B.2.2.16 IVP of \dfrac{x(x^2+1)}{4y^3}, y(0)=-\dfrac{1}{\sqrt{2}}

[karush]

Find the solution of $y^{\prime} = \dfrac{x(x^2+1)}{4y^3}, \quad y(0)=-\dfrac{1}{\sqrt{2}}$
in explicit form.

rewrite $\dfrac{y}{x}=\dfrac{x(x^2+1)}{4y^3}= y^3\, dy=\dfrac{x(x^2+1)}{4}\, dx$
integrate $\dfrac{y^4}{4}= \dfrac{1}{4}\left(\dfrac{x^4}{4} +\dfrac{ x^2}{2}\right)$
$y^4=\dfrac{x^4}{4} +\dfrac{ x^2}{2}+C$
$y=\left[\dfrac{x^4}{4} +\dfrac{ x^2}{2}+ C\right]^{1/4}$
obtain C $y(0)=\left[\dfrac{0}{4} +\dfrac{ 0}{2}+ C\right]^{1/4}=-\dfrac{1}{\sqrt{2}}
\implies =C^{1/4}=-\dfrac{1}{\sqrt{2}}
=C=-\left(\dfrac{1}{\sqrt{2}}\right)^4=-\dfrac{1}{4}$
finally $y=-\left[\dfrac{x^4}{4} +\dfrac{ x^2}{2}+\dfrac{1}{4}\right]^{1/4}
=-\left[ \dfrac{(x^2+ 1)^2}{4}\right]^{1/4}$
(b) Plot the graph of the solution.
(c)Determine the interval of the solution
book answer $y=−\sqrt{\dfrac{(x^2+1}{2}} \quad [−\infty<x<\infty]$
ok i think there are some bugs in this and probably some suggestions
I wanted to plot this in tikz but did know how to deal with the final eq
mahalo

 

[romsek]

Are you sure you've copied the problem correctly?

Mathematica is showing no solution. The problem is likely with the initial condition.

 

[topsquark]

Find the solution of $y^{\prime} = \dfrac{x(x^2+1)}{4y^3}, \quad y(0)=-\dfrac{1}{\sqrt{2}}$
in explicit form.

rewrite $\dfrac{y}{x}=\dfrac{x(x^2+1)}{4y^3}= y^3\, dy=\dfrac{x(x^2+1)}{4}\, dx$

What is this line? You can't possibly think that this is correct? I used to take points off for lines like this.

Write it out!
[math]\dfrac{dy}{dx} = \dfrac{x(x^2 + 1)}{4y^3}[/math]

[math]y^3 ~ dy = \dfrac{x(x^2 + 1)}{4} ~ dx[/math]

-Dan

 

[karush]

Are you sure you've copied the problem correctly?

Mathematica is showing no solution. The problem is likely with the initial condition.

here is the book answer

 

[karush]

What is this line? You can't possibly think that this is correct? I used to take points off for lines like this.

Write it out!
[math]\dfrac{dy}{dx} = \dfrac{x(x^2 + 1)}{4y^3}[/math]

[math]y^3 ~ dy = \dfrac{x(x^2 + 1)}{4} ~ dx[/math]

-Dan

ok it should of been and $\implies$ or a new line

 

[romsek]

We easily obtain that

$$\dfrac{dy}{dx} = \dfrac{x(x^2+1)}{4y^3}\\

4 y^3 dy = (x^3 + x) dx\\

y^4 = \dfrac 1 4 \left(x^4 + 2 x^2 + C\right)\\~\\

\text{It's easiest to evaluate the constant here}\\

(y(0))^4 = \dfrac 1 4 C = \left(-\dfrac{1}{\sqrt{2}}\right)^4 = \dfrac 1 4 \Rightarrow C = 1\\

y =\pm \left(\dfrac 1 4 \left(x^4 + 2x^2 + 1\right)\right)^{1/4} = \\

\pm \dfrac{1}{\sqrt{2}}\left(\left(x^2+1\right)^2\right)^{1/4} = \\

\pm \dfrac{1}{\sqrt{2}}\sqrt{(x^2+1)}\\

\text{and by the initial condition we see that we only keep the negative solution, i.e.}\\

y(x) = -\dfrac{1}{\sqrt{2}}\sqrt{x^2+1} = -\sqrt{\dfrac{x^2+1}{2}}
$$

 

[karush]

Mahalo
That helped a lot