Solveeit Logo
Login

Question

Mathematics Question on types of differential equations

Let y = y(x), y>0, be a solution curve of the differential equation (1 + x2) dy=y (x–y)dx. If y(0)=1 and y(2√2) , = b then

A

eβ−1=e−2(3+2√2)e β^{ − 1} = e^{-2} ( 3 + 2 √ 2 )

B

e3β−1=e(3+2√2)e ^3 β^{ − 1} = e ( 3 + 2 √ 2 )

C

eβ−1=e−2(5+√2)e β^{ − 1} = e^{-2} ( 5+ √ 2 )

D

e3β−1=e−2(3+2√2)e3 β^{ − 1} = e^{-2} ( 3+ 2√ 2 )

Answer

e3β−1=e−2(3+2√2)e3 β^{ − 1} = e^{-2} ( 3+ 2√ 2 )

Explanation

Solution

The correct option is(D): e3β−1=e−2(3+2√2)e3 β^{ − 1} = e^{-2} ( 3+ 2√ 2 )