{dx} = 0$

$x^4 + \dfrac{3y}{x} - \dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} - \dfrac{3}{x}y = x^4$       → linear in y
 

Hence,
$P = -\dfrac{3}{x}$

$Q = x^4$
 

Integrating factor:
$e^{\int P \, dx} = e^{\int \left( -\frac{3}{x} \right)dx}$

$e^{\int P \, dx} = e^{-3\int \frac{dx}{x}}$

Subscribe to RSS - linear differential equation