The order of the differential equation,whose | vedclass

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

Question

(B) Given the general solution: $y = (c_1 + c_2) \cos (x + c_3) - c_4 e^{x + c_5}$.We can simplify the expression by grouping the constants:Let $A = c

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Given the general solution: $y = (c_1 + c_2) \cos (x + c_3) - c_4 e^{x + c_5}$.We can simplify the expression by grouping the constants:Let $A = c_1 + c_2$ and $B = c_4 e^{c_5}$.Then the equation becomes $y = A \cos (x + c_3) - B e^x$.Expanding the cosine term: $y = A (\cos x \cos c_3 - \sin x \sin c_3) - B e^x$.$y = (A \cos c_3) \cos x - (A \sin c_3) \sin x - B e^x$.Let $K_1 = A \cos c_3$,$K_2 = -A \sin c_3$,and $K_3 = -B$.Thus,$y = K_1 \cos x + K_2 \sin x + K_3 e^x$.There are $3$ independent arbitrary constants $(K_1, K_2, K_3)$.The order of a differential equation is equal to the number of independent arbitrary constants in its general solution.Therefore,the order of the differential equation is $3$.

The order of the differential equation,whose general solution is given by $y = (c_1 + c_2) \cos (x + c_3) - c_4 e^{x + c_5}$,where $c_1, c_2, c_3, c_4$ and $c_5$ are arbitrary constants,is

Similar Questions

Determine the order and degree (if defined) of the differential equation $\frac{d^4y}{dx^4} + \sin(y''') = 0$.

The family of curves $y = Ax + A^3$ is represented by a differential equation of degree:

The order of the differential equation $\left(\frac{d^3 y}{d x^3}\right)^4+\left(\frac{d^2 y}{d x^2}\right)^2+\sin \left(\frac{d y}{d x}\right)+1=0$ is . . . . . . .

Which of the following equations is non-linear?

Determine the order and degree (if defined) of the differential equation: $(\frac{d^3y}{dx^3})^2 + (\frac{d^2y}{dx^2})^3 + (\frac{dy}{dx})^4 + y^5 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module