Let y = y(x) be the solution of the differential equation [x^3dy + (xy - 1)dx = 0,x 0,y(…

Question

Let $y = y(x)$ be the solution of the differential equation $${x^3}dy + (xy - 1)dx = 0,x > 0,y\left( {{1 \over 2}} \right) = 3 - \mathrm{e}$$. Then y (1) is equal to

Accepted Answer

$x^{3} d y+x y d x-d x=0$

$\Rightarrow \frac{d y}{d x}=\frac{1-x y}{x^{3}}$

$\Rightarrow \frac{d y}{d x}+\frac{y}{x^{2}}=\frac{1}{x^{3}}$

I.F. $=e^{\int \frac{d x}{x^{2}}}=e^{-\frac{1}{x}}$

$\therefore \quad y e^{-\frac{1}{x}}=\int \frac{e^{-\frac{1}{x}}}{x^{3}} d x$

For RHS put $-\frac{1}{x}=t \Rightarrow \frac{d x}{x^{2}}=d t$

$\therefore y e^{-\frac{1}{x}}=-\int t e^{t} d t$

$\Rightarrow y e^{-\frac{1}{x}}=-\left[t e^{t}-e^{t}\right]+c$

$\Rightarrow y e^{-\frac{1}{x}}=\frac{e^{-\frac{1}{x}}}{x}+e^{-\frac{1}{x}}+c$

$$
\downarrow y\left(\frac{1}{2}\right)=3-e
$$

$\Rightarrow(3-e) e^{-2}=2 e^{-2}+e^{-2}+c$

$\Rightarrow \quad c=-\frac{1}{e}$

For $y(1)$ put $x=1, c=-e^{-1}$ in equation (i) we get $y e^{-1}=e^{-1}+e^{-1}-e^{-1}$

$\Rightarrow y=1$

Solutions · 1