Show that the only value of \(\lambda \) for which \[ f(x)=\lambda \int _{0}^{1}xy(x+y)f(y)\,\mathrm{d}y \] has a non-trivial solution the roots of the equation\[ \lambda ^{2}+120\lambda -240=0 \] solution
The integral equation is a homogeneous Fredholm of the second kind. Normally it is written as\begin{equation} \varphi (x)=\lambda \int _{0}^{1}K\left ( x,y\right ) \varphi (y)\,\mathrm{d}y \tag{1} \end{equation} Where in our case the kernel \(K\left ( x,y\right ) =xy(x+y)\). Hence the kernel is separable and can be written as\begin{align*} K\left ( x,y\right ) & =x^{2}y+xy^{2}\\ & =\sum _{j}^{2}M_{j}\left ( x\right ) N_{j}\left ( y\right ) \end{align*}
Where \(M_{1}=x^{2},N_{1}=y,M_{2}=x,N_{2}=y^{2}\), therefore Eq. (1) can be written as\[ \varphi (x)=\lambda \int _{0}^{1}\left ( \sum _{j}^{2}M_{j}\left ( x\right ) N_{j}\left ( y\right ) \right ) \varphi (y)\,\mathrm{d}y \] Interchanging summation with integration\begin{equation} \varphi (x)=\lambda \sum _{j}^{2}M_{j}\left ( \int _{0}^{1}N_{j}\left ( y\right ) \varphi (y)\,\mathrm{d}y\right ) \tag{2} \end{equation} Now \(\int _{0}^{1}N_{j}\left ( y\right ) \varphi (y)\,\mathrm{d}y\) is a constant. Lets call it \(c_{j}\), \[ c_{j}=\int _{0}^{1}N_{j}\left ( y\right ) \varphi (y)\,\mathrm{d}y \] Therefore Eq. (2) becomes\[ \varphi (x)=\lambda \sum _{j}^{2}M_{j}c_{j}\] To find \(c_{i}\) we multiply both side by \(N_{i}\) and integrate w.r.t. \(x\) which gives\begin{equation} \int _{0}^{1}N_{i}\left ( x\right ) \varphi (x)\,\mathrm{d}x=\lambda \sum _{j}^{2}a_{ij}c_{j} \tag{3} \end{equation} Where \[ a_{ij}=\int _{0}^{1}N_{i}\left ( x\right ) M_{j}\left ( x\right ) \,\mathrm{d}x \] Which can be evaluated to a numerical values. Going back to Eq. (3) we see that the LHS is just \(c_{i}\), hence Eq. (3) can be written as\begin{equation} c_{i}=\lambda \sum _{j}^{2}a_{ij}c_{j} \tag{4} \end{equation} Writing this in matrix for\begin{align*} \begin{pmatrix} c_{1}\\ c_{2}\end{pmatrix} & =\lambda \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}\begin{pmatrix} c_{1}\\ c_{2}\end{pmatrix} \\\begin{pmatrix} c_{1}\\ c_{2}\end{pmatrix} \left ( I-\lambda \begin{pmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix} \right ) & =0\\\begin{pmatrix} c_{1}\\ c_{2}\end{pmatrix}\begin{pmatrix} 1-\lambda a_{11} & -\lambda a_{12}\\ -\lambda a_{21} & 1-\lambda a_{22}\end{pmatrix} & =0 \end{align*}
For a non-trivial solution to exist, \(\begin{pmatrix} c_{1}\\ c_{2}\end{pmatrix} \) can’t be zero. Hence the only possibility is that \[\begin{vmatrix} 1-\lambda a_{11} & -\lambda a_{12}\\ -\lambda a_{21} & 1-\lambda a_{22}\end{vmatrix} =0 \] Now we evaluate the \(a_{ij}\) coefficients. Recall from eariler that \(M_{1}\left ( x\right ) =x^{2},N_{1}\left ( y\right ) =y,M_{2}\left ( x\right ) =x,N_{2}\left ( y\right ) =y^{2}\), hence\begin{align*} a_{11} & =\int _{0}^{1}N_{1}\left ( x\right ) M_{1}\left ( x\right ) \,\mathrm{d}x=\int _{0}^{1}x^{3}\,\mathrm{d}x=\frac{1}{4}\\ a_{12} & =\int _{0}^{1}N_{1}\left ( x\right ) M_{2}\left ( x\right ) \,\mathrm{d}x=\int _{0}^{1}x^{2}\,\mathrm{d}x=\frac{1}{3}\\ a_{21} & =\int _{0}^{1}N_{2}\left ( x\right ) M_{1}\left ( x\right ) \,\mathrm{d}x=\int _{0}^{1}x^{4}\,\mathrm{d}x=\frac{1}{5}\\ a_{22} & =\int _{0}^{1}N_{2}\left ( x\right ) M_{2}\left ( x\right ) \,\mathrm{d}x=\int _{0}^{1}x^{3}\,\mathrm{d}x=\frac{1}{4} \end{align*}
Therefore \[\begin{vmatrix} 1-\lambda \frac{1}{4} & -\lambda \frac{1}{3}\\ -\lambda \frac{1}{5} & 1-\lambda \frac{1}{4}\end{vmatrix} =0 \] or\begin{align*} \left ( \frac{4-\lambda }{4}\right ) \left ( \frac{4-\lambda }{4}\right ) -\left ( \frac{-\lambda }{3}\right ) \left ( \frac{-\lambda }{5}\right ) & =0\\ -\frac{1}{240}\lambda ^{2}-\frac{1}{2}\lambda +1 & =0\\ \lambda ^{2}+120\lambda -240 & =0 \end{align*}
Therefore, the values of \(\lambda \) which satisfy this polynomial will allow a non-trivial solution. In otherwords, the roots of this polynomial. This is what was required to show.
Solve the following integral equation using Neumann series method\[ \varphi (x)=1+\frac{x^{2}}{2!}-\int _{0}^{x}\left ( x-t\right ) \varphi (t)\,\mathrm{d}t \] Solution:
The above is in the form \begin{equation} \varphi (x)=f\left ( x\right ) -\int _{0}^{x}K\left ( x,t\right ) \varphi (t)\,\mathrm{d}t \tag{1} \end{equation} Lets makes it in standard form, by changing the sign in front of the integral to \(+\) instead of \(-\)\[ \varphi (x)=f\left ( x\right ) +\int _{x}^{0}K\left ( x,t\right ) \varphi (t)\,\mathrm{d}t \] Hence it is a Volterra of the second kind. We start by approximating \(\varphi (x)\) to \(f\left ( x\right ) \)\[ \varphi _{0}(x)\simeq f\left ( x\right ) \] Then\begin{align*} \varphi _{1}(x) & \simeq \int _{x}^{0}K\left ( x,t\right ) \varphi _{0}(t)\,\mathrm{d}t\\ \varphi _{2}(x) & \simeq \int _{x}^{0}K\left ( x,t\right ) \varphi _{1}(t)\,\mathrm{d}t\\ & \vdots \\ \varphi _{n}(x) & \simeq \int _{x}^{0}K\left ( x,t\right ) \varphi _{n-1}(t)\,\mathrm{d}t \end{align*}
Therefore\begin{align*} \varphi _{0}(x) & =1+\frac{x^{2}}{2!}\\ \varphi _{1}(x) & =\int _{x}^{0}\left ( x-t\right ) \left ( 1+\frac{t^{2}}{2!}\right ) \,\mathrm{d}t=-\left ( \frac{x^{2}}{2}+\frac{x^{4}}{24}\right ) \\ \varphi _{2}(x) & =-\int _{x}^{0}\left ( x-t\right ) \left ( \frac{t^{2}}{2}+\frac{t^{4}}{24}\right ) \,\mathrm{d}t=\left ( \frac{x^{4}}{24}+\frac{x^{6}}{720}\right ) \\ \varphi _{3}(x) & =\int _{x}^{0}\left ( x-t\right ) \left ( \frac{t^{4}}{24}+\frac{t^{6}}{720}\right ) \,\mathrm{d}t=-\left ( \frac{x^{6}}{720}+\frac{x^{8}}{40320}\right ) \end{align*}
