Find the particular solution of the differential equation: `(1+e^(2x))dy+(1+y^2)e^x dx=0,` given that `y=1,\ ` when `x=0.`

Options

A. `cot^(-1)y-cot^(-1)e^(x)=(pi)/(2)`
B. `cot^(-1)y+cot^(-1)e^(x)=(pi)/(2)`
C. `tan^(-1)y-tan^(-1)e^(x)=(pi)/(2)`
D. `tan^(-1)y+tan^(-1)e^(x)=(pi)/(2)`

Answer

D

Explanation

Given differential equation is
`(1+e^(2x))dy+(1+y^(2))e^(x)dx=0`
`implies (dy)/(1+y^(2))+(e^(x)dx)/(1+e^(2x))=0`
`impliesint(dy)/(1+y^(2))+int(e^(x)dx)/(1+e^(2x))=C`
Let `t=e^(x)impliese^(x)dx=dt`
`:.tan^(-1)y+int(dt)/(1+t^(2))=C`
`implies tan^(-1)y+tan^(-1)t=C`
`implies tan^(-1)y+tan^(-1)e^(x)=C`
Now, put `x=0` and `y=1`,
`:. tan^(-1)1+tan^(-1)e^(0)=C`
`implies(pi)/(4)+(pi)/(4)=CimpliesC=(pi)/(2)`
put the value of `C` in equation `(1)`,
`tan^(-1)y+tan^(-1)e^(x)=(pi)/(2)`
which is the required particular solution.