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So for this one I'm having trouble isolating for y. If its not possible then in the form with dy with the y variable and x with the x variable.

$$\frac{dy}{dx}-2xy=e^{x^{2}}.$$

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    $\begingroup$ This isn't a separable equation. A better strategy is to look for an integrating factor. $\endgroup$ – Semiclassical Sep 18 '14 at 5:13
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If you have a differential equation of the form $$\frac{dy}{dx} + a(x) y(x) = b(x) \tag{1}$$ we call the equation a first-order linear ODE and we can obtain it's solution using the following method. First, we multiply both sides of $(1)$ by a function $f(x)$ (called the integrating factor) and we obtain $$f y' + fay = fb \tag{2}$$ Using the product rule $(fy)' = fy' + f'y$, we can rewrite $(2)$ as $$(fy)' - f'y + fay = fb \Longrightarrow (fy)' + y \left(f' + fa \right) = fb \tag{3}$$

If it is the case that $f' + fa = 0$, then the LHS of $(3)$ is just $(fy)'$, and integrating both sides would yield an expression for $fy$. So let's solve for the $f$ that guarantees that $f' + fa = 0$. Solving this separable differential equation for $f$, we get that $$\frac{df}{dx} = f'(x) = -f(x)a(x) \Longrightarrow \frac{df}{f(x)} = a(x) dx \Longrightarrow \log(f(x)) = \int a(x) dx \Longrightarrow f(x) = e^{\int a(x) dx} $$

Using the $f$ we just found, $(3)$ therefore reduces to $$(fy)' = fb \Longrightarrow fy = \int fb \: dx \Longrightarrow y = \frac{\int fb \: dx}{f}$$

Plugging in our formula for $f(x)$, we get that the solution to $(1)$ is $$y(x) = \displaystyle\frac{\displaystyle\int \left(b(x) \: e^{\int a(x) dx} \right) dx}{e^{\int a(x) dx}}$$

Now, noting that $a(x) = -2x$ and $b(x) = e^{x^2}$ in your example, we see that $$y(x) = \displaystyle\frac{\displaystyle\int \left(e^{x^2} \: e^{\int -2x dx} \right) dx}{e^{\int -2x dx}} = \frac{\displaystyle\int \left(e^{x^2} e^{-x^2}\right)dx}{e^{x^2}} = \frac{\int e^0 dx}{e{x^2}} = \frac{x+c}{e^{x^2}} \Longrightarrow \boxed{y(x) = xe^{-x^2} + c e^{-x^2}}$$

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Hint : Multiply throughout by the integrating factor, $e^{\int -2x dx} = e^{-x^2}$.

Then notice that $$\begin{align}\frac{d}{dx}(e^{-x^2}y) &= e^{-x^2}\frac{dy}{dx} - 2e^{-x^2}xy\\&=LHS\end{align}$$

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Rewrite the equation like this:

$$e^{-x^2}\frac{dy}{dx}-2xe^{-x^2}y=1$$

Notice that if we apply the product rule in differentiating $ye^{-x^2}$ with respect to $x$, that we get exactly the left hand side. In other words, the equation is equivalent to:

$$\frac{d(e^{-x^2}y)}{dx}=1$$

Integrating both sides yields:

$$e^{-x^2}y=x+c\implies y=xe^{-x^2}+ce^{-x^2}.$$

This kind of technique can be generalised to the method of 'integrating factors', however it happens to work out nicely enough here that you can just follow your nose.

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Hint

Because of the rhs, suppose that you define $y(x)=z(x)e^{x^2}$; then the differential equation write $$x z'(x)+z(x)=0$$ which is quite easy to integrate for $z(x)$.

I am sure that you can take from here.

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