Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Find the general solution of $\frac{\text{dy}}{\text{dx}}-3\text{y}=\sin2\text{x}.$
✓
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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