Question

If \[{{x}^{2}}+{{y}^{2}}=4\] , then \[y\dfrac{dy}{dx}+x\] is equal to:
(a) 4
(b) 0
(c) 1
(d) -1

Answer 1

Hint: In this question, we first need to differentiate the given expression with respect to x on both sides. Then write the differential terms accordingly and rearrange the terms on both the sides to get the value of the expression asked in the question.
\[\begin{align}
  & \dfrac{d}{dx}\left( {{x}^{2}} \right)=2x \\
 & \dfrac{d}{dx}\left( {{y}^{2}} \right)=2y\dfrac{dy}{dx} \\
 & \dfrac{d}{dx}\left( \text{constant} \right)=0 \\
\end{align}\]

Complete step by step answer:
Let us look at some of the definitions first.
An equation that involves an independent variable, dependent variable and differential coefficients of dependent variable with respect to the independent variable is called a differential equation.
Example:
\[\left( {{x}^{2}}+{{y}^{2}} \right)dx=\left( {{x}^{2}}-{{y}^{2}} \right)dy\]
ORDER AND DEGREE OF A DIFFERENTIAL EQUATION:
The order of a differential equation is the order of the highest derivative occurring in the equation. The order of a differential equation is always a positive integer.
The degree of a differential equation is the degree (exponent) of the derivative of the highest order in the equation, after the equation is free from negative and fractional powers of the derivatives.
Example: The order and degree of a differential equation
\[{{\left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)}^{2}}+2{{\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{3}}+3y=0\] are 3 and 2.
LINEAR AND NON-LINEAR DIFFERENTIAL EQUATIONS:
A differential equation is said to be linear, if the dependent variable and all of its derivatives occur in the first power and there are no products of these.
A linear differential equation is always of the first degree but every differential equation of the first degree need not be linear.
Example:
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \dfrac{dy}{dx} \right)}^{2}}+xy=0\] is not linear

SOLUTION OF DIFFERENTIAL EQUATIONS-
A solution of a differential equation is a relation between the variables, not involving the differential coefficients, such that this relation and the derivative obtained from it satisfy the given differential equation.
Now, let us consider the given equation.
\[\Rightarrow {{x}^{2}}+{{y}^{2}}=4\]
Now, on differentiating it with respect to x on both sides we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}+{{y}^{2}}=4 \\
 & \Rightarrow 2x+2y\dfrac{dy}{dx}=0 \\
\end{align}\]
 & \[ \begin{align}
  & \because \dfrac{d}{dx}\left( {{x}^{2}} \right)=2x \\
 & \dfrac{d}{dx}\left( {{y}^{2}} \right)=2y\dfrac{dy}{dx} \\
 & \dfrac{d}{dx}\left( \text{constant} \right)=0 \\
\end{align}\]
By rearranging the terms in the above equation we get,
\[\begin{align}
  & \Rightarrow 2y\dfrac{dy}{dx}=-2x \\
 & \Rightarrow y\dfrac{dy}{dx}=-x \\
 & \therefore y\dfrac{dy}{dx}+x=0 \\
\end{align}\]

Hence, the correct option is (b).

Note: Here, while differentiating with respect to x on both sides we need to be careful about the differential terms to be formed which affects the result of the solution.
\[\dfrac{d}{dx}\left( {{y}^{2}} \right)=2y\dfrac{dy}{dx}\]

If the solution of the differential equation contains as many independent arbitrary constants as the order of the differential equation, then it is called the general solution or the complete integral of the differential equation.

It is also important to note that the differentiation of a constant is zero because if we neglect that and keep it as 4 then the final result of the equation also becomes 4 which will be option (a).