Let $f$ be a real-valued differentiable function on $R$ (the set of all real numbers) such that $f(1

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

Professor PC Thomas Classes · Answer

(9) : The differential equation is\ $x \frac{d y}{d x}-y=-x^{3} \Rightarrow x d y-y d x=-x^{3} d x$\ $\Rightarrow \frac{x d y-y d x}{x^{2}}=-x d x$\ $\Rightarrow d\left(\frac{y}{x}ight)=-x d x \Rightarrow \frac{y}{x}=-\frac{x^{2}}{2}+k$\ $\Rightarrow y=-\frac{x^{3}}{2}+k x=-\frac{x^{3}}{2}+\frac{3}{2} x$ as $f(1)=1$\ $f(-3)=9$\