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Ordinary differentiation equations (ODE)-seperable

  • Thread starter naspek
  • Start date
  • #1
181
0
hey there..
i've got unsolved solution here..


Solve initial value problem..

xy’ – y = 2x2y ; y(1) = 1



x dy/dx = y + 2x2y

dy/dx = y/x + 2xy

dy/dx = y[(1/x) + 2x]

∫1/y dy = ∫ (1/x) + 2x dx

ln |y| = ln |x| + x2 + C

apply exp to both side

y = x + ex^2 + e^C



substitute the initial value

1 = 1 + e1 + e^C

e^(1+C) = 0

apply ln

1 + C = ln 0?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
6
ln |y| = ln |x| + x2 + C

apply exp to both side

y = x + ex^2 + e^C
You are forgetting your basic exponential rules:

[tex]y=e^{\ln(x)+x^2+c}=e^ce^{x^2}e^{\ln(x)}\neq x+e^{x^2}+e^c[/tex]
 

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