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 answer:
Answer:
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

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