# If y(x) is a solution of $\Large\frac{2-\sin x}{1-y}\bigg(\frac{dy}{dx}\bigg)\normalsize=-\cos x$ and y(0)=1,then find the value of $\frac{1}{2}.$

Toolbox:
• A linear differential equation of the form.
• $\large\frac{dy}{dx}$$=f(x), can be solved by seperating the variables and then integrating it. • \int \large\frac{dx}{x+a}$$=\log |x+a|+c$
Step 1:
$\large\frac{2-\sin x}{1-y}\bigg(\frac{dy}{dx}\bigg)$$=-\cos x Now seperating the variables we get, \large\frac{dy}{1-y}=\frac{-\cos x dx}{2 -\sin x} Let us integrate on both sides, \int \large\frac{dy}{1-y}=\int \frac{-\cos x dx}{2 -\sin x} we know \int \large\frac{dy}{1-y}$$=-\log (1-y)$
$=\log \bigg(\large\frac{1}{1-y}\bigg)$-----(1)
Consider $\int \large\frac{-\cos x dx}{2 -\sin x}$
Put $2-\sin x=t$. Differentiating with respect to t,
$-cos xdx=dt$
Now substituting this we get
$\int \large\frac{dt}{t}$$=\log t Now substituting for t we get \log |2 -\sin x|-----(2) Step 2: Hence combining equ (1) and equ(2) we get, \log \bigg( \large\frac{1}{1-y}\bigg)$$=\log |2-\sin x|+\log c$
$\log \bigg| \large\frac{1}{1-y}\bigg|$$=\log |c(2-\sin x)| => \large\frac{1}{1-y}$$=c(2-\sin x)$
Step 3:
Given $y(0)=-1$
=>When $x=0,y=1$
Substituting this in the above equation we get
$\large\frac{1}{1+1}$$=c(2-\sin 0) But \sin 0=0 =\large\frac{1}{2}$$=2c=>c=\large\frac{1}{4}$
Hence $\large\frac{1}{1-y}=\frac{1}{4}$$(2-\sin x) We are asked to find y(1/2) =>x=1/2 and we have to find the value of y \large\frac{1}{1-y}=\frac{1}{4}$$(2+1)$
$\large\frac{1}{1-y}=\frac{3}{4}$$=>3-3y=4 =>y=\large\frac{1}{3} Hence x=\large\frac{1}{2},$$y=\large\frac{1}{3}$