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I was given the equation

$y' = -xy$, where $y(0) = 1$.

My solution was as follows:

$$fracdydx = -xy $$

$$dy = -xy dx $$

$$int dy = int -xy dx $$

$$y = -fracx^22y + c $$

$$y + fracx^22y = c $$

$$y(1 + fracx^22) = c $$

$$y = fracc1 + fracx^22 $$

I tried plugging in $0$ and $1$ respectively to find $c$, and got $c = 0$.

I just know that what I did is wrong. Where did I go wrong, please? Thanks!

Hint:

This is a variable separable differential equation that is we can separate the differential equation such that for some functions $f(x)$ and $g(y)$: $f(x)mathrm dx =g(y)mathrm dy$. $$dfracmathrm dymathrm dx=-xy implies int dfrac1ymathrm dy =-int x mathrm dx$$

Now all that is left is to compute the integral and use the inital condition to figure out the value of constant. Can you proceed?

Note that you cannot simply integrate $-xy$ wrt $x$. Your computation assumes that $y$ is a constant which it most certainly is not.

(Too long for a comment.)

No one else has mentioned this, so I will...

Remember: $y$ is a function of $x$, not something that is constant with respect to variation of $x$. (Forgetting this is a fairly common error when first performing implicit differentiation.) It can help to write "$y(x)$" rather than just "$y$".

Your error is believing $int -x y(x) , mathrmdx = frac-12x^2 y$. But $y$ is not some constant in this integral; it is a function that depends on $x$. Otherwise, what is written could be the absurdity (when, say $y(x) = frac1x^2$), "$int -x cdot frac1x^2 ,mathrmdx = frac-12 x^2 cdot frac1x^2 + C_1 = C$" rather than the correct
$$ int -x cdot frac1x^2 ,mathrmdx = ln|x| + C text. $$

Quick way to verify this: Implicitly differentiate your antiderivative to see if you get the integrand and differential element back:beginalign*
left( frac-12x^2 y(x) right)'
&= frac-12x^2 (y(x))' + left( frac-12x^2 right)' y(x) \
&= frac-12x^2 ,mathrmdy(x) + left( frac-12 cdot 2 x ,mathrmdx right) y(x) \
&= frac-12x^2 ,mathrmdy(x) - x y(x) ,mathrmdx text,
endalign*
which isn't quite "$- x y,mathrmdx$".

(However, we have shown beginalign*
int left( frac-12x^2 fracmathrmdy(x)mathrmdx - x y(x) right) ,mathrmdx
&= int fracmathrmdmathrmdx left( frac-12x^2 y(x) right) , mathrmd x \
&= frac-12x^2 y(x) + C text.
endalign*
Familiarity with this kind of manipulation could be useful in the future.)

After writing

$$intfracmathrmdyy=int -x,mathrmdx,$$

you get $ln|y|=-fracx^22+c_1$, which implies

$$y(x)=cexp(-x^2/2)quadtextforquad c=pmexp(c_1).$$

After plugging in $y(0)=1$, you get $c=1$, so

$$y(x)=exp(-x^2/2).$$

Here is another method that does not use separation of variables and uses integrating factors. Write
$$
y'+xy=0
$$
and multiply both sides by $e^int x, dx=e^x^2/2$ to get that
$$
0=y'e^x^2/2+xye^x^2/2=fracddx(ye^x^2/2).
$$
So
$$
ye^x^2/2=cimplies y=ce^-x^2/2
$$
for some $c$. Since $y(0)=1$, we deduce that
$$
y=exp(-x^2/2).
$$