Find the P.I for the given Differential Equation.$16x^2 \frac{d^2 y}{dx^2}+(2x+4) \frac{dy}{dx}+4x(x+4) \frac{d^2 y}{dx^2}+2y=cos⁡(log⁡(2x+4))$

Question

TuteeHUB · Accepted Answer

$\frac{1}{2}×(-sin⁡(log⁡(2x+4))-cos⁡(log⁡(2x+4)))$

TuteeHUB · Answer

$\frac{1}{4}×(-sin⁡(log⁡(2x+4))+cos⁡(log⁡(2x+4)))$

TuteeHUB · Answer

$\frac{1}{4}×(-sin⁡(log⁡(2x+4))-cos⁡(log⁡(2x+4)))$

TuteeHUB · Answer

$\frac{1}{2}×(-sin⁡(log⁡(2x+4))-cos⁡(log⁡(2x+4)))$

TuteeHUB · Answer

$\frac{1}{2}×(-sin⁡(log⁡(2x+4))+cos⁡(log⁡(2x+4)))$

1.	Find the P.I for the given Differential Equation.\(16x^2 \frac{d^2 y}{dx^2}+(2x+4) \frac{dy}{dx}+4x(x+4) \frac{d^2 y}{dx^2}+2y=cos⁡(log⁡(2x+4))\)
A.	\(\frac{1}{4}×(-sin⁡(log⁡(2x+4))+cos⁡(log⁡(2x+4)))\)
B.	\(\frac{1}{4}×(-sin⁡(log⁡(2x+4))-cos⁡(log⁡(2x+4)))\)
C.	\(\frac{1}{2}×(-sin⁡(log⁡(2x+4))-cos⁡(log⁡(2x+4)))\)
D.	\(\frac{1}{2}×(-sin⁡(log⁡(2x+4))+cos⁡(log⁡(2x+4)))\)
Answer» C. \(\frac{1}{2}×(-sin⁡(log⁡(2x+4))-cos⁡(log⁡(2x+4)))\)

Find the P.I for the given Differential Equation.\(16x^2 \frac{d^2 y}{dx^2}+(2x+4) \frac{dy}{dx}+4x(x+4) \frac{d^2 y}{dx^2}+2y=cos⁡(log⁡(2x+4))\)