# Method to solve this DE

• May 8th 2011, 04:29 PM
Naples
Method to solve this DE
$2y" = 3y^2$

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 PM
TheEmptySet
Quote:

Originally Posted by Naples
$2y" = 3y^2$

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

Use the substition

$u=y' \implies \frac{du}{dy}\frac{dy}{dx}=y'' \iff u\frac{du}{dy}=y''$

This reduces the ODE to the first order separable ODE

$u\frac{du}{dy}=3y^2$

Can you finish from here?
• May 8th 2011, 05:01 PM
Naples
So..?
$2u du = 3y^2 dy$
$u^2 = y^3 + A$
$u = y^(^3^/^2^) + A^(^1^/^2^)$
$y' = y^(^3^/^2^) + A^(^1^/^2^)$
$A=0$
$y' = y^(^3^/^2^)$
$y = (2/5)y^(^5^/^2^) + B$
$B = 3/5$
$y = (2/5)y^(^5^/^2^) + (3/5)$
• May 8th 2011, 05:51 PM
TheEmptySet
Quote:

Originally Posted by Naples
So..?
$2u du = 3y^2 dy$
$u^2 = y^3 + A$
$u = y^(^3^/^2^) + A^(^1^/^2^)$
$y' = y^(^3^/^2^) + A^(^1^/^2^)$
$A=0$
$y' = y^(^3^/^2^)$
$y = (2/5)y^(^5^/^2^) + B$
$B = 3/5$
$y = (2/5)y^(^5^/^2^) + (3/5)$

Not quite. Unless you have some inital conditons I don't think this will simplify very nicely.
We have

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

But now we get

$\frac{dy}{dx}= \pm \sqrt{y^3+C} \implies \int \frac{dy}{\sqrt{y^3+C}}= \pm \int dx \implies x=\pm \int \frac{dy}{\sqrt{y^3+C}}$
• May 8th 2011, 05:56 PM
Naples
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 PM
TheEmptySet
Quote:

Originally Posted by Naples
I was given the initial conditions y(0)=1 and y'(0)=1
Sorry I forgot to put them in the original post

Yes this is good!

Now since

$y'=u$

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

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

Now we just have to solve the ODE

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

Now use the initial conditions and solve for y. The solution to this equation is not unique.
• May 8th 2011, 06:37 PM
Naples
• May 8th 2011, 06:46 PM
TheEmptySet
Quote:
Yes that is correct. You can verify the solution by plugging back into the original ODE.