Question

If we have an equation as ${{x}^{2}}+{{y}^{2}}=1$ then,
1). $yy''-2{{(y')}^{2}}+1=0$
2). $yy''+{{(y')}^{2}}+1=0$
3). $yy''-{{(y')}^{2}}-1=0$
4). $yy''+2{{(y')}^{2}}+1=0$

Answer 1

First of all differentiate the given term with respect to $x$ on both sides then take $2$common from the obtained differential to make it in simple form after that again differentiate the obtained differential with respect to $x$ on both the sides to check which option is correct in the given options.

Complete step-by-step solution:
The process of finding the differential coefficient of a function is called differentiation or we can say that differentiation is a process where we find the instantaneous rate of change in function based on one of its variables.
The derivative or differential coefficient of a function can be obtained directly by the definition of differentiation without using addition, multiplication and quotient formulae of differentiation.
The coefficient of $y$ with respect to $x$ is represented by $\dfrac{dy}{dx}$ .
We know that the meaning of $dx$ is the increment in $x$ and it may be positive or negative. Similarly $dy$means the increment in the value of $y$
If there is an increment in the value of $x$ and $y$ in the same direction either both positive or both negative then the value of $\dfrac{dy}{dx}$ is always positive. On the other hand if the increments in $x$ and $y$ are of positive direction that is one positive and other negative then the value of $\dfrac{dy}{dx}$ will be negative.
Let $x$ be a variable quantity, after some increment its value ${{x}_{1}}$ becomes ${{x}_{2}}$ . the difference between ${{x}_{1}}$ and ${{x}_{2}}$ is called the change in value of $x$. It is represented by $\delta x$. The change $\delta x$ may be positive or negative.
Let us suppose that $y$ is a function of $x$ and if $\delta x$ be the change in $x$ then the corresponding change in $y$ be denoted by $\delta y$.
Now according to the question:
We have given a term ${{x}^{2}}+{{y}^{2}}=1$
Differentiate the given term with respect to $x$ on both sides
$\Rightarrow \dfrac{d}{dx}({{x}^{2}})+\dfrac{d}{dx}({{y}^{2}})=\dfrac{d}{dx}(1)$
$\Rightarrow 2x+2y\dfrac{dy}{dx}=0$
We know that we can write differential of $y$ as $y'$
$\Rightarrow 2x+2yy'$
Now take $2$ common from left hand side we will get:
$\Rightarrow 2(x+yy')=0$
$\Rightarrow (x+yy')=0$
Now again differentiate the obtained differential with respect to $x$ on both sides:
$\Rightarrow \dfrac{d}{dx}x+\dfrac{d}{dx}yy'=\dfrac{dy}{dx}(0)$
Apply product rule in differentiation of $yy'$ :
$\Rightarrow 1+y'\cdot \dfrac{dy}{dx}+y\cdot \dfrac{d}{dx}(y')=0$
Differentiation of $y'=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$
$\Rightarrow 1+y'\cdot y'+y\cdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0$
You can write $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ as $y''$
$\Rightarrow 1+{{(y')}^{2}}+y\cdot y''=0$
$\Rightarrow y\cdot y''+{{(y')}^{2}}+1=0$
Hence we can say that option $(2)$ is correct.

Note: Students must understand the difference between $\dfrac{\delta y}{\delta x}$ and $\dfrac{dy}{dx}$ . Here $\dfrac{\delta y}{\delta x}$ is a fraction with $\delta y$ as a numerator and $\delta x$ as a denominator while $\dfrac{dy}{dx}$ is not a fraction. It is a limiting value of $\dfrac{\delta y}{\delta x}$ . The differential coefficients of those functions which start with $'co'$ like $\text{cosx,cotx,cosecx}$ are always negative.