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FInd the general solution of $(1+\tan y)(dx-dy)+2xdy=0$

$\begin{array}{1 1} (A)\;x(\sin y+\cos y)=\sin y+ce^{-y} \\ (B)\;x(\sin y-\cos y)=\sin y-ce^{-y}\\(C)\;x(\sin y-\cos y)=\sin y+ce^{-y} \\ (D)\;x(\sin y+\cos y)=\sin y-c\end{array}$

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  • A linear differential equation of the form $\large\frac{dy}{dx}$$=f(x)$ can be solved by seperating the variables and then integrating it.
  • A linear differential equation of the form $\large\frac{dx}{dy}$$+Px=Q$ . has the general solution $xe^{\int pdy}=\int Q e^{\int pdy}.dy+c$
  • $\int e^{ax}.\sin bx dx=\large\frac{e^{\Large ax}}{a^2+b^2}$$[a \sin bx-b \cos bx]+c$
Step 1:
Given $ (1+\tan y)(dx-dy)+2xdy=0$
This can be written as
$dx+\tan y dx-dy-\tan y dy+2x dy=0$
(ie) $-dy(1+\tan y -2x)+dx(1+\tan y)=0$
$dy(1+\tan y-2x)=(1+\tan y)dx$
$\large\frac{dy}{dx}=\frac{1+\tan y}{1+\tan y -2x}$
(or) $\large\frac{dx}{dy}=\frac{1+\tan y-2x}{1+\tan y }$
$\large\frac{dx}{dy}=\frac{1+\tan y}{1+\tan y }-\frac{2x}{1+\tan y}$
Therefore $ I.F=\sin y+\cos y.e^y$
Hence the required solution is
$x \times I.F=\int Q \times I.F \;dy+c$
$x. \sin y+\cos y.e^y=\int 1 \times \sin y+\cos y.e^y dy+c$
$x .e^y( \sin y+\cos y)=\int e^y( \sin y+\cos y) dy+c$
$x .e^y( \sin y+\cos y)=\int e^y \sin y dy +\int e^y \cos y dy+c$
Step 2:
Consider $\int e^y \sin y dy$
This is of the form
$ \int e^{ax} \sin bx dx =\large\frac{e^{\Large ax}}{a^2+b^2}$$(a \sin bx-b \cos bx)$
Here $ a=1\; and\; b=1$
Hence $\int e^y \sin y dy=\large\frac{\Large e^y}{2}$$(\sin x-\cos x)$
Similarly $ \int e^{ax} \cos bx dx=\large\frac{e^{ax}}{a^2+b^2}$$(a \cos bx +b \sin bx)$
Here $a=1\; and \; b=1$
Hence $\int e^y \cos y dy=\large\frac{\Large e^y}{2}$$(\cos y+\sin y)$
Now combining and substituting in the required solution we get,
$xe^y(\sin y+\cos y)=\large\frac{\Large e^y}{2}$$ (\sin y-\cos y)$$+\large\frac{\Large e^y}{2}$$(\cos y+\sin y)+c$
$x e^y(\sin y+\cos y)=\large\frac{\Large 2.e^y}{2}$$\sin y +c$
$x e^y(\sin y+\cos y)=e^y\sin y +c$
Dividing throughout by $e^y$ we get,
$x(\sin y+\cos y)=\sin y+ce^{-y}$
answered May 23, 2013 by meena.p
edited May 23, 2013 by meena.p
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