Differential Equation: y' = 2(3x + y)^2 - 1
How can i solve this problem y' = 2(3x+y)2-1 when x = 0, y =0. Can someone show me the solution ?
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$y' = 2(3x + y)^2 - 1$
$y' = 2(3x + y)^2 - 1$
z = 3x + y
dz = 3dx + dy
dz/dx = 3 + y'
y' = dz/dx - 3
$\dfrac{dz}{dx} - 3 = 2z^2 - 1$
$\dfrac{dz}{dx} = 2z^2 + 2$
$\dfrac{dz}{dx} = 2(z^2 + 1)$
$\dfrac{dz}{z^2 + 1} = 2\,dx$
$\arctan z = 2x + c$
$\arctan (3x + y) = 2x + c$
You can now apply the initial condition to solve for the constant of integration c.