Question

The differential equation of a curve $x=f\left(y\right)$ is given by $\frac{dy}{dx}=\frac{1}{y^2(y^3-x)}$ If the curve passes through the point $(-2,1)$, then the value of $f(-1)$ is

• A. 4
• B. 3
• C. -3
• D. -4

Answer 1

The correct option is D -4

$\begin{array}{c}\frac{dy}{dx}=\frac{1}{y^2\left(y^3-x\right)}\\ \frac{dx}{dy}+x{y}^2=y^5\\ x{e}^{\sqrt{y^2}dy}=\int e^{\frac{y^3}{3}}{y}^{5}dy\\ Now,let\frac{y^3}{3}=t{y}^2=\frac{dt}{dy}\\ x{e}^{\frac{y^3}{3}}=3\left(t{e}^t-e^t\right)+C\\ x{e}^{\frac{y^3}{3}}=3t{e}^t-3e^t+C\\ x{e}^{\frac{y^3}{3}}=y^3{e}^{\frac{y^3}{3}}-3e^{\frac{y^3}{3}}C\\ -2e^{\frac{1}{3}}=e^{\frac{1}{3}}-3e^{\frac{1}{3}}+C\\ \Rightarrow C=0\\ Now,x=y^3-3\\ f\left(y\right)=y^3-3\\ f\left(-1\right)=4\\ Hence,optionDiscorrectanswer.\end{array}$