Question

A solution of the differential equation ${\left(\frac{dy}{dx}\right)}^2-x\frac{dy}{dx}+y=0$ is

• A. $y=2x$
• B. $y=-2x$
• C. $y=2x-4$
• D. $y=2x+4$

Answer 1

Answer: (C)

Let $y=2x-4$ be a solution of the differential equation
${\left(\frac{dy}{dx}\right)}^2-x\left(\frac{dy}{dx}\right)+y=0$
Now, we cross check this solution
LHS ${\left(\frac{dy}{dx}\right)}^2-x\left(\frac{dy}{dx}\right)+y$
$={\left\{\frac{d}{dx}(2x-4)\right\}}^2-x\left\{\frac{d}{dx}(2x-4)\right\}+(2x-4)$
$=(2{)}^2-x(2)+2x-4$
$=4-2x+2x-4$
$=0=$ RHS
$\therefore y=2x-4$ be a required solution.