- #1
BOAS
- 552
- 19
Hello
Find the general solution of the following differential equations. In each case if
y = 2 when x = 1, fi nd y when x = 3.
[itex]x \frac{dy}{dx} = \frac{1}{y} + y [/itex]
[itex]x \frac{dy}{dx} = \frac{1}{y} + y [/itex]
[itex]x . \frac{1}{dx} =( \frac{1}{y} + y) \frac{1}{dy}[/itex]
[itex] \frac{1}{x} dx = y + \frac{1}{y} dy[/itex]
[itex]\int \frac{1}{x} dx = \int y + \frac{1}{y} dy[/itex]
[itex] ln|x| + c_{1} = \frac{y^{2}}{2} + ln|y| + c_{2}[/itex]
[itex] ln|x| - \frac{y^{2}}{2} - ln|y| = c_{2} - c_{1}[/itex]
[itex] ln|x| - \frac{y^{2}}{2} - ln|y| = c[/itex]
If y = 2 and x = 1, c = -2
(assuming I have found the general solution correctly...)
But I don't see how I can find y, when x = 3 because there are two terms involving y.
I think the most likely explanation is that I've done something wrong, so i'd really appreciate some help!
thanks,
BOAS :)
Homework Statement
Find the general solution of the following differential equations. In each case if
y = 2 when x = 1, fi nd y when x = 3.
[itex]x \frac{dy}{dx} = \frac{1}{y} + y [/itex]
Homework Equations
The Attempt at a Solution
[itex]x \frac{dy}{dx} = \frac{1}{y} + y [/itex]
[itex]x . \frac{1}{dx} =( \frac{1}{y} + y) \frac{1}{dy}[/itex]
[itex] \frac{1}{x} dx = y + \frac{1}{y} dy[/itex]
[itex]\int \frac{1}{x} dx = \int y + \frac{1}{y} dy[/itex]
[itex] ln|x| + c_{1} = \frac{y^{2}}{2} + ln|y| + c_{2}[/itex]
[itex] ln|x| - \frac{y^{2}}{2} - ln|y| = c_{2} - c_{1}[/itex]
[itex] ln|x| - \frac{y^{2}}{2} - ln|y| = c[/itex]
If y = 2 and x = 1, c = -2
(assuming I have found the general solution correctly...)
But I don't see how I can find y, when x = 3 because there are two terms involving y.
I think the most likely explanation is that I've done something wrong, so i'd really appreciate some help!
thanks,
BOAS :)