Therefore, we now see the sequence as\begin{align*} \varphi (x) & \simeq \varphi _{0}(x)+\varphi _{1}(x)+\varphi _{2}(x)+\varphi _{3}(x)+\cdots \\ & =\left ( 1+\frac{x^{2}}{2!}\right ) -\left ( \frac{x^{2}}{2}+\frac{x^{4}}{24}\right ) +\left ( \frac{x^{4}}{24}+\frac{x^{6}}{720}\right ) -\left ( \frac{x^{6}}{720}+\frac{x^{8}}{40320}\right ) +\cdots \\ & =1+\frac{x^{2}}{2!}-\frac{x^{2}}{2}-\frac{x^{4}}{4!}+\frac{x^{4}}{4!}+\frac{x^{6}}{6!}-\frac{x^{6}}{6!}-\frac{x^{8}}{8!}+\cdots \\ & =1 \end{align*}
Hence the solution is\[ \varphi (x)=1 \]
Solve the following integral equation using Neumann series method\[ \varphi (x)=x\cos x+\int _{0}^{x}t\varphi (t)\,\mathrm{d}t \] Solution:
The above is in the form \begin{equation} \varphi (x)=f\left ( x\right ) +\int _{0}^{x}K\left ( x,t\right ) \varphi (t)\,\mathrm{d}t \tag{1} \end{equation} Hence it is a Volterra of the second kind. We start by approximating \(\varphi (x)\) to \(f\left ( x\right ) \)\[ \varphi _{0}(x)\simeq f\left ( x\right ) =x\cos x \] Then\begin{align*} \varphi _{1}(x) & \simeq \int _{x}^{0}K\left ( x,t\right ) \varphi _{0}(t)\,\mathrm{d}t\\ \varphi _{2}(x) & \simeq \int _{x}^{0}K\left ( x,t\right ) \varphi _{1}(t)\,\mathrm{d}t\\ & \vdots \\ \varphi _{n}(x) & \simeq \int _{x}^{0}K\left ( x,t\right ) \varphi _{n-1}(t)\,\mathrm{d}t \end{align*}
Therefore\begin{align*} \varphi _{0}(x) & =x\cos x\\ \varphi _{1}(x) & =\int _{0}^{x}t\left ( t\cos t\right ) \,\mathrm{d}t\\ & =2x\cos x+\left ( -2+x^{2}\right ) \sin x=x^{2}\sin x+2x\cos x-2\sin x\\ & \\ \varphi _{2}(x) & =\int _{0}^{x}t\left ( 2\left ( t\cos t-\sin t\right ) +t^{2}\sin t\right ) \,\mathrm{d}t\\ & =-x^{3}\cos x+5x^{2}\sin x-12\sin x+12x\cos x\\ & \\ \varphi _{3}(x) & =\int _{0}^{x}t\left ( -t^{3}\cos t+5t^{2}\sin t-12\sin t+12t\cos t\right ) \,\mathrm{d}t\\ & =-x^{4}\sin x-9x^{3}\cos x+39x^{2}\sin x-90\sin x+90x\cos x\\ & \\ \varphi _{4}(x) & =\int _{0}^{x}t\left ( -t^{4}\sin t-9t^{3}\cos t+39t^{2}\sin t-90\sin t+90t\cos t\right ) \,\mathrm{d}t\\ & =x^{5}\cos x-14x^{4}\sin x-95x^{3}\cos x+375x^{2}\sin x-840\sin x+840x\cos x\\ & \\ \varphi _{5}(x) & =\int _{0}^{x}t\left ( t^{5}\cos t-14t^{4}\sin t-95t^{3}\cos t+375t^{2}\sin t-840\sin t+840t\cos t\right ) \,\mathrm{d}t\\ & =x^{6}\sin x+20x^{5}\cos x-195x^{4}\sin x-1155x^{3}\cos x+4305x^{2}\sin x-9450\sin x+9450x\cos x \end{align*}
Therefore, we now see the sequence as\begin{align*} \varphi (x) & \simeq \varphi _{0}(x)+\varphi _{1}(x)+\varphi _{2}(x)+\varphi _{3}(x)+\cdots \\ & =\left ( x\cos x\right ) \\ & +x^{2}\sin x+2x\cos x-2\sin x\\ & -x^{3}\cos x+5x^{2}\sin x-12\sin x+12x\cos x\\ & -x^{4}\sin x-9x^{3}\cos x+39x^{2}\sin x-90\sin x+90x\cos x\\ & +x^{5}\cos x-14x^{4}\sin x-95x^{3}\cos x+375x^{2}\sin x-840\sin x+840x\cos x\\ & +x^{6}\sin x+20x^{5}\cos x-195x^{4}\sin x-1155x^{3}\cos x+4305x^{2}\sin x-9450\sin x+9450x\cos x\\ & +\cdots \end{align*}
