Solving ODE dy/dx = (x+y)^2 , y(0)=1

  • #1
solve dy/dx = (x+y)^2 , y(0)=1

i let w = (x+y) and got the above equation rearranged to dw/dx - 1=w^2

after solving for C i got y=tan(x-pi/4) - x

just wanted to check my answer
 

Answers and Replies

  • #2
lanedance
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w = (x+y) then
dw/dx = 1+dy/dx
dw/dx = 1 + w^2

this part looks ok
 
  • #3
lanedance
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then
w = x+y=x+(tan(x-pi/4)-x) = tan(x-pi/4)
dw/dx = sec(x-pi/4)^2 = 1+tan(x-pi/4)^2
 
  • #4
lanedance
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just remains to check IC
w(0) = tan(-pi/4)

does that look correct?
 

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