Question

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

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

Answer 1

The correct option is A $y=2x-4$

${\left(\frac{dy}{dx}\right)}^2-x\left(\frac{dy}{dx}\right)+y=0$

Equation-1: $y=2x-4$

$\Rightarrow \frac{dy}{dx}=2\Rightarrow {\left(\frac{dy}{dx}\right)}^2-x\left(\frac{dy}{dx}\right)+y=2^2-2\ast x+y=4-2x+2x-4=0$

Equation-2: $y=2x^2-4$

$\Rightarrow \frac{dy}{dx}=4x\Rightarrow {\left(\frac{dy}{dx}\right)}^2-x\left(\frac{dy}{dx}\right)+y={4x}^2-\left(4x\right)\ast x+y=16x^2-4x^2+2x^2-4=14x^2-4$

not zero.

Equation-3: $y=2$

$\Rightarrow \frac{dy}{dx}=0\Rightarrow {\left(\frac{dy}{dx}\right)}^2-x\left(\frac{dy}{dx}\right)+y=0^2-0\ast x+y=y=2$

not zero.

Hence, only A is the solution of given differential equation.