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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
The general solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}\ \text{g}(\text{x})=\text{g}(\text{x})\ \text{g}'(\text{x})$ is a given function of $x,$ is :
  • A
    $\text{g}(\text{x})+\log(1+\text{y}+\text{g}(\text{x}))=\text{C}$
  • ✓
    $\text{g}(\text{x})+\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$
  • C
    $\text{g}(\text{x})-\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$
  • D
    None of these.
✓

Answer

Correct option: B.
$\text{g}(\text{x})+\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$
We have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\ \text{g}'(\text{x})=\text{g}(\text{x})\ \text{g}'(\text{x})\ ...(\text{i})$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=\text{g}'(\text{x}), \text{Q}=\text{g}(\text{x})\ \text{g}'(\text{x})$
$\text{I.F}=\text{e}^{\int\text{P}\text{dx}}$
$=\text{e}^{\int\text{g}'(\text{x})\text{dx}}$
$=\text{e}^{\text{g}(\text{x})}$
Multiplying both sides, we get
$\text{e}^{\text{g}(\text{x})}\Big(\frac{\text{dy}}{\text{dx}}+\text{yg}'({\text{x}})\Big)=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})$
$\text{e}^{\text{g}(\text{x})}\frac{\text{dy}}{\text{dx}}+\text{e}^{\text{g}(\text{x})}\text{yg}'({\text{x}})=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})$
Integrating both sides with respect to $x, $ we get
$\text{y}\ \text{e}^{\text{g}(\text{x})}=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})\ \text{dx}+\text{K}$
$\text{y}\ \text{e}^{\text{g}(\text{x})}=\text{I}+\text{K}$
Where, $\text{I}=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})\ \text{dx}$
Now,
$\text{I}=\text{e}^{\text{g}(\text{x})}\text{g}(\text{x})\ \text{g}'(\text{x})\ \text{dx}$
Putting $\text{g}'(\text{dx})=\text{dt}$
$\text{I}=\int\text{t}\ \text{e}^{\text{t}}\ \text{dt}$
$=\text{t}\int\ \text{e}^{\text{t}}\ \text{dt}-\int\Big[\frac{\text{d}}{\text{dx}}(\text{t})\int\text{e}^{\text{t}}\ \text{dt}\Big]\text{dt}$
$=\text{t}\text{e}^{\text{t}}-\text{e}^{\text{t}}$
$=\text{g}(\text{x})\text{e}^{\text{g}(\text{x})}-\text{e}^{\text{g}(\text{x})}$
$\Rightarrow \text{y}\ \text{e}^{\text{g}(\text{x})}=\text{g}(\text{x})\text{e}^{\text{g}(\text{x})}-\text{e}^{\text{g}(\text{x})}+\text{K}$
$\Rightarrow \text{y}\ \text{e}^{\text{g}(\text{x})}=\text{g}(\text{x})\text{e}^{\text{g}(\text{x})}-\text{e}^{\text{g}(\text{x})}=\text{K}$
Taking $\log$ on both sides, we get
$\log\Big[\text{y+1}-\text{g}(\text{x})\Big]=-\text{g}(\text{x})+\log\text{K}$

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