JEE Main 2021 (Online) 1st September Evening ShiftQuestion 1
If $y=y(x)$ is the solution curve of the differential equation $x^{2}dy+\left(y-\frac1x\right)dx=0$; $x>0$ and $y(1)=1$, then $y\left(\frac12\right)$ is equal to :
Differential Equation
Solution
Concept. Divide by $x^2\,dx$ to get a linear equation in $y$. Formulas used. (1) $y'+Py=Q$: $y\,e^{\int P\,dx}=\int Q\,e^{\int P\,dx}\,dx+C$(2) $\int u\,e^u\,du=(u-1)e^u$Step 1 - Linear form. $$\frac{dy}{dx}+\frac{y}{x^2}=\frac{1}{x^3},\qquad\text{IF}=e^{\int x^{-2}dx}=e^{-1/x}.$$ Step 2 - Integrate. Put $u=-\frac1x$, $du=\frac{dx}{x^2}$, $\frac1x=-u$: $$\int\frac{e^{-1/x}}{x^3}\,dx=\int(-u)e^u\,du=(1-u)e^u=\left(1+\frac1x\right)e^{-1/x}.$$ So $y\,e^{-1/x}=\left(1+\frac1x\right)e^{-1/x}+C$, i.e. $$y=1+\frac1x+C\,e^{1/x}.$$ Step 3 - Use $y(1)=1$. $1=2+Ce$, so $C=-\frac1e$. Step 4 - Value. $y\left(\frac12\right)=1+2-\frac{e^2}{e}=3-e$. Answer: (B)