Differential Equations Problem #1

By KS Teng
Question 2a

Question 2a

To solve a DE with substitution, we first observe that we are given an equation with x and y variables, that is x \frac{dy}{dx} + 2 y = xy^2 then we introduce u = x^2 y. Here, our intention is to remove our y variables and replace it with u instead.

So we should proceed to find \frac{du}{dx}.
\frac{du}{dx} = 2x y + x^2 \frac{dy}{dx}
Since x \frac{dy}{dx} + 2 y = xy^2, we have that \frac{dy}{dx} + 2 \frac{y}{x} = y^2.
\Rightarrow \frac{du}{dx} = 2x y + x^2 (y^2 - 2 \frac{y}{x})
\frac{du}{dx} = 2x y + x^2 y^2 - 2 x y
\frac{du}{dx} = x^2 y^2
\frac{du}{dx} = \frac{u^2}{x^2}
\frac{1}{u^2} \frac{du}{dx} = \frac{1}{x^2}
\int \frac{1}{u^2} du = \int \frac{1}{x^2} dx
-\frac{1}{u} = -\frac{1}{x}+c
-\frac{1}{x^2 y} = -\frac{1}{x}+c
\frac{1}{x^2 y} = \frac{1}{x}+c
All we need to do now, is to make y in terms of x.

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