linear equation (general sol'n): (y+1)dx + (4x-y)dy = 0

[T_TEngineer_AdamT_T]

find the general solution of:

$\displaystyle (y+1)dx + (4x - y)dy = 0$

then i put in standard form ...

$\displaystyle \frac{dx}{dy} + \frac{4x}{1+y} = \frac{y}{y+1}$

now im stuck... i know the integrating factor is $\displaystyle (1+y)^4$...
which i multiply the equation with
and im confused what to do next... i cant really hit the answer at the back of the book

$\displaystyle 20x = 4y - 1 + c(y+1)^{-4}$

 

[royhaas]

Re: linear equation (general solution)

When you multiply by the integrating factor, the left hand side can now be written as the derivative of a product, while the right hand side is a function of one variable only.

 

[T_TEngineer_AdamT_T]

Re: linear equation (general solution)

thank you very much

 

[Deleted member 4993]

Re: linear equation (general solution)

find the general solution
$\displaystyle (y+1)dx + (4x - y)dy = 0$


then i put in standard form ...
$\displaystyle \frac{dx}{dy} + \frac{4x}{1+y} = \frac{y}{y+1}$

now im stuck... i know the integrating factor is $\displaystyle (1+y)^4$...
which i multiply the equation with
and im confused what to do next... i cant really hit the answer at the back of the book

Multiplying both sides by the integrating factor

$\displaystyle \frac{dx}{dy} \cdot\ (1+y)^4+ \frac{4x}{1+y} \cdot\ (1+y)^4 = \frac{y}{y+1}\cdot\ (1+y)^4$

$\displaystyle \frac{dx}{dy} \cdot\ (1+y)^4+ \frac{4x}{1+y} \cdot\ (1+y)^4 = {y}\cdot\ (1+y)^3$

Integrating both sides

$\displaystyle x \cdot\ (1+y)^4 = \frac {(1+y)^5}{5} - \frac {(1+y)^4}{4} + c$

Simplifying further

$\displaystyle 20x = 4y - 1 + c(y+1)^{-4}$