Solve the separable initial value problem. 1. y' = ln(x)(1 + y2), y(1) = 3 = y= tan(xlnx-x+1+arctan(3) 2. y' = 9x? v1 + x? (1 + y2), y(0) = 3 = y=

1) y = tan(xlnx-x+tan⁻¹(3) + 1)

2)

Step-by-step explanation:

1) y' = ln(x)(1 + y²),  y(1) = 3    Ans y= tan(x lnx-x+1+tan⁻¹(3))

solution:

y' = ln(x)(1 + y²)

using condition y(1) = 3

tan⁻¹(3)=-1+c

c = tan⁻¹(3) + 1

now,

tan⁻¹(y)=xlnx-x+tan⁻¹(3) + 1

y = tan(xlnx-x+tan⁻¹(3) + 1)

2)  y' = 9x² √(1 + x)³ (1 + y²),   y(0) = 3

let 1+x³=u    

inserting value of 'u' in equation above

inserting value y(0) = 3

c =  tan⁻¹(3) - 2

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer

i think it is c, not positive though

step-by-step explanation: