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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The general solution of $e^x\cos y\;dx-e^x\sin y\;dy=0$ is:\begin{array}{1 1}(A)\;e^x\cos y=c & (B)\;e^x\sin y=c\\(C)\;e^x=c\;\cos y & (D)\;e^x=c\;\sin y\end{array}

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Toolbox:
  • If a linear differential equation is of the form $\large \frac{dy}{dx}$$=f(x)$ then it can be solved by seperating the variables and then integrating
Given $e^x \cos y \;dx-e^x \sin y\; dy=0$
Dividing by $e^x$ we get
$\cos y \;dx=\sin y \; dy$
$=>\large\frac{\sin y}{\cos y}$$dy=dx$
(ie)$ \tan y \;dy=dx$
Integrating on both sides we get,
$\int \tan y \;dy=\int dx$
$=> -\log_e |\cos y|=x+c$
$=>\log_e \bigg(\large\frac{1}{\cos y}\bigg)$$=x+c$
$=>e^x=\large\frac{1}{\cos y}+c$
$\cos y.e^x=c$
Hence $A$ is the correct answer
answered May 16, 2013 by meena.p
 

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