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I need help understand this solution. I don't grasp the steps in the integration and have never seen this approach before. I thought a non-homogeneous 2nd order linear equations had a general solution $y=y_p(x)+y_h(x)$.

$$ \begin{cases} -u''(x)=f(x), \quad 0<x<1 \\ u(0)=u(1)=0. \end{cases} $$ First integration: $$-u'(x)=\int_0^xf(y)\mathrm{d}y + C_1$$ Second integration: $$-u(x)=\int_0^x \int_0^zf(y)\mathrm{d}y\mathrm{d}z+C_1x+C_2$$

How is the first integration done? Why is there a integration constant when it's not a indefinite integral? And why are the bounds not between $0$ and $1$?

I thought I should solve it like: $$ -u'(x)=\int f(x)\mathrm{d}x=F(x)+C_1 $$ and then $$ -u(x)=\int (F(x)+C_1) \mathrm{d}x= F(x)x+C_1x+C_2 $$

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  • $\begingroup$ Think of a water tank. The integral of the flow gives you the volume of water that flowed into the tank over an interval of time. The integration constant gives you the initial volume of water in the tank. $\endgroup$ – Rodrigo de Azevedo Feb 2 '17 at 11:52
  • $\begingroup$ @RodrigodeAzevedo Hi! I grasp the physical interpretation, but not the math-steps. $\endgroup$ – JDoeDoe Feb 2 '17 at 11:54
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Let's look at these two solutions line by line.

Quoted solution:

First integration: $$ -u'(x)=\int_0^xf(y)\mathrm{d}y + C_1 $$

Your solution:

$$ -u'(x)=\int f(x)\mathrm{d}x=F(x)+C_1 $$

These are the same statement. The Fundamental Theorem of Calculus tells us that, if $f$ is continuous, then $g(x) = \int_0^x f(y)\,dy$ is an antiderivative/indefinite integral of $f$.

Quoted solution:

Second integration: $$-u(x)=\int_0^x \int_0^zf(y)\mathrm{d}y\mathrm{d}z+C_1x+C_2$$

Your solution:

$$-u(x)=\int (F(x)+C_1) \mathrm{d}x= F(x)x+C_1x+C_2$$

Not quite. You integrated a function of $x$ by multiplying it by $x$. That typically only works for constant functions. For instance, if $f(x) = \cos x$, then $F(x) = \sin x$, and the integral of $\sin x$ is not $x \sin x$.

The quoted solution is applying the same FTC principle again. To find an antiderivative of $F(x) + C_1$, we integrate from $0$ to $x$: \begin{align*} \int(F(x) + C_1)\,dx &= \int_0^x (F(z) + C_1) \,dz + C_2 \\ &= \int_0^x \int_0^z f(y)\,dy\,dx + \left.C_1 z\right|^x_0 + C_2 \\ &= \int_0^x \int_0^z f(y)\,dy\,dx + C_1 x + C_2 \\ \end{align*}

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  • $\begingroup$ Thanks for the answer! In the solution the integral bounds was from $0$ to $x$ and from $0$ to $z$. Why is the bounds not from $0$ to $1$? $\endgroup$ – JDoeDoe Feb 18 '17 at 19:08
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    $\begingroup$ @JDoeDoe The bounds are not $0$ to $1$ because we are looking for a function of $x$. If you integrate from $0$ to $1$ you will get the number $-u(1)$. $\endgroup$ – Matthew Leingang Feb 18 '17 at 21:25
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Using the concept of differentiation under the integral sign, we get, $$\frac {d}{dx}(-u'(x)) = \frac {d}{dx}(\int_{0}^{x} f (y) \mathrm {d}y + C_1) $$ $$\Leftrightarrow-u''(x) = f (x) \mathrm {d}x \frac {d}{dx}(x) - 0 + 0$$ $$\Leftrightarrow-u''(x) = f (x) \tag {1}$$ We can similarly proceed to write the secone form back into $(1)$. Hope it helps.

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First step: the function $F$, given by $F(x)=\int_0^xf(y)\mathrm{d}y$ is an antiderivative of $f$.

Since $-u''=f$, $-u'$ is also an antiderivative of $f$.

Hence there is some constant $C_1$ such that

$-u'(x)=F(x)+C_1$

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