Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Find the general solution of (1 + tany) (dx - dy) + 2xdy = 0.
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Answer
Given difference equation is (1 + tan y) (dx - dy) + 2xdy = 0
Dividing both sides of above equation by dy, we get
$(1+\tan\text{y})\Big(\frac{\text{dx}}{\text{dy}}-1\Big)+2\text{x}=0$
$\Rightarrow(1+\tan\text{y})\frac{\text{dx}}{\text{dy}}-(1+\tan\text{y})+2\text{x}=0$
$\Rightarrow(1+\tan\text{y})\frac{\text{dx}}{\text{dy}}+2\text{x}=(1+\tan\text{y})$
$\Rightarrow\frac{\text{dx}}{\text{dy}}+\frac{2\text{x}}{1+\tan\text{y}}=1$
This is a linear differential equation.
On comparing it with $\frac{\text{dx}}{\text{dy}}+\text{Px}=\text{Q}, $ we get
$\text{P}-\frac{2}{1+\tan\text{y}},\text{Q}=1$
$\text{I.F.}=\text{e}^{\int\frac{2}{1+\tan\text{y}}\text{dy}}=\text{e}^{\int\frac{2\cos\text{y}}{\cos\text{y}+\sin\text{y}}\text{dy}}$
$=\text{e}^{\frac{\cos\text{y}6\sin\text{y}+\cos\text{y}-\sin\text{y}}{\cos\text{y}+\sin\text{y}}\text{dy}}$
$=\text{e}^{\Big(1+\frac{\cos\text{y}-\sin\text{y}}{\cos\text{y}+\sin\text{y}}\Big)\text{dy}}$
$=\text{e}^{\text{y}+\log(\cos\text{y}+\sin\text{y})}$
$=\text{e}^\text{y}.(\cos\text{y}+\sin\text{y})$
Thus, the general solution is,
$\text{x}.\text{e}^\text{y}(\cos\text{y}+\sin\text{y})=\int1.\text{e}^\text{y}(\cos\text{y}+\sin\text{y})\text{dy}+\text{C}$
$\Rightarrow\text{x}.\text{e}^\text{y}(\cos\text{y}+\sin\text{y})=\int\text{e}^\text{y}(\sin\text{y}+\cos\text{y})\text{dy}+\text{C}$
$\Rightarrow\text{x}.\text{e}^\text{y}(\cos\text{y}+\sin\text{y})=\text{e}^\text{y}\sin\text{y}+\text{C}$ $\Big[\because\int\text{e}^\text{x}\left\{\text{f}(\text{x})+\text{f}'(\text{x})\right\}\text{dx}=\text{e}^{\text{x}}\text{f}(\text{x})\Big]$
$\Rightarrow\text{x}(\sin\text{y}+\cos\text{y})=\sin\text{y}+\text{C}\text{e}^{-\text{y}}$
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