Let us consider a curve, y = f(x) passing through the point (-2, 2) and the slope of the t…

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) + xf'(x) = x2. Then :

Accepted Answer

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

$$ \Rightarrow {{dy} \over {dx}} + {y \over x} = x$$

If $${e^{\int {{1 \over x}dx} }} = x$$

Solution of DE

$$ \Rightarrow y\,.\,x = \int {x\,.\,x\,dx} $$

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

Passes through (-2, 2), so

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

$$\therefore$$ 3xy = x3 - 4

i.e. 3x . f(x) = x3 - 4

Solutions · 1