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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
Question
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:
  1. $\text{g}(\text{x})+\log(1+\text{y}+\text{g}(\text{x}))=\text{C}$
  2. $\text{g}(\text{x})+\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$
  3. $\text{g}(\text{x})-\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$
  4. None of these.
✓

Answer

  1. $\text{g}(\text{x})+\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$

Solution:

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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