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

[mat331760298]

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

 

[lanedance]

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

this part looks ok

 

[lanedance]

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

 

[lanedance]

just remains to check IC
w(0) = tan(-pi/4)

does that look correct?