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Find the particular solution to the following second order linear ordinary differential equation that satisfies the $ϕ$

$y'' = 3x + 1$, $ϕ(0) = 1, ϕ'(0) = 2$

My solution:

$y' = \frac{3x^2}{2} + x + c_1$

$y = \frac{x^3}{2} + \frac{x^2}{2} + c_1x + c_2$

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

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

I checked the back of my textbook and it has a different answer than mine. Am I doing something wrong here? PLEASE HELP!

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    Why are we using both $\phi$ and $y$ for the same function? – 2017-02-05
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    @Moo $ϕ(x) = \frac{x^3}{2} + \frac{x^2}{2} + 2x + 1$ – 2017-02-05
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    I get the same as the book. – 2017-02-05
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    After your first step you can conclude that $c_1=\phi^\prime(0)=2$. In your third step you can only conclude that $c_2=1$. What is this variable $c$ which you use in step $3$? – 2017-02-05
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    @Moo how? ............................. – 2017-02-05
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    don't ask the same question twice. – 2017-02-05

1 Answers 1

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Here's my derivation of your book's solution: $$ y''=3x+1\implies y'=3/2x^2+x+c\implies y=\frac{x^3}{2}+\frac{x^2}{2}+cx+d $$ then boundary conditions yield $d=1$ since $$ y(0)=\frac{0^3}{2}+\frac{0^2}{2}+c*0+d=d=1 $$ and similarly $$ y'(0)=2=c $$ and thus $$ y(x)=\frac{x^3}{2}+\frac{x^2}{2}+2x+1 $$