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$$\frac{dy}{dx} = \sin^{-1} (y)$$ The above equation is a form of $\frac{dy}{dx} = f(y)$, so degree should be $1$. But if I write it as $$y = \sin\left(\frac{dy}{dx}\right)$$ then degree is not defined as it is not a polynomial in $\frac{dy}{dx}$. Please explain?

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The explanation is simple: they are not the same equations. Even if two equations are equivalent, they are not exactly the same. For example:

$$\frac{dy}{dx}=\sqrt[3]{x}\tag1$$ $$\left(\frac{dy}{dx}\right)^3=x\tag2$$

The equation $(1)$ is not the same as equation $(2)$ even if they do have exactly the same solutions (in $\mathbb R$ to be clear). You can see that $(1)$ has degree $1$ and $(2)$ has degree $3$.

The problem is when you try to find degree of i.e. $$y=e^{y'} \quad\text{or}\quad y=\sin\left(\frac{dy}{dx}\right)\tag{a,b}$$ There exist a formula that allow you define a degree of non-polynomials, namely $$\deg\;f(x)=\lim_{x\to\infty}\frac{\log|f(x)|}{\log(x)}$$ but in some cases, such as $(b)$, is unlikely to work, whereas for other cases it allows to define a degree of non-polynomial functions. For example equation $(a)$ may be degree $\infty$.

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    $\begingroup$ What is $f(x)$ in $a$ ? $\endgroup$
    – user312097
    Sep 23, 2017 at 9:01
  • $\begingroup$ @A---B $f(x)$ is any function we want to compute degree of. In case $(a)$ we have to substitute $y' = x$ to get $f(x)=e^x$ (because we want to find a degree of the term $y'$). $\endgroup$ Sep 24, 2017 at 8:37
  • $\begingroup$ Actually the definition of equivalence they must be exactly the same. Power relations are never equivalence relations. $\endgroup$ Jul 23, 2018 at 14:12
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Degree is not defined for terms of the form $\sin(\frac{dy}{dx})$ because, on expanding the sinusoid, the degree of the highest power goes to infinity.

In order to find the degree corresponding to the differential equation, first bring it to its standard form. Try to express the equation as a polynomial function of the derivatives. Then the power corresponding highest order is called the degree of the equation. Hence the degree of the equation you mentioned is 1.

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  • $\begingroup$ But if we try to reduce the differential equation into polynomial in derivative then we get the different differential equation. $\endgroup$
    – neelkanth
    Dec 26, 2015 at 11:46
  • $\begingroup$ so give the degree of different differential equation , not the original one...please explane my confusion... $\endgroup$
    – neelkanth
    Dec 26, 2015 at 11:47
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    $\begingroup$ @neela, You always have to reduce the given differential equation to the standard form. Only then can we talk about the degree of the equation. The degree of the new equation is undefined (because if you expand the sinusoid, the degree tends to infinity). $\endgroup$
    – TheChetan
    Dec 26, 2015 at 14:02
  • $\begingroup$ please can you solve my this similar problem $\endgroup$
    – neelkanth
    Dec 26, 2015 at 14:17
  • $\begingroup$ math.stackexchange.com/questions/1589320/… $\endgroup$
    – neelkanth
    Dec 26, 2015 at 14:17
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To make sense of the definition of the order of a differential equation, you need to first isolate the highest order derivative first. For example, $\frac{dy}{dx} = y, \frac{dy}{dx} = y^2 + 1$ are of the first order. $ \frac{d^2 y}{dx^2} = y, \frac{d^2 y}{dx^2} = (\frac{dy}{dx})^3 + 1$ are of second order.

Your differential equation $\frac{dy}{dx} = \sin^{-1}y$ is first order even if it can be written in the equivalent form $y = \sin(\frac{dy}{dx}).$

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    $\begingroup$ i actually want to know about the degree.. $\endgroup$ Jan 24, 2015 at 14:33
  • $\begingroup$ @AryamanBansal. First order (just as abel answered). $\endgroup$ Jan 24, 2015 at 14:35
  • $\begingroup$ @AryamanBansal, even if you are talking about degree, what is the degree of $\sin.$ these are called transcendental functions, that is, they are like polynomials but of infinite degree. $\endgroup$
    – abel
    Jan 24, 2015 at 14:38
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    $\begingroup$ @abel Sir the problem was concerning about degree of a differential equation and not about the order of the differential equation. There is difference between order and degree. Please see. $\endgroup$
    – Singh
    Dec 22, 2015 at 3:01
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    $\begingroup$ @YoTengoUnLCD Sir I think you are confused with the terms order and degree because in many cases we use both the terms as same but in case of differential equations we have precise definition of order and degree. One more thing I want to say is that definitions are not bounded by the boundaries of nations definitions are same everywhere. Please see (www3.ul.ie/cemtl/pdf%20files/bm2/DegreeOrder.pdf) $\endgroup$
    – Singh
    Dec 22, 2015 at 3:42

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