A) \[x+y+ln\,\,(x+y)=c\]
B) \[x-y+ln\,\,(x+y)=c\]
C) \[y-x+ln\,\,(x+y)=c\]
D) \[y-x-ln\,\,(x-y)=c\]
Correct Answer: C
Solution :
| [c] Differential equation is |
| \[(x+y)(dx-dy)=dx+dy\] |
| dividing by dx on both the sides |
| \[(x+y)\left( 1-\frac{dy}{dx} \right)=1+\frac{dy}{dx}\] |
| Putting \[x+y=v\] |
| \[1+\frac{dy}{dx}=\frac{dv}{dx}\] and \[\frac{dy}{dx}=\frac{dv}{dx}-1\] |
| The equation changes to |
| \[v\left\{ 1-\left( \frac{dv}{dx}-1 \right) \right\}=\frac{dv}{dx};\,\,\,v\left( 2-\frac{dv}{dx} \right)=\frac{dv}{dx}\] |
| \[2v-v\frac{dv}{dx}=\frac{dv}{dx};\,\,\,2v=(1+v)\frac{dv}{dx}\] |
| \[\left( \frac{1+v}{v} \right)dv=2dx\] or, \[\left( \frac{1}{v}+1 \right)dv=2dx\] |
| Integrating on both the sides. |
| \[\int{\frac{dv}{v}+\int{dv=2\int{dx+c}}}\] |
| \[\log v+v=2x+c\] |
| Putting \[v=x+y\] |
| \[\log (x+y)+x+y=2x+c\] |
| or, \[\log (x+y)+y-x=c\] |
| or, \[y-x+\log (x+y)=c\] |
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