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Question

Question: Which of the following differential equations is not linear? A. \(\dfrac{{{d^2}y}}{{d{x^2}}} + x\d...

Which of the following differential equations is not linear?
A. d2ydx2+xdydx+2y=0\dfrac{{{d^2}y}}{{d{x^2}}} + x\dfrac{{dy}}{{dx}} + 2y = 0
B. d2ydx2+ydydx+x=0\dfrac{{{d^2}y}}{{d{x^2}}} + y\dfrac{{dy}}{{dx}} + x = 0
C. d2ydx2+yx+siny=x2\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{y}{x} + \sin y = {x^2}
D. (1+x)dydxxy=1(1 + x)\dfrac{{dy}}{{dx}} - xy = 1

Explanation

Solution

In this question, we are given four differential equations and we have to tell out of these four which one is not a differential equation.
We know that a linear equation consisting of derivatives of the dependent variable to the independent variable is known as a linear differential equation.
Linear means, the power of the dependent variable must be one and any product of the dependent variable and its derivative should not be present in that equation. Also, the dependent variable should not be present as an angle of any function.

Complete step by step solution:
We are given four differential equations.
We have to find the differential equation which is not linear.
Now, as we know, a linear differential equation has the degree of the dependent variable to be one and it should not contain any product of the dependent variable and its derivative.
So, we’ll check these points in each of the given equations. So, our first equation is d2ydx2+xdydx+2y=0\dfrac{{{d^2}y}}{{d{x^2}}} + x\dfrac{{dy}}{{dx}} + 2y = 0 .
In this, the dependent variable is yy and its degree in each term is one and there is no product of yy and its derivatives present in the equation.
So, it is a linear differential equation.
Now, the second equation is d2ydx2+ydydx+x=0\dfrac{{{d^2}y}}{{d{x^2}}} + y\dfrac{{dy}}{{dx}} + x = 0 .
In this equation also dependent variable is yy and its degree in each term is one but the product of yy and its derivative i.e., dydx\dfrac{{dy}}{{dx}} is present. So, it is not a linear differential equation.
Next, the equation is d2ydx2+yx+siny=x2\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{y}{x} + \sin y = {x^2} .
In this equation also dependent variable is yy and its degree in each term is one but the dependent variable is present as an angle of a trigonometric function.
So, it is not a linear differential equation.
Now, our last equation is (1+x)dydxxy=1(1 + x)\dfrac{{dy}}{{dx}} - xy = 1 .
In this equation also dependent variable is yy and its degree in each term is one and there is no product of yy and its derivatives present in the equation.
So, it is a linear differential equation.
Hence, d2ydx2+ydydx+x=0\dfrac{{{d^2}y}}{{d{x^2}}} + y\dfrac{{dy}}{{dx}} + x = 0 and d2ydx2+yx+siny=x2\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{y}{x} + \sin y = {x^2} are not linear differential equations and thus, option (B)(B) and (C)(C) are correct.

Therefore, the correct option is B and C

Note: One must know the proper definition and properties of linear and non-linear differential equations to solve such problems.
We can solve these equations using different methods to get the value of yy i.e., dependent variable in terms of independent variable i.e., xx .
A question can have more than one correct option, unless stated, so it is important to check every option while solving an MCQ.