Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS4 Marks
Question
Find the general solution of $\frac{\text{dy}}{\text{dx}}-3\text{y}=\sin2\text{x}.$
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Answer
We have, $\frac{\text{dy}}{\text{dx}}-3\text{y}=\sin2\text{x}$ This is a linear differential equation. On comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get $\text{P}=-3,\text{Q}=\sin2\text{x}$ $\text{I.F.}=\text{e}^{-3\text{x}}$ So, the general solution is, $\text{y}.\text{e}^{-3}\text{x}=\int\text{e}^{-3\text{x}}\sin2\text{xdx}+\text{c}\ ......(\text{i})$ Now $\int\text{e}^{-3\text{x}}\sin2\text{xdx}=-\text{e}^{-3\text{x}}\frac{\cos2\text{x}}{2}-\int3\text{e}^{-3\text{x}}\frac{\cos2\text{x}}{2}\text{dx}$ $=\text{e}^{-3\text{x}}\frac{\cos2\text{x}}{2}-\frac{3}{2}\Big[\text{e}^{-3\text{x}}\frac{\sin2\text{x}}{2}+\int3\text{e}^{-3\text{x}}\frac{\sin2\text{x}}{2}\text{dx}\Big]+\text{C}$ $=\text{e}^{-3\text{x}}\frac{\cos2\text{x}}{2}-\frac{3}{4}\text{e}^{-3\text{x}}\sin2\text{x}-\frac{9}{4}\int\text{e}^{-3\text{x}}\sin2\text{xdx}+\text{C}$ $\Rightarrow-\log(1-\text{y})^2=\log\text{x}+\log\text{C}$ $\Rightarrow-\log\Big(1-\frac{\text{y}^2}{\text{x}^2}\Big)=\log\text{Cx}$ $\Rightarrow-\log\Big(\frac{\text{x}^2-\text{y}^2}{\text{x}^2}\Big)=\log\text{Cx}$ $\Rightarrow-\log\Big(\frac{\text{x}^2}{\text{x}^2-\text{y}^2}\Big)=\log\text{Cx}$ $\Rightarrow\frac{\text{x}^2}{\text{x}^2-\text{y}^2}=\text{Cx}\ ......(\text{ii})$ since the curve passes through the point (2, 1), we have $\frac{4}{4-1}=2\text{C}$ $\Rightarrow\text{C}=\frac{2}{3}$ So, the required solution is $2(\text{x}^2-\text{y}^2)=3\text{x}.$
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