Question

For the differential equation ${\left(\frac{dy}{dx}\right)}^2-x\left(\frac{dy}{dx}\right)+y=0$, which one of the following is not its solution?

• A. $y=x-1$
• B. $4y=x^2$
• C. $y=x$
• D. $y=-x-1$

Answer 1

The correct option is C $y=x$

Given differential equation is ${\left(\frac{dy}{dx}\right)}^2-x\frac{dy}{dx}+y=0$,

To solve this we just need to substitute the options,

$A)$ $y=x-1⟹\frac{dy}{dx}=1$,

$1-x+x-1=0$,

Therefore $A$ is a solution of the given differential equation.

$B)$ $y=\frac{x^2}{4}⟹\frac{dy}{dx}=\frac{x}{2}$,

$\frac{x^2}{4}-\frac{x^2}{2}+\frac{x^2}{4}=0$,

Therefore $B$ is a solution to the given differential equation.

$C)$ $y=x⟹\frac{dy}{dx}=1⟹1-x+x=1\ne 0$,

Therefore $C$ is not a solution of the given differential equation.

$D)$ $y=-x-1⟹\frac{dy}{dx}=-1$

$1+x-1-x=0$,

Therefore $D$ is a solution of the given differential equation.