What should be used?

Relevant wiki: First Order Linear Differential Equations

$f\left( x \right)+ \displaystyle\int_{0}^{x}{t f\left( t\right) }\mathrm{d}t +x^2=0$

Differentiating the given equation, we get

$f'\left( x \right)+ x f\left( x\right) +2x=0 \\ \Rightarrow \dfrac{f'\left( x \right)}{f\left( x\right) +2}=-x$

Now integrating both the sides, we get

$\displaystyle\int{\dfrac{f'\left( x \right)}{f\left( x\right) +2}}\mathrm{d}x=\displaystyle\int{-x }\mathrm{d}x \\ \Rightarrow \ln{\left(f\left( x\right) +2 \right)}=\dfrac{-x^2}{2}+C \\ \Rightarrow f\left( x\right) +2 =ke^{-x^2/2} \\ \Rightarrow f\left(x\right)=ke^{-x^2/2}-2$

Now, using the original equation, if we put $x=0$ , then $f\left(x\right)=0$

Putting these values in the last equation we get $k=2$ .

Hence $\boxed{ f\left(x\right)=2e^{-x^2/2}-2}$

By differentiating $f\left(x\right)$ we can easily find out that the function has only one maxima at $x=0$ . This means that the function touches the x axis at only one point.

Also, by putting -x in place of x, we can easily find out that $f\left(x\right)$ = $f\left(-x\right)$ , which means that this is an even function.

Now $\displaystyle\lim _{x \rightarrow \infty }{f\left( x \right) } =\displaystyle\lim _{x \rightarrow \infty }{2e^{-x^2/2}-2}=0-2=\boxed{-2}$ .

Hence, the option $\displaystyle\lim _{x \rightarrow \infty }{f\left( x \right) } =-2$ is correct.

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Alternate Solution

Differentiating the given equation, we get

$f'\left( x \right)+ x f\left( x\right) +2x=0$

Let $y=f\left(x\right) \Rightarrow f'\left(x\right)=y'=\dfrac{\mathrm{d}y}{\mathrm{d}x}$ and the equation becomes

$\dfrac{\mathrm{d}y}{\mathrm{d}x}+ xy +2x=0$

Now using integrating factor method of solving the first order linear differential equations (of type $\dfrac{\mathrm{d}y}{\mathrm{d}x}+Py=Q$ ; P and Q are constants or functions of x) , we get:

$\text{Integrating factor}=\exp\left({\displaystyle\int{P}\mathrm{d}x}\right)=\exp\left({\displaystyle\int{x}\mathrm{d}x}\right)=\exp\left(\dfrac{x^2}{2}\right)$

The general solution to these types of differential equations is

$y=\exp\left( \displaystyle\int{-P}\mathrm{d}x \right) \cdot \displaystyle\int{\left( Q\cdot \exp(\displaystyle \left( \int{P} \mathrm{d}x \right) \right)}\mathrm{d}x+C$

Hence $y=f\left(x\right)=\exp\left(\dfrac{-x^2}{2}\right) \displaystyle\int{\left(-2x\cdot \exp\left(\dfrac{-x^2}{2}\right) \right)}\mathrm{d}x+C$

Integrating by substituting $x^2=t$ , we get:

$\boxed{f\left(x\right)=Ce^{-x^2/2}-2}$

Now, as done before, using the original equation, if we put $x=0$ , then $f\left(x\right)=0$

Putting these values in the last equation we get $C=2$ .

$\boxed{f\left(x\right)=2e^{-x^2/2}-2}$

Then follow the same procedure as done in the first solution to get to the correct answer.