The differential equation whose solution is $y = {Ax}^{5} + {Bx}^{4}$ is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

$x^{2} \frac{d^{2} y}{{dx}^{2}} + 8 x \frac{dy}{dx} + 20 y = 0$

$x^{2} \frac{d^{2} y}{{dx}^{2}} + 8 x \frac{dy}{dx} - 20 y = 0$

$x^{2} \frac{d^{2} y}{{dx}^{2}} - 8 x \frac{dy}{dx} + 20 y = 0$

$x^{2} \frac{d^{2} y}{{dx}^{2}} - 8 x \frac{dy}{dx} - 20 y = 0$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} (1) G i v e n y = A x^{5} + B x^{4} - - (1) \\ y_{1} = 5 A x^{4} + 4 B x^{3} - - (2) \\ y_{2} = 20 A x^{3} + 12 B x^{2} - - (3) \\ (2) \times 4 x - (3) \times x^{2} \Rightarrow 4 x y_{1} - x^{2} y_{2} = 4 B x^{4} \\ \Rightarrow B x^{4} = \frac{4 x y_{1} - x^{2} y_{2}}{4} \\ (3) \times x^{2} - (2) \times 3 x \Rightarrow x^{2} y_{2} - 3 x y_{1} = 5 A x^{5} \\ \Rightarrow A x^{5} = \frac{x^{2} y_{2} - 3 x y_{1}}{5} \\ F r o m (1) \\ y = \frac{x^{2} y_{2} - 3 x y_{1}}{5} + \frac{4 x y_{1} - x^{2} y_{2}}{4} \\ \Rightarrow 20 y = 4 x^{2} y_{2} - 12 x y_{1} + 20 x y_{1} - 5 x^{2} y_{2} \\ \Rightarrow x^{2} y_{2} - 8 x y_{1} + 20 y = 0 \end{array}$