2 problems found
Explain diagrammatically, or otherwise, why \[ \frac{\mathrm{d}}{\mathrm{d}x}\int_{a}^{x}\mathrm{f}(t)\,\mathrm{d}t=\mathrm{f}(x). \] Show that, if \[ \mathrm{f}(x)=\int_{0}^{x}\mathrm{f}(t)\,\mathrm{d}t+1, \] then \(\mathrm{f}(x)=\mathrm{e}^{x}.\) What is the solution of \[ \mathrm{f}(x)=\int_{0}^{x}\mathrm{f}(t)\,\mathrm{d}t? \] Given that \[ \int_{0}^{x}\mathrm{f}(t)\,\mathrm{d}t=\int_{x}^{1}t^{2}\mathrm{f}(t)\,\mathrm{d}t+x-\frac{x^{5}}{5}+C, \] find \(\mathrm{f}(x)\) and show that \(C=-2/15.\)
The functions \(\mathrm{x}\) and \(\mathrm{y}\) are related by \[ \mathrm{x}(t)=\int_{0}^{t}\mathrm{y}(u)\,\mathrm{d}u, \] so that \(\mathrm{x}'(t)=\mathrm{y}(t)\). Show that \[ \int_{0}^{1}\mathrm{x}(t)\mathrm{y}(t)\,\mathrm{d}t=\tfrac{1}{2}\left[\mathrm{x}(1)\right]^{2}. \] In addition, it is given that \(\mbox{y}(t)\) satisfies \[ \mathrm{y}''+(\mathrm{y}^{2}-1)\mathrm{y}'+\mathrm{y}=0,\mbox{ }(*) \] with \(\mathrm{y}(0)=\mathrm{y}(1)\) and \(\mathrm{y}'(0)=\mathrm{y}'(1)\). By integrating \((*)\), prove that \(\mathrm{x}(1)=0.\) By multiplying \((*)\) by \(\mathrm{x}(t)\) and integrating by parts, prove the relation \[ \int_{0}^{1}\left[\mathrm{y}(t)\right]^{2}\,\mathrm{d}t=\tfrac{1}{3}\int_{0}^{1}\left[\mathrm{y}(t)\right]^{4}\,\mathrm{d}t. \] Prove also the relation \[ \int_{0}^{1}\left[\mathrm{y}'(t)\right]^{2}\,\mathrm{d}t=\int_{0}^{1}\left[\mathrm{y}(t)\right]^{2}\,\mathrm{d}t. \]
Solution: Consider \(\frac12 x(t)^2\) then differentiating we obtain \(x(t)x'(t) = x(t)y(t)\). Also note that \(x(0) = \int_0^0 y(u) \d u = 0\) Therefore, \begin{align*} \int_0^1 x(t)y(t) \d t &= \left [ \frac12 x(t)^2 \right]_0^1 \\ &= \frac12[x(1)]^2 \end{align*} \begin{align*} && 0 &= y'' + (y^2-1)y' + y \\ \Rightarrow && 0 &= \int_0^1 \l y'' + (y^2-1)y' + y \r \d t \\ &&&= \left [y'(t) + \frac13y^3-y+x \right]_0^1 \\ &&&= x(1) \end{align*} Therefore \(x(1) = 0\). \begin{align*} && 0 &= xy'' + (y^2-1)y' x+ yx \\ \Rightarrow && 0 &= \int_0^1 \l xy'' + (y^2-1)y'x + xy \r \d t \\ &&&= \left [ x y' +(\frac13 y^3-y)x \right]_0^1 - \int_0^1 yy'+\frac13y^4-y^2 \d t \\ &&&= 0 - \frac13 \int_0^1 [y(t)]^4 \d t - \int_0^1 [y(t)]^2 \d t \\ \Rightarrow && \int_0^1 [y(t)]^2 \d t &= \frac13 \int_0^1 [y(t)]^4 \d t \end{align*} \begin{align*} && 0 &= yy'' + (y^2-1)y' y+ y^2 \\ \Rightarrow && 0 &= \int_0^1 \l yy'' + (y^2-1)y'y + y^2 \r \d t \\ &&&= \left [ y y' +(\frac14 y^4-\frac12y^2) \right]_0^1 - \int_0^1 [y'(t)]^2 \d t + \int_0^1 y^2 \d t \\ &&&= 0 - \int_0^1 [y'(t)]^2 \d t + \int_0^1 y^2 \d t \\ \Rightarrow && \int_0^1 [y'(t)]^2 \d t &= \int_0^1 [y(t)]^2 \d t \end{align*}