Differential Equations Problem #1

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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KS Teng

KS has been teaching H2/H1 Mathematics and IB mathematics for the more than 10 years. Having taught students from all various Junior Colleges, KS adapts to students' abilities and help them better understand the topics. As someone who loves to teach mathematics and sees it as a truly useful tool in life, KS seeks to enable students to appreciate math. Therefore, his tuition mission is to motivate and cultivate students to be independent and confident thinkers.

Differential Equations Problem #1

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