JEE Mains Model Paper 4

dy

dt

- (

t

1+t

) y =

1

1+t

This is linear differential equation
On comparing it with

dy

dt

+ Py = Q We get
P = −

t

1+t

, Q =

1

1+t

I.F. = e−∫

t

1+t

dt = e−∫(1−

1

1+t

)dt
⇒ e^{-t}. elog(1+t) ⇒ e−t (1 + t)
General solution is y (t)

1+t

et

= −e−t + C ..... (i)
Given when t = 0 and y = -1, then C=0
So, equation (i) reduces to
y (t)

1+t

et

= −e−t = y (t) = −

1+t

et

= y (1) = −

1

2