# Differential Equation: y' = 2(3x + y)^2 - 1

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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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Jhun Vert

$y' = 2(3x + y)^2 - 1$

Let
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.

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