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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!

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    Your calculation is correct. You already know that $c_2=0$ so you are free to substitute that into the last equation. – 2017-02-05
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    @lulu I got $c_2 = 0 $ for the first equation. I cannot just substitute that, can I? – 2017-02-05
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    You absolutely can! Think of it as a system of two linear equations: $c_2=0$ and $c_1+c_2=2$. – 2017-02-05
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    this is almost identical to the question you already asked here: http://math.stackexchange.com/questions/2130824/find-that-solution-%CF%95-which-satisfies-%CF%950-1-%CF%950-2-for-y-3x-1/2130838#2130838 – 2017-02-05
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    Possible duplicate of [Find that solution $ϕ$ which satisfies $ϕ(0) = 1, ϕ'(0) = 2$ for $y'' = 3x + 1$.](http://math.stackexchange.com/questions/2130824/find-that-solution-%cf%95-which-satisfies-%cf%950-1-%cf%950-2-for-y-3x-1) – 2017-03-26

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

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    This is not really an answer. Explain the steps to the OP! – 2017-02-05
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    @TheCount I was just guiding them. At that moment I was unable to comment due to low reputation, otherwise, I would have used the comment section. A Similar question was asked previously by the same user. – 2017-02-06