Collecting\begin{align*} \varphi (x) & =\cos x\left ( x+2x+12x+90x+840x+9450x+\cdots \right ) \\ & +\cos x\left ( -x^{3}-9x^{3}-95x^{3}-1155x^{3}\cdots \right ) \\ & +\cos x\left ( x^{5}+20x^{5}+\cdots \right ) \\ & \vdots \\ & +\sin x\left ( -2-12-90-840-9450\cdots \right ) \\ & +\sin x\left ( x^{2}+5x^{2}+39x^{2}+375x^{2}+4305x^{2}+\cdots \right ) \\ & +\sin x\left ( -14x^{4}-195x^{4}-\cdots \right ) \\ & +\sin x\left ( x^{6}+\cdots \right ) \end{align*}
or\begin{align*} \varphi (x) & =\cos x\left ( \begin{array} [c]{c}x+2x+12x+90x+840x+9450x+\cdots \\ -x^{3}-9x^{3}-95x^{3}-1155x^{3}\cdots \\ +x^{5}+20x^{5}+\cdots \end{array} \right ) \\ & +\sin x\left ( \begin{array} [c]{c}-2-12-90-840-9450\cdots \\ +x^{2}+5x^{2}+39x^{2}+375x^{2}+4305x^{2}+\cdots \\ -14x^{4}-195x^{4}-\cdots \\ x^{6}+\cdots \end{array} \right ) \end{align*}
But \(\cos x=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}+\cdots \) and \(\sin x=x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\cdots \) then the above can be written as\begin{align*} \varphi (x) & =\left ( 1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}+\cdots \right ) \left ( \begin{array} [c]{c}x+2x+12x+90x+840x+9450x+\cdots \\ -x^{3}-9x^{3}-95x^{3}-1155x^{3}\cdots \\ +x^{5}+20x^{5}+\cdots \end{array} \right ) \\ & +\left ( x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\cdots \right ) \left ( \begin{array} [c]{c}-2-12-90-840-9450\cdots \\ +x^{2}+5x^{2}+39x^{2}+375x^{2}+4305x^{2}+\cdots \\ -14x^{4}-195x^{4}-\cdots \\ x^{6}+\cdots \end{array} \right ) \end{align*}
Looking at few terms\begin{align*} \varphi (x) & =\left ( \begin{array} [c]{c}x+2x+12x+90x+\\ -x^{3}-9x^{3}-95x^{3}\cdots \\ +x^{5}+20x^{5}+\cdots \end{array} \right ) -\frac{x^{2}}{2}\left ( \begin{array} [c]{c}x+2x+12x+90x+\\ -x^{3}-9x^{3}-95x^{3}\cdots \\ +x^{5}+20x^{5}+\cdots \end{array} \right ) +\frac{x^{4}}{24}\left ( \begin{array} [c]{c}x+2x+12x+90x+\\ -x^{3}-9x^{3}-95x^{3}\cdots \\ +x^{5}+20x^{5}+\cdots \end{array} \right ) \cdots \\ & +x\left ( \begin{array} [c]{c}-2-12-90-\cdots \\ +x^{2}+5x^{2}+39x^{2}\cdots \\ -14x^{4}-195x^{4}-\cdots \\ x^{6}+\cdots \end{array} \right ) -\frac{x^{3}}{6}\left ( \begin{array} [c]{c}-2-12-90-\cdots \\ +x^{2}+5x^{2}+39x^{2}\cdots \\ -14x^{4}-195x^{4}-\cdots \\ x^{6}+\cdots \end{array} \right ) +\frac{x^{5}}{120}\left ( \begin{array} [c]{c}-2-12-90-\cdots \\ +x^{2}+5x^{2}+39x^{2}\cdots \\ -14x^{4}-195x^{4}-\cdots \\ x^{6}+\cdots \end{array} \right ) \cdots \end{align*}
