Find that solution $ϕ$ which satisfies $ϕ(0) = 0, ϕ(1) = 3$ for $y'' = 3x + 1$. Find the particular solution to the following second order linear ordinary differential equation that satisfies the $ϕ$

$y'' = 3x + 1$, $ϕ(0) = 0, ϕ(1) = 3$

My solution:

$y' = \frac{3x^2}{2} + x + c_1$
$y = \frac{x^3}{2} + \frac{x^2}{2} + c_1x + c_2$
$ϕ(0) = 0 => 0 + 0 + 0 + c_2 = 0 => c_2 = 0$
$ϕ(1) = 3 => \frac{1}{2} + \frac{1}{2} + c_1 + c_2 = 3$

I am in trouble in step 4 where I need to find the value of $c_1$. How do I go about finding this? Should I do this: $\frac{1}{2} + \frac{1}{2} + c_1 + c_2 = 3 => 1 + 2c = 3 => c_1 = 1$? Please help me!

You already have c2=0 substitute it and find c1.
