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How do I solve y'= 4y, y(0)=1?

When y=4?

Guest Apr 13, 2017
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If  y'  =   4y   .....  then.......

 

dy / dx   =  4y       this is a separable equation

 

dy / y  =  4 dx      integrate both sides

 

∫ 1/y    dy     =   ∫ 4 dx

 

ln y    =   4x  + C          exponentiate both sides

 

e^(ln y)  =  e^(4x + C)  

 

y   =   e^C * e^(4x)  = Ce^(4x)

 

And we have that  y(0)  = 1 .....so.....

 

1  =  Ce^(4 * 0)

1  = C

 

And  when y  =4

 

  4 =  e^(4*x)       take the Ln  of both sides

 

Ln 4   =  Ln e^(4x)

 

Ln 4  = (4x) Ln e

 

Ln 4  =  4x     divide both sides by 4

 

Ln 4 / 4   =  x   ≈  .34657  

 

 

 

cool cool cool

CPhill  Apr 13, 2017

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