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So I am attempting to solve this equation

$$y^{\prime\prime\prime} + y^{\prime\prime} +y^{\prime} = x+1. $$ Is it vaild to take the indefinite integral on both sides in order to simplify the expression and make it easier to solve?

\begin{align*} \int (y^{\prime\prime\prime} + y^{\prime\prime} +y^{\prime})\, dx&= \int (x+1)\, dx\\ y^{\prime\prime} +y^{\prime} + y &= \frac{x^2}{2} +x + c \end{align*}

Thanks!

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You can, but for the purpose of solving your specific equation, you may want to reserve that for the last step. In other words, let $z = y'$ then $$z''+z'+z=x+1$$ which is exact and does not have any constants floating around etc. It is also simpler since RHS is linear, not quadratic.

Now solve the equation and once you know what $z = z(x)$ is, you can find $$ y(x) = \int y'(x) dx = \int z(x) dx, $$ which will have the constant on the RHS...

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Yes

This is exactly what you will find yourself doing in higher classes on calculus and differential equations.

It works even in case of different variables on the RHS and LHS:

$$\int y^2dy= \int x^3 dx$$ becomes: $$y^3/3= x^4/4+C$$

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Actually there is a bit more to that. You have to use Fundamental Theorem of Integral Calculus. After first integration (say from $a$ to $x$ ) you should write $$y''(x)-y''(a)+y'(x)-y'(a)+y(x)-y(a)=\frac{x^2}{2}+x-\frac{a^2}{2}-a$$ In order to write in the form you have written $y(a), y'(a), y''(a)$ should be well defined.

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