JEE Main 2021 (Online) 27th August Evening ShiftQuestion 1
If the solution curve of the differential equation $\left(2x-10y^{3}\right)dy+y\,dx=0$, passes through the points $(0,1)$ and $(2,\beta)$, then $\beta$ is a root of the equation :
Differential Equation
Solution
Concept. The equation is linear in $x$ when $y$ is the independent variable. Formulas used. (1) $\dfrac{dx}{dy}+Px=Q$: $x\,e^{\int P\,dy}=\int Q\,e^{\int P\,dy}\,dy+C$Step 1 - Linear form. $$\frac{dx}{dy}=\frac{10y^3-2x}{y}\ \Longrightarrow\ \frac{dx}{dy}+\frac{2}{y}x=10y^2,\qquad\text{IF}=y^2.$$ Step 2 - Integrate. By (1), $xy^2=\int10y^4\,dy=2y^5+C$. Step 3 - Through $(0,1)$. $0=2+C$, so $C=-2$ and $xy^2=2y^5-2$. Step 4 - Through $(2,\beta)$. $2\beta^2=2\beta^5-2$, i.e. $\beta^5-\beta^2-1=0$. Answer: (C)