Hence\begin{align*} \varphi (x) & =\left ( \begin{array} [c]{c}x+2x+12x+90x+\\ -x^{3}-9x^{3}-95x^{3}\cdots \\ +x^{5}+20x^{5}+\cdots \end{array} \right ) +\left ( \begin{array} [c]{c}-\frac{x^{3}}{2}-x^{3}-6x^{3}-45x^{3}+\\ +\frac{x^{5}}{2}+\frac{9}{2}x^{5}+\frac{95}{2}x^{5}\cdots \\ -\frac{x^{7}}{2}-10x^{7}+\cdots \end{array} \right ) +\left ( \begin{array} [c]{c}\frac{x^{5}}{24}+\frac{x^{5}}{12}+\frac{x^{5}}{2}+\frac{45}{12}x^{5}+\\ -\frac{x^{7}}{24}-\frac{9}{24}x^{7}-\frac{95}{24}x^{7}\cdots \\ +\frac{x^{9}}{24}+\frac{20}{24}x^{9}+\cdots \end{array} \right ) \cdots \\ & +\left ( \begin{array} [c]{c}-2x-12x-90x-\cdots \\ +x^{3}+5x^{3}+39x^{3}\cdots \\ -14x^{5}-195x^{5}-\cdots \\ x^{7}+\cdots \end{array} \right ) +\left ( \begin{array} [c]{c}+\frac{1}{3}x^{3}+2x^{3}+15x^{3}-\cdots \\ -\frac{x^{5}}{6}-\frac{5}{6}x^{5}-\frac{39}{6}x^{5}\cdots \\ +\frac{14}{6}x^{7}+\frac{195}{6}x^{7}-\cdots \\ -\frac{x^{9}}{6}+\cdots \end{array} \right ) +\left ( \begin{array} [c]{c}-\frac{x^{5}}{60}-\frac{1}{10}x^{5}-\frac{3}{4}x^{5}-\cdots \\ +\frac{1}{120}x^{7}+\frac{5}{120}x^{7}+\frac{39}{120}x^{7}\cdots \\ -\frac{14}{120}x^{9}-\frac{195}{120}x^{9}-\cdots \\ \frac{x^{10}}{120}+\cdots \end{array} \right ) \cdots \end{align*}
Hence, collecting terms. In this we start with the smallest coefficients of each power of \(x\) found in all the above, and add them\begin{align*} \varphi (x) & =x\left ( 1+2+12+90+\cdots -2-12-90-\cdots \right ) \\ & +x^{3}\left ( \frac{1}{3}-\frac{1}{2}-1-1+2+5-6+15+39-45-\cdots \right ) \\ & +x^{5}\left ( -\frac{1}{60}+\frac{1}{24}+\frac{1}{12}+\frac{45}{12}-\frac{1}{10}\right ) \end{align*}
or\[ \varphi (x)=x+x^{3}\left ( -\frac{1}{6}+O\left ( h\right ) \right ) +x^{5}\left ( \frac{451}{120}+O\left ( h\right ) \right ) \] May be a \[ \fbox{$\sin \left ( x\right ) $}\] since \(\sin \left ( x\right ) =x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\cdots \)? This problem is too hard but the above is my final answer.
Solve \[ 2\cosh \left ( x\right ) -\sinh \left ( x\right ) -\left ( 2-x\right ) =1+\int _{0}^{x}\left ( 2-x+t\right ) \varphi \left ( t\right ) dt \] Solution:
Taking derivative w.r.t. \(x\) gives\begin{align} 2\sin \left ( x\right ) -\cosh \left ( x\right ) +1 & =\int _{0}^{x}\frac{d}{dx}\left ( 2-x+t\right ) \varphi \left ( t\right ) dt+\frac{dx}{dx}\left ( \left ( 2-x+x\right ) \varphi \left ( x\right ) \right ) -0\nonumber \\ 2\sin \left ( x\right ) -\cosh \left ( x\right ) +1 & =\int _{0}^{x}-\varphi \left ( t\right ) dt+2\varphi \left ( x\right ) \nonumber \\ 2\sinh \left ( x\right ) -\cosh \left ( x\right ) +1-2\varphi \left ( x\right ) & =-\int _{0}^{x}\varphi \left ( t\right ) dt\nonumber \\ \cosh \left ( x\right ) -2\sinh \left ( x\right ) -1+2\varphi \left ( x\right ) & =\int _{0}^{x}\varphi \left ( t\right ) dt \tag{1} \end{align}
Differneting again w.r.t. \(x\)\begin{align*} \sinh \left ( x\right ) -2\cosh \left ( x\right ) +2\varphi ^{\prime }\left ( x\right ) & =\int _{0}^{x}\frac{d}{dx}\varphi \left ( t\right ) dt+\frac{dx}{dx}\left ( \varphi \left ( x\right ) \right ) -0\\ & =\varphi \left ( x\right ) \end{align*}
