The general solution of the differential equation dy dx +y g(x)=g(x) g'(x) is a given function of x, is: 1. g(x)+ (1+y+g(x))=C 2. g(x)+ (1+y-g(x))=C 3. g(x)- (1+y-g(x))=C 4. None of these.

2. g(x)+ (1+y-g(x))=C Solution: We have, dy dx +y g'(x)=g(x) g'(x) ...(i) Clearly, it is a linear differential equation of the form dy dx +Py=Q Where P=g'(x), Q=g(x) g'(x) I.F=e^ =e^ '(x)dx =e^ g(x) Multiplying both sides, we get e^ g(x) ( dy dx +yg'( x ) )=e^ g(x) g(x) g'(x) e^ g(x) dy dx +e^ g(x) yg'( x )=e^ g(x) g(x) g'(x) Integrating both sides with respect to x, we get y e^ g(x) =e^ g(x) g(x) g'(x) dx+K y e^ g(x) =I+K Where, I=e^ g(x) g(x) g'(x) dx Now, I=e^ g(x) g(x) g'(x) dx Putting g'(dx)=dt I= e^ t dt =t e^ t dt- [ d dx (t) ^ t dt ]dt =te^ t -e^ t =g(x)e^ g(x) -e^ g(x) y e^ g(x) =g(x)e^ g(x) -e^ g(x) +K y e^ g(x) =g(x)e^ g(x) -e^ g(x) =K Taking log on both sides, we get [y+1-g(x) ]=-g(x)+

The general solution of the differential equation dy dx +y g(x)=g…

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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:

$\text{g}(\text{x})+\log(1+\text{y}+\text{g}(\text{x}))=\text{C}$
$\text{g}(\text{x})+\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$
$\text{g}(\text{x})-\log(1+\text{y}-\text{g}(\text{x}))=\text{C}$
None of these.

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Answer

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

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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