Question

If the solution of $y=x\frac{dy}{dx}+\frac{dy}{dx}-{\left(\frac{dy}{dx}\right)}^2;\frac{d^{2}y}{d{x}^2}\ne 0$ is $y=f\left(x\right)$ and it is defined as $f:R\to R$ then $f\left(x\right)$ is

• A. one-one
• B. into
• C. even
• D. onto

Answer 1

The correct option is B into

$y=x\frac{dy}{dx}+\frac{dy}{dx}-{\left(\frac{dy}{dx}\right)}^2$
Differentiating w.r.t. $x$,
$y^{\prime }=x{y}^{\prime \prime }+y^{\prime }+y^{\prime \prime }-2y^{\prime }{y}^{\prime \prime }\Rightarrow 0=(x+1-2y^{\prime })y^{\prime \prime }\Rightarrow y^{\prime }=\frac{x+1}{2}$
Putting it in the given differential equation we get,
$y=(x+1)\times \frac{(x+1)}{2}-\frac{(x+1{)}^2}{4}\Rightarrow y=\frac{(x+1{)}^2}{4}$
So,
$f\left(x\right)=\frac{(x+1{)}^2}{4}$
The function is many one.

As $f\left(x\right)\ne f(-x)$ so, it is not even.

The range of function is $[0,\infty )$
Therefore, the function is into.