Solve the differential equation $y'''=x$ I want to solve $y'''=x$.
I integrated it and I got
$$y''=x^2/2+c_1$$
Integrating again
$$y'=x^3/12+c_2$$
Integrating for the third time
$$y=x^4/48+c_3$$
but the answer on my book is $y=x^2/24+c_1 \cdot x^2/2+c_2 \cdot x+c_3$... where am I wrong?
 A: You should know that $\displaystyle \int ndx=nx$ where $n$ is a constant.
Your first step was correct when you got
$y''=\dfrac{x^2}{2}+C_1$
But when you integrate that you should get
$y'=\dfrac{x^3}{2\times 3}+C_1x+C_2$
And likewise for the third step you should get again by integrating
$y=\dfrac{x^4}{2\times 3\times 4}+\dfrac{C_1x^2}{2}+C_2x+C_3$
A: Should be
$$
y''=\frac{x^2}{2}+c_1
$$
$$
y'=\frac{x^3}{6}+xc_1+c_2
$$
$$
y=\frac{x^4}{24}+\frac{x^2c_1}{2}+xc_2+c_3
$$
A: When we integrate a function $f$, we find a differentiable function $F$ such that $F'=f$. However, if $F$ is an antiderivative, then it is easy to see that $F+C$ is also an antiderivative for every constant $C$. Hence we should obtain a constant of integration every time we integrate.
In your case, we want to integrate the function $y'''=x$ three times. First time we get
$$y''=\frac{x^2}{2}+C_1.$$
Second time we get
$$y'=\frac{x^3}{6}+C_1x+C_2.$$
Third time we get
$$y=\frac{x^4}{24}+C_1\frac{x^2}{2}+C_2x+C_3.$$
Observe the power of $x$ in the first term.
