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

Answer:

1) y = tan(xlnx-x+tan⁻¹(3) + 1)

2) $tan^{-1}y=2(1+x^3)^{\frac{3}{2}}+ tan^{-1}(3) - 2$

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)

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

$\frac{dy}{1+y^2}={lnx}{dx}\\\int\frac{dy}{1+y^2}=\int \{lnx}dx\\tan^{-1}y=xlnx-x+c$

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

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

$\dfrac{dy}{1+y^2} = 9x^2\sqrt{1+x^3}\\\int\dfrac{dy}{1+y^2} = \int9x^2\sqrt{1+x^3}\\ tan^{-1}y=  \int9x^2\sqrt{1+x^3}$

let 1+x³=u    ${u}'= 3x^2$

$tan^{-1}y=\int 3\sqrt{u}du\\tan^{-1}y=2u^{\frac{3}{2}}$

inserting value of 'u' in equation above

$tan^{-1}y=2(1+x^3)^{\frac{3}{2}}+c$

inserting value y(0) = 3

c =  tan⁻¹(3) - 2

$tan^{-1}y=2(1+x^3)^{\frac{3}{2}}+ tan^{-1}(3) - 2$