Solving the linear first order differential equation?

Question

I was given the equation

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

My solution was as follows:

$$\frac{dy}{dx} = -xy $$

$$dy = -xy \ dx $$

$$\int {dy} = \int {-xy dx} $$

$$y = -\frac{x^2}{2}y + c $$

$$y + \frac{x^2}{2}y = c $$

$$y(1 + \frac{x^2}{2}) = c $$

$$y = \frac{c}{1 + \frac{x^2}{2}} $$

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!

Answer 1

Hint:

This is a variable separable differential equation that is we can separate the differential equation such that for some functions $f(x)$ and $g(y)$: $f(x)\mathrm dx =g(y)\mathrm dy$. $$\dfrac{\mathrm dy}{\mathrm dx}=-xy \implies \int \dfrac{1}{y}\mathrm dy =-\int x \mathrm dx$$

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?

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

Answer 2

(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 $\int -x y(x) \, \mathrm{d}x = \frac{-1}{2}x^2 y$. 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 $y(x) = \frac{1}{x^2}$), "$\int -x \cdot \frac{1}{x^2} \,\mathrm{d}x = \frac{-1}{2} x^2 \cdot \frac{1}{x^2} + C_1 = C$" rather than the correct $$ \int -x \cdot \frac{1}{x^2} \,\mathrm{d}x = \ln|x| + C \text{.} $$

Quick way to verify this: Implicitly differentiate your antiderivative to see if you get the integrand and differential element back:\begin{align*} \left( \frac{-1}{2}x^2 y(x) \right)' &= \frac{-1}{2}x^2 (y(x))' + \left( \frac{-1}{2}x^2 \right)' y(x) \\ &= \frac{-1}{2}x^2 \,\mathrm{d}y(x) + \left( \frac{-1}{2} \cdot 2 x \,\mathrm{d}x \right) y(x) \\ &= \frac{-1}{2}x^2 \,\mathrm{d}y(x) - x y(x) \,\mathrm{d}x \text{,} \end{align*} which isn't quite "$- x y\,\mathrm{d}x$".

(However, we have shown \begin{align*} \int \left( \frac{-1}{2}x^2 \frac{\mathrm{d}y(x)}{\mathrm{d}x} - x y(x) \right) \,\mathrm{d}x &= \int \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{-1}{2}x^2 y(x) \right) \, \mathrm{d} x \\ &= \frac{-1}{2}x^2 y(x) + C \text{.} \end{align*} Familiarity with this kind of manipulation could be useful in the future.)

Answer 3

After writing

$$\int\frac{\mathrm{d}y}{y}=\int -x\,\mathrm{d}x,$$

you get $\ln|y|=-\frac{x^2}{2}+c_1$, which implies

$$y(x)=c\exp(-x^2/2)\quad\text{for}\quad c=\pm\exp(c_1).$$

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

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

Answer 4

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 $$ 0=y'e^{x^2/2}+xye^{x^2/2}=\frac{d}{dx}(ye^{x^2/2}). $$ So $$ ye^{x^2/2}=c\implies y=ce^{-x^2/2} $$ for some $c$. Since $y(0)=1$, we deduce that $$ y=\exp(-x^2/2). $$