Answer to Question #127110 in Differential Equations for Muhammad Hilal

Question

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

Answer 1

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

 

ODE classification:

first-order nonlinear ordinary differential equation.

Step-by-step solution:

1-step:

Solve $x+2y(x)-1=(x+2y(x)+1)\tfrac{dy(x)}{dx}$

2-step:

Let $v(x)=x+2y(x),$ which gives $\tfrac{dv(x)}{dx}=2\tfrac{dy(x)}{dx}+1:$

$v(x)-1=\tfrac{1}{2}(\tfrac{dv(x)}{dx}-1)(v(x)+1)$

3-step:

Solve for $\tfrac{dv(x)}{dx}:$

$\tfrac{dv(x)}{dx}=\tfrac{3v(x)-1}{v(x)+1}$

4-step:

Divide both sides by $\tfrac{3v(x)-1}{v(x)+1}$ :

$\tfrac{\tfrac{dv(x)}{dx}(v(x)+1)}{3v(x)-1}=1$

5-step:

Integrate both sides with respect to x:

$\int$ $\tfrac{\tfrac{dv(x)}{dx}(v(x)+1)}{3v(x)-1}dx=\int1dx$

6-step:

Evaluate the integrals:

$\tfrac{1}{9}(4log(3v(x)-1)+3v(x)-1)=x+C_{1}$

where $C_{1}$ is an arbitrary constant.

Note:$W$ is the product log function.

7-step:

Solve for $v(x):$

$v(x) = \tfrac{1}{3}(4W(- \tfrac{1}{4} \sqrt[4]{e^{9(x+C_{1})}})+1$ )

or

$v(x) = \tfrac{1}{3}(4W(- \tfrac{1}{4}i \sqrt[4]{e^{9(x+C_{1})}})+1)$

or

$v(x) = \tfrac{1}{3}(4W( \tfrac{1}{4}i \sqrt[4]{e^{9(x+C_{1})}})+1)$

or

$v(x) = \tfrac{1}{3}(4W( \tfrac{1}{4} \sqrt[4]{e^{9(x+C_{1})}})+1)$

8-step:

Substitute back for $y(x) =$ $\tfrac{1}{2}(-x+v(x)):$

Answer:

$y(x) = \tfrac{1}{6}(-3x+4W(- \tfrac{1}{4} \sqrt[4]{e^{9(x+C_{1})}})+1)$

or

$y(x) = \tfrac{1}{6}(-3x+4W(- \tfrac{1}{4}i \sqrt[4]{e^{9(x+C_{1})}})+1)$

or

$y(x) = \tfrac{1}{6}(-3x+4W( \tfrac{1}{4}i \sqrt[4]{e^{9(x+C_{1})}})+1)$

or

$y(x) = \tfrac{1}{6}(-3x+4W(\tfrac{1}{4} \sqrt[4]{e^{9(x+C_{1})}})+1)$

Differential equation solution:

$y(x)=\tfrac{2}{3}(W(-e^{\tfrac{9x}{4+C_{1}-1}})+1)+\tfrac{1}{2}(-x-1).$