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. `tan^(-1)y+tan^(-1)e^(x)=pi/2`
B. ` tan^(-1)x+tan^(-1)e^(y)=pi/2`
C. ` tan^(-1) x + tan^(-1)e^(y)=pi/4`
D. ` tan^(-1)y +tan^(-1)e^(x)=pi/3`

Answer

a

Explanation

Given differential equation is
` 1+e^((2x)) dy + ( 1+y^(2)) e^(x)dx = 0 `
Separating the variables , we get ` (dy)/(1+y^(2)) +(e^(x)dx)/(1+e^(2x))=0`
On intergrating both sides , we get ` int (dy)/(1+y^(2)) + int (e^(x)dx)/(1+e^(2x)) = C`
Put `e^(x) = t rArr e^(x) dx = dt `
` rArr tan^(-1) y + int (dt)/(1+t^(2))=C`
` rAr tan^(-1) ty + tan^(-1) t = C`
` rArr tan^(-1) + tan^(-1) e^(x) = C " "` ... (i) ltbRgt Now, put x = 0 and y = 1
` :. tan^(-1) + tan^(-1) e^(0) = C`
` rArr pi/4 +pi/4 C rArr C = pi/2 `