A) \[y=cx{{e}^{-x}}\]
B) \[y=cy{{e}^{-x}}\]
C) \[y+{{e}^{-x}}=cx\]
D) \[y{{e}^{x}}=cx\]
Correct Answer: D
Solution :
Given, \[ydx-xdy=xydx\] \[\Rightarrow \] \[ydx-xydx=xdy\] \[\Rightarrow \] \[y(1-x)dx=xdy\] \[\Rightarrow \] \[\left( \frac{1-x}{x} \right)dx=\frac{dy}{y}\] \[\Rightarrow \] \[\left( \frac{1}{x}-1 \right)dx=\frac{1}{y}dy\] On integrating, we get \[\log x-x=\log y-\log c\] \[\Rightarrow \] \[x=\log \frac{xc}{y}\] \[\Rightarrow \] \[{{e}^{x}}=\frac{xc}{y}\] \[\Rightarrow \] \[y{{e}^{x}}=xc\]
You need to login to perform this action.
You will be redirected in
3 sec
