Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS3 Marks
Question
For each of the differential equations given in find the general solution:$\text{y dx}+(\text{x}-\text{y}^{2})\ \text{dy}=0$
✓
Answer
$\text{y dx}+(\text{x}-\text{y}^{2})\text{dy}=0$$\Rightarrow\frac{\text{dx}}{\text{dy}}=\frac{\text{y}^2-\text{x}}{\text{y}}=\text{y}-\frac{\text{x}}{\text{y}}$
$\Rightarrow\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{\text{y}}=\text{y}$
This is a linear differential equation of the form:
$\frac{\text{dy}}{\text{dx}}+\text{px}=\text{Q}\ (\text{where p}=\frac{1}{\text{y}\ }\ \text{and}\ \text{Q}=\text{y})$
$\text{Now, I.F}=\text{e}^{\int\text{pdy}}=\text{e}^{\int\frac{1}{\text{y}}\text{dy}}=\text{e}^{\log\text{y}}=\text{y}.$
The general solution of the given differential equation is given by the relation,
$\text{y(I.F)}=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow\text{xy}=\int(\text{y}\cdot\text{y})\text{dy}+\text{C}$
$\Rightarrow\text{xy}=\int\text{y}^2\text{dy}+\text{C}$
$\Rightarrow\text{xy}=\frac{\text{y}^3}{3}+\text{C}$
$\Rightarrow\text{x}=\frac{\text{y}^2}{3}+\frac{\text{C}}{\text{y}}$
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