The order and degree of the following differential equation are, respectively: \[ \frac{d^4y}{dx^4} + 2 \frac{d^2y}{dx^2} + y^2 = 0. \]

Question

Given that the solution of \[ \frac{d^2y}{dx^2} + \alpha \frac{dy}{dx} + \beta y = -e^{-x} \] is \[ y(x) = C_1 e^{-x} + C_2 e^{2x} + x e^{-x}, \] find the values of $\alpha$ and $\beta$.

Answer 1

The given equation is: \[ \frac{d^4y}{dx^4} + 2 \frac{d^2y}{dx^2} + y^2 = 0. \] - Order: The order of a differential equation is determined by its highest derivative. In this equation, the highest derivative is $\frac{d^4y}{dx^4}$, hence the order is 4. - Degree: The degree is the power of the highest order derivative once fractional and radical terms are eliminated. For the highest derivative, $\frac{d^4y}{dx^4}$, the exponent is 1. Thus, the degree is 1. Consequently, the order is 4 and the degree is 1.

The order and degree of the following differential equation are, respectively: \[ \frac{d^4y}{dx^4} + 2 \frac{d^2y}{dx^2} + y^2 = 0. \]

Show Hint

The Correct Option is D

Solution and Explanation

Top Questions on Differential Equations

Questions Asked in CBSE Class XII exam