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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find a particular solution satisfying the given condition$\large\frac{dy}{dx}=$$y\tan x;\;y=1\;when\;x=0$

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  • $\tan x$ can be found by substitution method.
Step 1:
Integration of tanx can be written as integration of $\large\frac{\sin x}{\cos x}.$
Let $\cos x$ be = t,so $dt = -\sin xdx$
Hence integration of $\tan x =\int\large\frac{ dt}{t }$$=\log t. $
Now substituting the value of $t$ we get, -$\log|\cos x| = \log|\sec x$|
Given $\large\frac{dy}{dx }=$$ y \tan x$
Seperating the variables we get
$\large\frac{dy}{y }=$$ \tan xdx$
Step 2:
Integrating on both sides , and using the information in the tool box we get
$\log y =\log|\sec x| +\log c$
$\log y -\log|\sec x| = \log c$
$y\sec x = C$
Step 3:
To find the values of C, let us substitute the values of $x = 0$ and $ y = 1$ we get
$1.\log|\sec 0| = C$, but $\sec 0$ is 1
$y\sec x$ = 1 is the required solution.
answered Aug 14, 2013 by sreemathi.v
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