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Thread: Taking a Differential Equations Class; Can't remember/understand anything.

  1. #1
    Aug 2010

    Taking a Differential Equations Class; Can't remember/understand anything.

    I'm currently enrolled in a Differential Equations class at my local community college. I'm trying to get through the first problem, but I don't even remember/know what I'm doing.

    Right now, I'm having trouble with classification by linearity. I can't tell (without guessing and looking at the back of the book ) whether a DE is linear or non-linear.

    An problem I'm having trouble understanding goes as:

    Determine whether the given first-order DE is linear in the indicated dependent variable by matching it with the first differential equation given in (7)
    $\displaystyle [(y^2)-1]dx + x dy = 0; in y; in x$.

    The Answer Book said:
    Writing the DE in the form x(dy/dx) = Y^2 = 1, we see that it is nonlinear in y because of y^2. However, writing it in the form of (y^2 - 1)(dx/dy) + x = 0, we see that it is linear in x.

    I can't remember anything really about dx or dy and how the question got to the solution. If anyone can direct me to the subject online I should review (I have the 6e stewart book if anyone else knows/uses it).

    I'm sorry for my ignorance; hope someone can help.
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  2. #2
    Member Haven's Avatar
    Jul 2009
    Linear basically means it has the same properties as a line. The easiest classification of this is that the equation only contain terms that are singular powers (i.e., $\displaystyle x^1$).

    So $\displaystyle x^2+1$ is not linear
    $\displaystyle \sqrt{x} + x$ is also not linear.
    however, $\displaystyle 3x+2y$ is linear. (no powers, exponentials, logarithms, roots, just plain old ordinary variables)

    In general, if we haves variables $\displaystyle x_1 \dots x_n$, then a linear equation is of the form.

    $\displaystyle c_1x_1 + \dots + c_nx_n = d$

    Where $\displaystyle c_1 \dots c_n$ and d are all constants.

    Now a first order ordinary differential equation is always of this form

    $\displaystyle \frac{dy}{dx} = f(y)$

    where f(y) is some expression involving y, as a function of x.

    So, $\displaystyle \frac{dy}{dx} = 3y$ is a differential equation.
    and $\displaystyle \frac{dy}{dx} = e^y$ is also a differential equation.

    We call a differential equation linear, if f(y) is linear. So the first one is linear, but the second one isn't.

    How the book got it's answers was through simple algebra. In order to determine if the differential equation is linear in x, we needed the derivative of x on one side and an expression involving x on the other.

    So, we do this $\displaystyle (y^2-1)dx + xdy = 0 \Rightarrow (y^2-1)dx = -xdy \Rightarrow (y^2-1)\frac{dx}{dy}= -x$

    Since the right hand side is a linear expression, we say the differential equation is linear in x.

    However, similar algebra tells us that the differential equation is NOT linear in y.
    i.e., $\displaystyle (y^2-1)dx + xdy = 0 \Rightarrow (y^2-1)dx = -xdy \Rightarrow x\frac{dy}{dx}= 1-y^2$. Since the right hand side is not linear, the differential equation is not linear.
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