Solve the Following Differential Equation!

Question

anonymous · Answer

$$x^{2} y ^{''}(x)=x^{2}+1 , y^{'}(1)=0 , y(1)=\frac{ 1 }{ 2 }$$

anonymous · Answer

I don't get what I'm supposed to do with the x^2 in front of y double prime

zepdrix · Answer

$$\large x^2y''=x^2+1$$We'll start by dividing both sides by x^2.$$\large y''=\frac{x^2+1}{x^2}$$Rewrite the right side as two fractions,$$\large y''=\frac{x^2}{x^2}+\frac{1}{x^2} \qquad = \qquad 1+x^{-2}$$

zepdrix · Answer

Understand how to integrate from here? :)
It will involve the Power Rule for Integration.

anonymous · Answer

Oh okay! I think I can go from there, thanks!!

anonymous · Answer

okay I went down through the steps and I got to $$dy=\int\limits_{}^{}  x dx - \int\limits_{}^{} x ^{-1}dx$$
but like x^0/0???????? what do I do for that second integral

zepdrix · Answer

Before we deal with that, did you remember to add a +C after integrating a first time? :)

anonymous · Answer

ya my c =0 :(

zepdrix · Answer

$x^{-1}$ or $\dfrac{1}{x}$ is the ONE power that we can't apply the power rule to! :)

It's an important one to remember! ^^

$$\large (\ln x)'=\frac{1}{x}$$

anonymous · Answer

Oh yeah!!!!! okay perfect!

zepdrix · Answer

Hmm it looks like your constants end up being 0 both times you integrate.
What a weird problem. lol