Let f be a real valued differentiable function on R (the set of all real numbers) such that f (1) = 1. If the y‑intercept of the tangent at any point P(x, y) on the curve y = f (x) is equal to the cube of the abscissa of P, then the value of f (

Engineering

Mathematics

Linear Differential Equation

Question

Let f be a real valued differentiable function on R (the set of all real numbers) such that f (1) = 1. If the y‑intercept of the tangent at any point P(x, y) on the curve y = f (x) is equal to the cube of the abscissa of P, then the value of f (–3) is equal to

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$(Y - y) = \frac{dy}{dx} (X - x)$

Put $X = 0, Y = y - \frac{xdy}{dx}$

Now $y - \frac{xdy}{dx} - y = - x^{3}$

$\Rightarrow \frac{xdy}{dx} - y = - x^{3} \Rightarrow \frac{dy}{dx} - \frac{1}{x} \cdot y = - x^{2}$ .....(1)

$∴ I . F . = \frac{1}{x}$

Now general solution is

$\frac{y}{x} = - \int x^{2} \times \frac{1}{x} dx + C \Rightarrow \frac{y}{x} = \frac{- x^{2}}{2} + C$

As $f (1) = 1 \Rightarrow C = \frac{3}{2}$

$∴ y = - \frac{1}{2} x^{3} + \frac{3}{2} x$

Hence $f (- 3) = \frac{27}{2} - \frac{9}{2} = \frac{18}{2} = 9$

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