How we can solved 2nd part to obtain unique solution for the above scheme
i start firstly integrating both sides to get $u'(a)-u'(b)=\int_{a}^{b} f(x)dx=0$ Now how i will continue ??
How we can solved 2nd part to obtain unique solution for the above scheme
i start firstly integrating both sides to get $u'(a)-u'(b)=\int_{a}^{b} f(x)dx=0$ Now how i will continue ??
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we have $-u''(x)=f(x)$ then $-\int_{a}^{b}u''(x)=\int_{a}^{b}f(x)dx=0$
so that, $-(u'(x))_{a}^{b}=0$ it follows that, $u'(a)=u'(b)$
integrate again to get $u(a)=c_{1}$ and $u(b)=c_{2}$