Solving Differential Equations with Initial Conditions

[BOAS]

Hello

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.

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

Homework Equations

The Attempt at a Solution

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

$x . \frac{1}{dx} =( \frac{1}{y} + y) \frac{1}{dy}$

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

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

$ln|x| + c_{1} = \frac{y^{2}}{2} + ln|y| + c_{2}$

$ln|x| - \frac{y^{2}}{2} - ln|y| = c_{2} - c_{1}$

$ln|x| - \frac{y^{2}}{2} - ln|y| = c$

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 :)

 

[vela]

$$\left(y+\frac1y\right)^{-1} \ne \frac{1}{y} + y$$

 

[Ray Vickson]

Hello

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.

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

Homework Equations

The Attempt at a Solution

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

$x . \frac{1}{dx} =( \frac{1}{y} + y) \frac{1}{dy}$

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

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

$ln|x| + c_{1} = \frac{y^{2}}{2} + ln|y| + c_{2}$

$ln|x| - \frac{y^{2}}{2} - ln|y| = c_{2} - c_{1}$

$ln|x| - \frac{y^{2}}{2} - ln|y| = c$

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 :)

Always substitute your alleged 'solution' into the original DE, to see if it works. My guess is that you have not done that yet.

 

[BOAS]

Always substitute your alleged 'solution' into the original DE, to see if it works. My guess is that you have not done that yet.

I've only just started learning DE's and this may sound silly, but how do I substitute it in?

 

[BOAS]

$$\left(y+\frac1y\right)^{-1} \ne \frac{1}{y} + y$$

$( \frac{1}{y} + y)^{-1} = \frac{1}{\frac{1}{y} + y}$

I'll see if I can solve it now

 

[vela]

No, it wouldn't. You can't distribute the exponent over the sum.

Just try it with y=1. On the lefthand side, you end up with 1/2; on the righthand side you end up with 2.

 

[Ray Vickson]

I've only just started learning DE's and this may sound silly, but how do I substitute it in?

In the obvious way: try it and see. Your solution says that x and y are related by
$$ln|x| - \frac{y^{2}}{2} - ln|y| = c,$$
while the DE says they are related by
$$x dy = \left( y + \frac{1}{y} \right) dx .$$
Do these two statements agree? Can you take the first form and show that it satisfies the second?

 

[BOAS]

No, it wouldn't. You can't distribute the exponent over the sum.

Just try it with y=1. On the lefthand side, you end up with 1/2; on the righthand side you end up with 2.

I think I have identified and fixed the problem, or the second problem that was stopping me from making sense of your help :)

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

$x . \frac{1}{dx} =( \frac{1}{y} + y) \frac{1}{dy}$

$\frac{1}{x} dx = ( \frac{1}{y} + y)^{-1} dy$

$\int \frac{1}{x} dx = \int ( \frac{1}{y} + y)^{-1} dy$

$ln|x| + c_{1} = \frac{ln|y^{2} + 1|}{2} + c_{2}$

$ln|x| - \frac{ln|y^{2} + 1|}{2} = c_{2} - c_{1}$

$ln|x| - \frac{ln|y^{2} + 1|}{2}| = c$

(whilst trying to implement your help, I was trying to differentiate that section when I needed to be integrating it)