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How the following equation is not linear?

$ \frac {dy}{dx}+xy=xy^2$

is it not linear because its not in the below given form?

$ \frac {dy}{dx}+p(x)y=q(x)$

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    You just answered your own question. $xy^{2}$ is not a function of only x.2012-11-23

3 Answers 3

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Hint: It is not of the right form, so it is not linear. But even nicer, it is separable.

Added: Alternately, since you are interested in a linear equation, divide everything by $y^2$. The right substitution will get you a linear equation. You will lose the solution $y=0$.

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    @Andre : Ever met someone who told you : Oh I am strong on minus signs! I wait for that day, still.2012-11-23
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Precisely because of that. A differential equation is called linear when it can be re-written as $ \mathcal L(y) = y_0 $ where $\mathcal L(y)$ is a linear differential operator, i.e. $ \mathcal L(\alpha y_1 + y_2) = \alpha \mathcal L(y_1) + \mathcal (y_2) $ for all $\alpha \in \mathbb R$ and $y_1, y_2$ are sufficiently differentiable functions for your operator to work. Think of $\mathcal L$ as some function of $y$ and possibly $x$ (i.e. $y(x)$, <- that $x$). The function $y_0$ here would be something not depending on $y$.

Hope that helps,

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Your equation is equivalent to: $\frac{{dy}}{{dx}} + xy - x{y^2} = 0.$ The left-hand side of the equation is not a linear function of $y$ and its derivative.