Differential Equations Question 1
Question 1 - 24 January - Shift 1
Let $y=y(x)$ be the solution of the differential equation $x^{3} d y+(x y-1) d x=0, x>0$, $y(\frac{1}{2})=3-e$. Then $y(1)$ is equal to
(1) 1
(2) e
(3) 2-e
(4) 3
Show Answer
Answer: (1)
Solution:
Formula: Linear differential equations of first order
$\frac{d y}{d x}=\frac{1-x y}{x^{3}}=\frac{1}{x^{3}}-\frac{y}{x^{2}}$
$\frac{d y}{d x}+\frac{y}{x^{2}}=\frac{1}{x^{3}}$
IF $=e^{\int \frac{1}{x^{2}} d x}=e^{-\frac{1}{x}}$
$y \cdot e^{-\frac{1}{x}}=\int e^{-\frac{1}{x}} \cdot \frac{1}{x^{3}} d x(.$ put $.-\frac{1}{x}=t)$
$y . e^{-\frac{1}{x}}=-\int e^{t} \cdot t d t$
$y=\frac{1}{x}+1+Ce^{\frac{1}{x}}$
Where $C$ is constant
Put $x=\frac{1}{2}$
$3-e=2+1+Ce^{2}$
$C=-\frac{1}{e}$
$y(1)=1$
