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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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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 $\frac{1}{2}.$

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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}$
answered May 9, 2013 by meena.p
 
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