Hence the ODE is\[ \varphi ^{\prime }\left ( x\right ) -\frac{1}{2}\varphi \left ( x\right ) =\cosh \left ( x\right ) -\frac{1}{2}\sinh \left ( x\right ) \] Integrating factor is \(e^{-\frac{x}{2}}\)hence the solution is\begin{align*} d\left ( e^{-\frac{x}{2}}\varphi \left ( x\right ) \right ) & =\int e^{-\frac{x}{2}}\cosh \left ( x\right ) -\frac{1}{2}\int e^{-\frac{x}{2}}\sinh \left ( x\right ) \\ e^{-\frac{x}{2}}\varphi \left ( x\right ) & =\cosh \left ( \frac{x}{2}\right ) -\frac{1}{3}\cosh \left ( \frac{3x}{2}\right ) +\sinh \left ( \frac{x}{2}\right ) +\frac{1}{3}\sinh \left ( \frac{3x}{2}\right ) \\ & -\frac{1}{2}\left [ \sinh \left ( \frac{x}{2}\right ) -\frac{1}{3}\sinh \left ( \frac{3x}{2}\right ) +\cosh \left ( \frac{x}{2}\right ) +\frac{1}{3}\cosh \left ( \frac{3x}{2}\right ) \right ] +C \end{align*}
Simplfying RHS\begin{align*} e^{-\frac{x}{2}}\varphi \left ( x\right ) & =\cosh \left ( \frac{x}{2}\right ) -\frac{1}{3}\cosh \left ( \frac{3x}{2}\right ) +\sinh \left ( \frac{x}{2}\right ) +\frac{1}{3}\sinh \left ( \frac{3x}{2}\right ) -\frac{1}{2}\sinh \left ( \frac{x}{2}\right ) \\ & +\frac{1}{6}\sinh \left ( \frac{3x}{2}\right ) -\frac{1}{2}\cosh \left ( \frac{x}{2}\right ) -\frac{1}{6}\cosh \left ( \frac{3x}{2}\right ) +C\\ & \\ & =\frac{1}{2}\cosh \left ( \frac{1}{2}x\right ) -\frac{1}{2}\cosh \left ( \frac{3}{2}x\right ) +\frac{1}{2}\sinh \left ( \frac{1}{2}x\right ) +\frac{1}{2}\sinh \left ( \frac{3}{2}x\right ) +C \end{align*}
Therefore\[ \varphi \left ( x\right ) =\frac{e^{\frac{x}{2}}}{2}\left ( \cosh \left ( \frac{1}{2}x\right ) -\cosh \left ( \frac{3}{2}x\right ) +\sinh \left ( \frac{1}{2}x\right ) +\sinh \left ( \frac{3}{2}x\right ) \right ) +Ce^{\frac{x}{2}}\] Where the constant of integration \(C.\) To find this constant, looking at Eq. (1) above, repeated below\[ \cosh \left ( x\right ) -2\sinh \left ( x\right ) -1+2\varphi \left ( x\right ) =\int _{0}^{x}\varphi \left ( t\right ) dt \] We see that at \(x=0\) the above gives\begin{align*} 1-1+2\varphi \left ( 0\right ) & =0\\ \varphi \left ( 0\right ) & =0 \end{align*}
Therefore, we let \(x=0\) in the solution itself and find \(C\) which gives\begin{align*} \varphi \left ( 0\right ) & =0=\frac{1}{2}\left ( 1-1+0+0\right ) +C\\ C & =0 \end{align*}
Hence the solution is\[ \varphi \left ( x\right ) =\frac{e^{\frac{x}{2}}}{2}\left ( \cosh \left ( \frac{1}{2}x\right ) -\cosh \left ( \frac{3}{2}x\right ) +\sinh \left ( \frac{1}{2}x\right ) +\sinh \left ( \frac{3}{2}x\right ) \right ) \]
Solve \[ u\left ( x\right ) =\cos \left ( x\right ) -x-2+\int _{0}^{x}\left ( t-x\right ) u\left ( t\right ) dt \]
solution
Taking derivative w.r.t. \(x\)\begin{align*} u^{\prime }\left ( x\right ) & =-\sin \left ( x\right ) -1+\int _{0}^{x}\frac{d}{dx}\left ( t-x\right ) u\left ( t\right ) dt+\left [ \frac{dx}{dx}\left ( x-x\right ) u\left ( x\right ) -0\right ] \\ & =-\sin \left ( x\right ) -1-\int _{0}^{x}u\left ( t\right ) dt \end{align*}
