# Thread: need help witth this diff. eq.

1. ## need help witth this diff. eq.

y' + xy = x/y

work done:

y = vx => y' = v +v'x

v +v'x + v(x^2) = 1/v =>

v^2 + v'vx + (vx)^2 = 1 => v'vx + (v^2)(1+x^2) = 1

... it tappers off here i 'm afraid basically i thought i could turn it into a Bernouli eq. but i can't...

i'v tried a few other substitutions bu i am hitting a wall,

just a clue please

2. Originally Posted by pepsi
y' + xy = x/y

work done:

y = vx => y' = v +v'x

v +v'x + v(x^2) = 1/v =>

v^2 + v'vx + (vx)^2 = 1 => v'vx + (v^2)(1+x^2) = 1

... it tappers off here i 'm afraid basically i thought i could turn it into a Bernouli eq. but i can't...

i'v tried a few other substitutions bu i am hitting a wall,

just a clue please

Does this question require you to apply the Bernouli's method ?

If not , I think we should use an other way : separating variables

$y' + xy = \frac{x}{y}$

$y' = \frac{dy}{dx} = x \left( \frac{1}{y} - y \right)$

by separating the variables ,

$\frac{y dy}{ 1-y^2 } = x dx$

integrating ,

$\frac{\ln( 1- y^2)}{2}= \frac{-x^2 + K }{2}$

$\ln(1 - y^2 ) =- x^2 + K$

$y = \sqrt{ 1 - e^{-x^2 + K }}$

Write $e^K = c$

$y = \sqrt{ 1 - ce^{-x^2}}$

3. Originally Posted by simplependulum
Does this question require you to apply the Bernouli's method ?

If not , I think we should use an other way : separating variables

$y' + xy = \frac{x}{y}$

$y' = \frac{dy}{dx} = x \left( \frac{1}{y} - y \right)$

by separating the variables ,

$\frac{y dy}{ 1-y^2 } = x dx$

integrating ,

$\frac{\ln( 1- y^2)}{2}= \frac{-x^2 + K }{2}$

$\ln(1 - y^2 ) =- x^2 + K$

$y = \sqrt{ 1 - e^{-x^2 + K }}$

Write $e^K = c$

$y = \sqrt{ 1 - ce^{-x^2}}$
embarassed and thankful