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solve y(1+x2)y'-x(1+y2)=0 and find a particular solution that satisfies the initial condition y(0)=√3

Guest Aug 4, 2017
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y(1+x^2)y'-x(1+y^2) = 0     this is a separable equation

 

y ( 1 + x^2)y'  =  x ( 1 + y^2)   rearrange as

 

[ y] / [ 1 + y^2 ] y'  =  [x] / [ 1 + x^2]   integrate both sides

 

∫ y / [ 1 + y^2] dy  = ∫ x / [ 1 + x^2]  dx

 

(1/2) ln ( 1 + y^2)  =   (1/2) ln ( 1 + x^2) + C

 

ln (1 + y^2)  =  ln ( 1 + x^2 ) + C    exponentiate both sides

 

e^[ln (1 + y^2) ] =  e ^ [  ln ( 1 + x^2 ) + C]

 

1 + y^2  =  e^[ ln ( 1 + x^2)] * e^C       →     let e^C   =  C2

 

1 + y^2  =  (1 + x^2) * C2

 

y^2  = ( 1 + x^2) *C2  - 1

 

y = ± √  [   ( 1 + x^2) *C2  - 1 ]

 

y (x) =   ± √  [   C2x^2 + C - 1 ]

 

Now....since the intial condition is that   y(0)  = √3  .....the positive root must be correct....solving for C2, we have that

 

√3  =   √  [   C2(0)^2 + C2  - 1 ]

 

√3 = √ [ C2 - 1]

 

3  = C2 - 1

 

C2 = 4

 

So....the particular solution is   y(x)  = √  [   4x^2 + 4  - 1 ]  =   √  [  4x^2 + 3 ]

 

 

 

cool cool cool

CPhill  Aug 4, 2017

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