Taking another derivative w.r.t. \(x\)\begin{align*} u^{\prime \prime }\left ( x\right ) & =-\cos \left ( x\right ) -\left ( \int _{0}^{x}\frac{d}{dx}u\left ( t\right ) dt+\frac{dx}{dx}u\left ( x\right ) -0\right ) \\ & =-\cos \left ( x\right ) -u\left ( x\right ) \end{align*}
The differential equation is\begin{equation} u^{\prime \prime }\left ( x\right ) +u\left ( x\right ) =-\cos x \tag{1} \end{equation} The homogeneous solution is\[ u_{h}=Ae^{\lambda _{1}x}+Be^{\lambda _{2}x}\] Where \(\lambda _{1,2}=\frac{\pm \sqrt{-4}}{2}=\pm i\), hence \[ u_{h}=A\cos x+B\sin x \] To find the particular solution, let \[ u_{p}=c_{1}x\cos x+c_{2}x\sin x \] hence \begin{align*} u_{p}^{\prime } & =c_{1}\cos x-c_{1}x\sin x+c_{2}\sin x+c_{2}x\cos x\\ u_{p}^{\prime \prime } & =-c_{1}\sin x-c_{1}\sin x-c_{1}x\cos x+c_{2}\cos x+c_{2}\cos x-c_{2}x\sin x \end{align*}
Substituting all the above in the original ODE Eq. (1)\begin{align*} -c_{1}\sin x-c_{1}\sin x-c_{1}x\cos x+c_{2}\cos x+c_{2}\cos x-c_{2}x\sin x+\left ( c_{1}x\cos x+c_{2}x\sin x\right ) & =-\cos x\\ 2c_{2}\cos x-2c_{1}\sin x & =-\cos x \end{align*}
Hence \(c_{1}=0\) and \(c_{2}=-\frac{1}{2}\), therefore, the particular solution is\[ u_{p}=-\frac{1}{2}x\sin x \] This gives the complete solution\begin{equation} u=A\cos x+B\sin x-\frac{1}{2}x\sin x \tag{1} \end{equation} At \(x=0\), the original integral equation becomes\begin{align*} u\left ( 0\right ) & =\cos \left ( 0\right ) -0-2+\int _{0}^{0}\left ( t-x\right ) u\left ( t\right ) dt\\ & =1-2=-1 \end{align*}
Hence Eq.(1) when \(x=0\) gives\[ -1=A \] Hence the solution now is\begin{equation} u=-\cos x+B\sin x-\frac{1}{2}x\sin x \tag{2} \end{equation} Now to find \(B\) we take derivative of the integral equation and evaluate it at \(x=0\) which gives\begin{align*} u^{\prime }\left ( 0\right ) & =-\sin \left ( 0\right ) -1-\int _{0}^{0}u\left ( t\right ) dt\\ & =-1 \end{align*}
But the derivative of Eq (2) is\[ u^{\prime }=\sin x+B\cos x-\frac{1}{2}\left ( \sin x+x\cos x\right ) \] And at \(x=0\) it gives\[ u^{\prime }\left ( 0\right ) =B \] Hence \(B=-1\) and the solution is\[ u=-\cos x-\sin x-\frac{1}{2}x\sin x \]
Solve \(u\left ( x\right ) =x+\lambda \int _{0}^{1}\left ( 1+x+t\right ) u\left ( t\right ) dt\)
solution
Taking derivative w.r.t. \(x\)\begin{align*} u^{\prime }\left ( x\right ) & =1+\lambda \int _{0}^{1}\frac{d}{dx}\left ( 1+x+t\right ) u\left ( t\right ) dt\\ & =1+\lambda \int _{0}^{1}u\left ( t\right ) dt \end{align*}
Taking derivative w.r.t. \(x\) again\begin{align*} u^{\prime \prime }\left ( x\right ) & =\lambda \int _{0}^{1}\frac{d}{dx}u\left ( t\right ) dt\\ & =0 \end{align*}
