Question

Verify that y = log x + c is a solution of the differential equation $x\frac{d^{2}y}{dx^{2}}+ \frac{dy}{dx}=0$.

Answer 1

To verify that y = log x + c is a solution of the differential equation $x\frac{d^{2}y}{dx^{2}}+ \frac{dy}{dx}=0$.
1) differentiate equation once

2) differentiate twice ,from equation ,1st order derivative try to bring final DE

$y = log x + c \\ \\ \frac{dy}{dx} = \frac{1}{x} \\ \\ \frac{ {d}^{2} y}{ {dx}^{2} } = - \frac{1}{ {x}^{2} } \\ \\ \frac{ {d}^{2} y}{ {dx}^{2} } = \frac{1}{ {x} } ( \frac{ - 1}{x} ) \\ \\ \frac{ {d}^{2} y}{ {dx}^{2} } = \frac{dy}{dx} ( \frac{ - 1}{x} ) \\ \\ x\frac{ {d}^{2} y}{ {dx}^{2} } = - \frac{dy}{dx} \\ \\ x\frac{ {d}^{2} y}{ {dx}^{2} } + \frac{dy}{dx} = 0$
hence proved

Answer 2

Answer:

Step-by-step explanation:

Final answer is dy/dx= 10^y ln10