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Anna007 [38]
3 years ago
5

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=
Mathematics
1 answer:
Romashka [77]3 years ago
6 0

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

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