Let us consider a curve
Question:

Let us consider a curve, $y=f(x)$ passing through the point $(-2,2)$ and the slope of the tangent to the curve at any point $(x, f(x))$ is given by $f(x)+x f^{\prime}(x)=x^{2}$. Then :

1. $x^{2}+2 x f(x)-12=0$

2. $x^{3}+x f(x)+12=0$

3. $x^{3}-3 x f(x)-4=0$

4. $x^{2}+2 x f(x)+4=0$

Correct Option: , 3

Solution:

$\mathrm{y}+\frac{\mathrm{xdy}}{\mathrm{dx}}=\mathrm{x}^{2}$ (given)

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{x}$

If $=e^{\int \frac{1}{x} d x}=x$

Solution of DE

$\Rightarrow \mathrm{y} \cdot \mathrm{x}=\int \mathrm{x} \cdot \mathrm{x} \mathrm{dx}$

$\Rightarrow \mathrm{xy}=\frac{\mathrm{x}^{3}}{3}+\frac{\mathrm{c}}{3}$

Passes through $(-2,2)$, so

$-12=-8+c \Rightarrow c=-4$.

$\therefore 3 x y=x^{3}-4$

ie. $3 x \cdot f(x)=x^{3}-4$