Question

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=

Answer 1

1) y = tan(xlnx-x+tan⁻¹(3) + 1)

2)
`tan^(-1)y=2(1+x^3)^{(3)/(2)}+ tan^(-1)(3) - 2`

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)

`\frac{\mathrm{d} y}{\mathrm{d} x}= lnx\ (1+y^2)`

`(dy)/(1+y^2)={lnx}{dx}\\\int(dy)/(1+y^2)=\int \{lnx}dx\\tan^(-1)y=xlnx-x+c`

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

`\frac{\mathrm{d} y}{\mathrm{d} x}=9x^2√(1+x^3)(1+y^2)`

`(dy)/(1+y^2) = 9x^2√(1+x^3)\\\int(dy)/(1+y^2) = \int9x^2√(1+x^3)\\ tan^(-1)y=  \int9x^2√(1+x^3)`

let 1+x³=u
`{u}'= 3x^2`

`tan^(-1)y=\int 3√(u)du\\tan^(-1)y=2u^{(3)/(2)}`

inserting value of 'u' in equation above

`tan^(-1)y=2(1+x^3)^{(3)/(2)}+c`

inserting value y(0) = 3

c = tan⁻¹(3) - 2

`tan^(-1)y=2(1+x^3)^{(3)/(2)}+ tan^(-1)(3) - 2`