1. The problem statement, all variables and given/known data Given a strong form boundary value problem, derive the weak form using weighted residuals. 2. Relevant equations ##(2-x)u''(x)-u'(x)+u(x)=f(x)## for ##x\in(0,1)## with $$u(0)=u(1)=0$$ 3. The attempt at a solution I must multiply both sides of this equation by an arbitrary test function ##v(x)## where $$v(0)=v(1)=0$$ to eliminate boundary terms under integration. $$\int_0^1\left[(2-x)u''v-u'v+uv\right]dx=\int_0^1fvdx$$ Regroup to one side so I have the differential equation's residual in the integrand. $$\int_0^1\left[(2-x)u''v-u'v+uv-fv\right]dx$$ Expand the product on the left, separate out the ##v(x)##. $$\int_0^1\left[2u''-xu''-u'+u-f\right]vdx$$ I separated this out into separate integrals and integrated by parts individually. I wanted to reduce the order of the differential equation and end up with everything being multiplied by ##v'(x)## so I could pull that out and end up with my weak form. For organization, label each product, from left to right, 1 to 5. 1:$$2\int_0^1u''vdx=-2\int_0^1u'v'dx$$ 2:$$-\int_0^1xu''vdx=\int_0^1\left(xu'-u\right)v'dx$$ 3:$$-\int_0^1u'vdx=\int_0^1uv'dx$$ 4:$$\int_0^1uvdx=$$ ??? 5:$$-\int_0^1fvdx=$$ ??? I am blanking out on how to integrate by parts on the last two functions. I have to integrate one side while differentiating the other, but I end up with a bunch of (indefinite?) integrals that I do not know what to do with. How else should I approach this problem?