The differential equation frac{d^3y}{dx^3}-5y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) differential equation is said to be linear if the dependent variable $y$ and its derivatives occur only in the first degree and are not multiplied

Vedclass · Accepted Answer

Non-linear equation of degree $1$

Vedclass · Answer

(B) differential equation is said to be linear if the dependent variable $y$ and its derivatives occur only in the first degree and are not multiplied together.In the given equation $\frac{d^3y}{dx^3}-5y \frac{dy}{dx}+xy=0$,the term $y \frac{dy}{dx}$ involves the product of the dependent variable $y$ and its derivative $\frac{dy}{dx}$.Because of this product,the equation is non-linear.The order of the highest derivative $\frac{d^3y}{dx^3}$ is $3$.The degree of the highest derivative is $1$.Therefore,it is a non-linear differential equation of order $3$ and degree $1$.

The differential equation $\frac{d^3y}{dx^3}-5y \frac{dy}{dx}+xy=0$ represents :-

Similar Questions

The differential equation $\frac{d^3y}{dx^3} + 2\left[ 1 + \frac{d^2y}{dx^2} \right] = 1$ has order and degree as:

The order and the degree of the differential equation $\sqrt{1 + (\frac{d^2y}{dx^2})^2} = \sqrt[3]{x + (\frac{dy}{dx})^3}$ are respectively . . . . . . and . . . . . . .

The differential equation representing the family of curves ${y^2} = 2c(x + \sqrt{c}),$ where $c$ is a positive parameter,is of

The sum of the order and degree of the differential equation $x\left(\frac{d^2 y}{d x^2}\right)^{\frac{1}{2}}=\left(1+\frac{d y}{d x}\right)^{\frac{4}{3}}$ is

The differential equation representing the family of curves $y^{2}=2 d(x+\sqrt{d})$, where $d$ is a parameter, is of

Vedclass Test Series

Exam Paper Generator

Online Exam Module