Asked by maath
Solve the following differential equation algebraically
dy/dx = xy^2 + 4x y(0) = 0
I reached the point dy/(y^2+4) = xdx but I don't know how to solve the left side of the equation.
dy/dx = xy^2 + 4x y(0) = 0
I reached the point dy/(y^2+4) = xdx but I don't know how to solve the left side of the equation.
Answers
Answered by
oobleck
dy/dx = xy^2 + 4x
dy/dx = x(y^2+4)
dy/(y^2+4) = x dx
now just integrate both sides.
dy/dx = x(y^2+4)
dy/(y^2+4) = x dx
now just integrate both sides.
Answered by
oobleck
1/2 arctan(y/2) = 1/2 x^2 + c/2
y/2 = tan(x^2 + c)
y = 2 tan(x^2+c)
y/2 = tan(x^2 + c)
y = 2 tan(x^2+c)
Answered by
maath
I got that, but I wasn't sure how to integrate dy/(x^2+4)
Answered by
maath
sorry I responded too early, thank you so much
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