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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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Yes.${}{}{}{}{}$ –  Gerry Myerson Nov 23 '12 at 4:44
    
Thanks for the confirmation :) –  TPSstar Nov 23 '12 at 4:44
    
You just answered your own question. $xy^{2}$ is not a function of only x. –  Ryan Nov 23 '12 at 4:45

3 Answers 3

up vote 4 down vote accepted

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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Can it be converted to linear form? –  TPSstar Nov 23 '12 at 5:04
    
I am sure it can. One can cheat, solve the equation, and use the solution to decide on how to transform to linear form. Or one can make a clever guess. –  André Nicolas Nov 23 '12 at 5:11
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I should add a little. Divide everything by $y^2$. You will see that $\frac{1}{y^2}\frac{dy}{dx}$ looks an awful lot like the derivative of $\frac{1}{y}$. –  André Nicolas Nov 23 '12 at 5:14
    
Sorry but i think Iam still doing it wrong, dividing by $$\frac 1y^2$$ gives $$\frac{1}{y^2}\frac{dy}{dx}=x-\frac {x}{y}$$ ? –  TPSstar Nov 23 '12 at 6:27
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Let $z=\frac{1}{y}$. Then $\frac{dz}{dx}=-\frac{1}{y^2}\frac{dy}{dx}$. We get $-\frac{dz}{dx}=x-xz$. This gives $\frac{dz}{dx}-xz=-x$. Nicely linear! Don't trust my calculation, I am weak on minus signs. –  André Nicolas Nov 23 '12 at 6:50

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.

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