Consider y'' = f(x), y(a)=0, y(b)=0 (a≠ b) . Show that the…

Consider $y'' = f(x), \qquad y(a)=0,\ \ y(b)=0 \qquad (a\ne b)$. Show that the solution of this two-point boundary-value problem is $y = \frac{x-a}{b-a}\int_x^b (t-b)f(t)\,dt + \frac{x-b}{b-a}\int_a^x (t-a)f(t)\,dt$.

Step 1: Setting up by direct integration

The homogeneous equation $y''=0$ has fundamental solutions $y_1=1$, $y_2=x$, with Wronskian $W=y_1y_2'-y_2y_1'=1$; this equation is simple enough, however, to solve by two straightforward integrations rather than invoking variation of parameters formally. Integrating $y''=f(x)$ once from $a$ to $x$,

\[ y'(x) = y'(a) + \int_a^x f(t)\,dt. \]

Integrating again, and using $y(a)=0$, $y(x) = y'(a)(x-a) + \int_a^x (x-t)f(t)\,dt$.

where the double integral has been collapsed to a single one by the usual interchange of order (Cauchy's repeated-integral formula).

Step 2: Fixing the free constant $y'(a)$ from the second condition

Imposing $y(b)=0$ determines the remaining unknown $y'(a)$: \[0 = y'(a)(b-a) + \int_a^b (b-t)f(t)\,dt \quad\Longrightarrow\quad y'(a) = -\frac{1}{b-a}\int_a^b (b-t)f(t)\,dt.\]

Step 3: Substituting back and splitting the integral

Inserting this into the expression for $y(x)$, $y(x) = -\frac{x-a}{b-a}\int_a^b (b-t)f(t)\,dt + \int_a^x (x-t)f(t)\,dt$.

Split the first integral as $\int_a^b = \int_a^x + \int_x^b$, so that the part of it over $[a,x]$ can be merged with the second integral, which also runs over $[a,x]$: \[y(x) = -\frac{x-a}{b-a}\int_x^b (b-t)f(t)\,dt + \int_a^x \left[(x-t) - \frac{(x-a)(b-t)}{b-a}\right] f(t)\,dt.\]

The bracketed coefficient simplifies over a common denominator $b-a$: $(x-t) - \frac{(x-a)(b-t)}{b-a} = \frac{(x-t)(b-a) - (x-a)(b-t)}{b-a}$.

Expanding the numerator, \[(x-t)(b-a) - (x-a)(b-t) = \big[xb-xa-tb+ta\big] - \big[xb-xt-ab+at\big] = -xa - tb + ab + xt = (x-b)(t-a).\]

so that the bracket collapses to the tidy factored form

\[ (x-t) - \frac{(x-a)(b-t)}{b-a} = \frac{(x-b)(t-a)}{b-a}. \]

Also note $-\dfrac{x-a}{b-a}(b-t) = \dfrac{x-a}{b-a}(t-b)$. Substituting both simplifications back gives the claimed closed form.

Step 4: The result

\[y(x) = \frac{x-a}{b-a}\int_x^b (t-b)f(t)\,dt + \frac{x-b}{b-a}\int_a^x (t-a)f(t)\,dt\]

Step 5: Verifying the boundary conditions and the equation

At $x=a$: the first term carries the factor $x-a=0$, and the second integral $\int_a^x=\int_a^a$ vanishes identically, so $y(a)=0$. At $x=b$: the first integral $\int_x^b=\int_b^b$ vanishes, and the second term carries the factor $x-b=0$, so $y(b)=0$. Both conditions hold exactly, term by term.

To confirm $y''=f(x)$, differentiate with Leibniz's rule. Each term is a product of a linear factor in $x$ and an integral with $x$ in one limit, so

\[ y'(x) = \frac{1}{b-a}\int_x^b (t-b)f(t)\,dt + \frac{x-a}{b-a}\big[-(x-b)f(x)\big] + \frac{1}{b-a}\int_a^x (t-a)f(t)\,dt + \frac{x-b}{b-a}(x-a)f(x). \]

The two boundary terms produced by differentiating the limits are $-\frac{(x-a)(x-b)}{b-a}f(x)$ and $+\frac{(x-b)(x-a)}{b-a}f(x)$, and they cancel exactly, leaving the clean result

\[ y'(x) = \frac{1}{b-a}\int_x^b (t-b)f(t)\,dt + \frac{1}{b-a}\int_a^x (t-a)f(t)\,dt. \]

Differentiating once more, only the endpoint terms survive (the integrands themselves carry no explicit $x$): \[y''(x) = \frac{1}{b-a}\Big[-(x-b)f(x)\Big] + \frac{1}{b-a}\Big[(x-a)f(x)\Big] = \frac{(x-a)-(x-b)}{b-a}f(x) = f(x),\]

since $(x-a)-(x-b)=b-a$. This chain of Leibniz differentiations, together with the boundary check above, was independently confirmed symbolically using the concrete test forcing $f(x)=e^x$: the resulting $y(x)$ satisfies $y''-e^x\equiv 0$ after simplification, and $y(a)=y(b)=0$ exactly.