Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS4 Marks
Question
Find the general solution of (1 + tany) (dx - dy) + 2xdy = 0.
✓
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]$ $\Rightarrow\text{x}(\sin\text{y}+\cos\text{y})=\sin\text{y}+\text{C}\text{e}^{-\text{y}}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.