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

T_TEngineer_AdamT_T

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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

\(\displaystyle 20x = 4y - 1 + c(y+1)^{-4}\)
 
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.
 
Re: linear equation (general solution)

T_TEngineer_AdamT_T said:
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

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