$\displaystyle 2y" = 3y^2$

Basically I don't have any idea about what method I should use to solve this DE...I've tried several ways but they don't seem to work

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- May 8th 2011, 04:29 PMNaplesMethod to solve this DE
$\displaystyle 2y" = 3y^2$

Basically I don't have any idea about what method I should use to solve this DE...I've tried several ways but they don't seem to work - May 8th 2011, 04:40 PMTheEmptySet
- May 8th 2011, 05:01 PMNaples
So..?

$\displaystyle 2u du = 3y^2 dy$

$\displaystyle u^2 = y^3 + A$

$\displaystyle u = y^(^3^/^2^) + A^(^1^/^2^)$

$\displaystyle y' = y^(^3^/^2^) + A^(^1^/^2^)$

$\displaystyle A=0$

$\displaystyle y' = y^(^3^/^2^)$

$\displaystyle y = (2/5)y^(^5^/^2^) + B$

$\displaystyle B = 3/5$

$\displaystyle y = (2/5)y^(^5^/^2^) + (3/5)$ - May 8th 2011, 05:51 PMTheEmptySet
Not quite. Unless you have some inital conditons I don't think this will simplify very nicely.

We have

$\displaystyle 2u\frac{du}{dy}=3y^2 \iff \int udu = \int 3y^2dy \implies u^2=y^3+C \implies u=\pm \sqrt{y^3+C}$

But now we get

$\displaystyle \frac{dy}{dx}= \pm \sqrt{y^3+C} \implies \int \frac{dy}{\sqrt{y^3+C}}= \pm \int dx \implies x=\pm \int \frac{dy}{\sqrt{y^3+C}}$ - May 8th 2011, 05:56 PMNaples
I was given the initial conditions y(0)=1 and y'(0)=1

Sorry I forgot to put them in the original post - May 8th 2011, 06:14 PMTheEmptySet
Yes this is good!

Now since

$\displaystyle y'=u$

We can use the equation above and the intial condions to get

$\displaystyle u^2= y^3+C \implies 1^2 = 1^3 +0 \implies C=0$

Now we just have to solve the ODE

$\displaystyle (y')^2=y^3 \implies \frac{dy}{dx} = \pm y^{\frac{3}{2}} \implies y^{-\frac{3}{2}}dy = \pm dx \implies \frac{-2}{\sqrt{y}}= \pm x+D$

Now use the initial conditions and solve for y. The solution to this equation is not unique. - May 8th 2011, 06:37 PMNaples
- May 8th 2011, 06:46 PMTheEmptySet