Solving for y in a logarithmic equation involving |y|

[schniefen]

Homework Statement

Solve for $y$ in the below equation and find the value of the constant of integration.

Relevant Equations

Find the equation of the curve $y'(x)\frac{1}{y}=\frac{1}{2x^3}$ in terms of $y=y(x)$ if it passes through the point $(1,1)$.

Integrating both sides of the equation yields

$\ln{|y|}=-\frac{1}{4x^2}+C$
$\iff \ln{|y|}=-\frac{1}{4x^2}+\ln{D}$
$\iff |y|=De^{-\frac{1}{4x^2}}$
​

At $(1,1)$, $D=e^{\frac{1}{4}}$. So for $y>0$, $y=e^{ \frac{1}{4}-\frac{1}{4x^2}}$, and for $y<0$, $y=-e^{\frac{1}{4}-\frac{1}{4x^2}}$.

Apparently one can solve for $y$ so that $y=e^{ \frac{1}{4}-\frac{1}{4x^2}}$ for all $y$. How?

 

[WWGD]

I don't see how you go from the 2nd line to the 3rd. Why are you grouping as a product. That holds when you use ln, but you're going in the opposite direction, taking exponentials: $e^{a+b} \neq e^a+e^b$

 

[schniefen]

I don't see how you go from the 2nd line to the 3rd. Why are you grouping as a product. That holds when you use ln, but you're going in the opposite direction, taking exponentials: $e^{a+b} \neq e^a+e^b$

$\ln{|y|}=-\frac{1}{4x^2}+\ln{D}$
$\iff \ln{|y|}-\ln{D}=-\frac{1}{4x^2}$
$\iff \ln{\frac{|y|}{D}}=-\frac{1}{4x^2}$
$\iff \frac{|y|}{D}=e^{-\frac{1}{4x^2}}$
$\iff |y|=De^{-\frac{1}{4x^2}}$
​

Is this incorrect?

 

[WWGD]

$\ln{|y|}=-\frac{1}{4x^2}+\ln{D}$
$\iff \ln{|y|}-\ln{D}=-\frac{1}{4x^2}$
$\iff \ln{\frac{|y|}{D}}=-\frac{1}{4x^2}$
$\iff \frac{|y|}{D}=e^{-\frac{1}{4x^2}}$
$\iff |y|=De^{-\frac{1}{4x^2}}$
​

Is this incorrect?

Looks fine. Now use the condition that the solution passes through (1,1) to find D and sub in the last equation.

 

[schniefen]

At $(1,1)$, $D=e^{\frac{1}{4}}$, so $|y|=e^{\frac{1}{4}-\frac{1}{4x^2}}$. How can one conclude from here that $y=e^{\frac{1}{4}-\frac{1}{4x^2}}$?

 

[WWGD]

At $(1,1)$, $D=e^{\frac{1}{4}}$, so $|y|=e^{\frac{1}{4}-\frac{1}{4x^2}}$. How can one conclude from here that $y=e^{\frac{1}{4}-\frac{1}{4x^2}}$?

Notice that $e^x>0$ for all x.

 

[Fred Wright]

$$\int \frac {dy}{y}= \int \frac{1}{2}x^{-3}dx $$​

$$ log(y) = -\frac {1}{x^2} + C $$​

$$y= exp(-\frac{1} {x^2} + C) $$​

$$ log(1) = -1 +C $$​

$$0=C-1 $$​

$$C=1$$​

 

[WWGD]

$$\int \frac {dy}{y}= \int \frac{1}{2}x^{-3}dx $$​

$$ log(y) = -\frac {1}{x^2} + C $$​

$$y= exp(-\frac{1} {x^2} + C) $$​

$$ log(1) = -1 +C $$​

$$0=C-1 $$​

$$C=1$$​

Sorry, I don't see your point. How does this relate to the OP?

 

[archaic]

Notice that $e^x>0$ for all x.

yes, but that says nothing about $y$ doesn't it? of course it'll be positive since $|y|$ is.

 

[WWGD]

yes, but that says nothing about $y$ doesn't it? of course it'll be positive since $|y|$ is.

But we know |y|=y and that is all we need, isn't it?

 

[archaic]

This is a first order linear differential equation, the solution is $y=a\exp{(\int\frac{1}{2x^3}dx)}$.

 

[WWGD]

This is a first order linear differential equation, the solution is $y=a\exp{(\int\frac{1}{2x^3}dx)}$.

But the response given also checks out.Edit: The given expression may be equivalent to yours.

 

[Mark44]

$$\int \frac {dy}{y}= \int \frac{1}{2}x^{-3}dx $$​

$$ log(y) = -\frac {1}{x^2} + C $$​

You have a mistake in the line above. Check your integration.

$$y= exp(-\frac{1} {x^2} + C) $$​

$$ log(1) = -1 +C $$​

$$0=C-1 $$​

$$C=1$$​