Transformation of RV: Finding PDF

Question

I am trying to work through this example problem in my textbook but I keep getting the wrong final answer.

My Notation:
PDF X : p_X(x)
CDF X : F_X(x)

Question:
Consider the transform Y=X²
if p_X(x) = o.5e^(-|x|), find the PDF of Y.

My Solution

F_Y(y) = P(Y$\le$y) = P(X²$\le$y) = P(X$\le\pm\sqrt y$) = [P(X$\le\sqrt y$)+P(X$\le-\sqrt y$)] = [F_X($\sqrt y$)+ F_X($-\sqrt y$)]

F_X($\sqrt y$) = $\int_0^\sqrt y$ p_X(x)dx = $\int_0^\sqrt y$ o.5e^(-|x|) dx = 0.5 - o.5e^{(-$\sqrt y$)}

F_X($-\sqrt y$) = $\int_{-\sqrt y}^0 $p_X(x)dx = $\int_{-\sqrt y}^0 $ o.5e^(-|x|)dx = o.5e^{(-$\sqrt y$)} - 0.5

After this point I figured I would add F_X($\sqrt y$) and F_X($-\sqrt y$) to get F_Y(y), then finally take the derivative of F_Y(y) to get p_Y(y)... but after adding the two value together I get F_Y(y) = 0 hence p_Y(y) = 0

However, the text book (without showing a complete solution) said p_Y(y) = $\frac{exp(-\sqrt y)}{2\sqrt y}$ Could anyone explain how they got this answer? Thanks in advance.

EDIT: The book also mentions that Y cannot be negative (Y>0). Im still not sure how that helps.

Answer 1

\begin{align}F_Y(y) &= P(Y\le y)\\ &= P(X^2\le y)\\ &= P(-\sqrt y \le X\le \sqrt y)\\ &= [P(X\le\sqrt y)-P(X\le-\sqrt y)]\\ &= [F_X(\sqrt y)- F_X(-\sqrt y)]\\ \end{align}

Differentiating, we get

\begin{align}f_Y(y) &= \frac{f_X(\sqrt{y})}{2\sqrt{y}}+\frac{f_X(-\sqrt{y})}{2\sqrt{y}}\\ &=\frac{f_X(\sqrt{y})}{2\sqrt{y}}+\frac{f_X(\sqrt{y})}{2\sqrt{y}}\\ &=\frac{f_X(\sqrt{y})}{\sqrt{y}}\\ &=\frac{\exp(-\sqrt{y})}{2\sqrt{y}}\end{align}