Mathematics - Differential Equation Question with Solution | TestHub

The equation of the tangent at $(x,y)$ to the given curve $y=f(x)$ is $$\begin{array}{c}Y-y=\frac{dy}{dx}(X-x)\\ Y\text{-intercept }=y-x\frac{dy}{dx}\end{array}$$ According to the question $$x^3=y-x\frac{dy}{dx}$$ $$\Rightarrow \frac{dy}{dx}-\frac{y}{x}=-x^2$$ which is linear in $x$. $$\text{ IF }=e^{\int \frac{-1}{x}dx}=\frac{1}{x}$$ $\therefore$ Required solution is $$\begin{array}{rl}& y\cdot \frac{1}{x}=\int -x^2\cdot \frac{1}{x}dx\\ & \Rightarrow \frac{y}{x}=\frac{-x^2}{2}+c\\ & \Rightarrow y=\frac{-x^3}{2}+cx\\ & \text{ At }x=1,y=1\text{, }\\ & 1=\frac{-1}{2}+c\\ & \Rightarrow c=\frac{3}{2}\\ & \end{array}$$ Now, $\begin{array}{rl}f(-3)& =\frac{27}{2}+\frac{3}{2}(-3)\\ & =\frac{27-9}{2}=9\end{array}$