Hence the solution is\[ u\left ( x\right ) =Ax+B \] We can rewrite the integral equation as\[ u\left ( x\right ) =x+\lambda \left ( \int _{0}^{x}\left ( 1+x+t\right ) u\left ( t\right ) dt+\int _{x}^{1}\left ( 1+x+t\right ) u\left ( t\right ) dt\right ) \] And it will give the same answer ofcourse. \(x\) is some point between \(0\) and \(1.\) Now we processed as before. Taking derivative w.r.t. \(x\)\begin{align*} u^{\prime }\left ( x\right ) & =1+\lambda \left ( \int _{0}^{x}\frac{d}{dx}\left ( 1+x+t\right ) u\left ( t\right ) dt+\frac{dx}{dx}\left ( 1+x+x\right ) u\left ( x\right ) -0\right ) \\ & +\lambda \left ( \int _{x}^{1}\frac{d}{dx}\left ( 1+x+t\right ) u\left ( t\right ) dt+0-\frac{dx}{dx}\left ( 1+x+x\right ) u\left ( x\right ) \right ) \\ & \\ & =1+\lambda \left ( \int _{0}^{x}u\left ( t\right ) dt+\left ( 1+2x\right ) u\left ( x\right ) \right ) +\lambda \left ( \int _{x}^{1}u\left ( t\right ) dt-\left ( 1+2x\right ) u\left ( x\right ) \right ) \end{align*}
Taking derivative w.r.t. \(x\) one more time\begin{align*} u^{\prime \prime }\left ( x\right ) & =\lambda \left ( \int _{0}^{x}\frac{d}{dx}u\left ( t\right ) dt+\left [ \frac{dx}{dx}u\left ( x\right ) -0\right ] +\frac{d}{dx}\left ( 1+2x\right ) u\left ( x\right ) \right ) \\ & +\lambda \left ( \int _{x}^{1}\frac{d}{dx}u\left ( t\right ) dt+\left [ 0-\frac{dx}{dx}u\left ( x\right ) \right ] -\frac{d}{dx}\left ( 1+2x\right ) u\left ( x\right ) \right ) \\ & \\ & =\lambda \left ( u\left ( x\right ) +2u\left ( x\right ) +\left ( 1+2x\right ) u^{\prime }\left ( x\right ) \right ) +\lambda \left ( -u\left ( x\right ) -2u\left ( x\right ) -\left ( 1+2x\right ) u^{\prime }\left ( x\right ) \right ) \\ & \\ & =\lambda \left ( \left ( 1+2x\right ) u^{\prime }\left ( x\right ) \right ) +\lambda \left ( -\left ( 1+2x\right ) u^{\prime }\left ( x\right ) \right ) \\ & =0 \end{align*}
Hence the solution is\begin{equation} u\left ( x\right ) =Ax+B \tag{1} \end{equation} To find the constants of integrations, we see that at \(x=0\) the integral equation gives\[ u\left ( 0\right ) =0+\lambda \int _{0}^{0}\left ( 1+x+t\right ) u\left ( t\right ) dt=0 \] Hence from Eq. (1) at \(x=0\) we find \(B=0\), hence the solution is\begin{equation} u\left ( x\right ) =Ax \tag{2} \end{equation} Taking derivative \(u^{\prime }\left ( x\right ) =A\), but from original integral equation we found that\[ u^{\prime }\left ( x\right ) =1+\lambda \int _{0}^{1}u\left ( t\right ) dt \] Hence at \(x=0\) \[ u^{\prime }\left ( 0\right ) =1 \] Therefore \(A=1\) and the solution is\[ u\left ( x\right ) =x \]
Solve \(\frac{dy\left ( t\right ) }{dt}=2-\frac{t^{2}}{2}-\frac{1}{4}\int _{0}^{t}y\left ( \tau \right ) d\tau \)
Solution
If we integrate from \(0\) to \(t\), then\begin{align*} \int _{0}^{t}\frac{dy\left ( \tau \right ) }{d\tau }d\tau & =\int _{0}^{t}2d\tau -\int _{0}^{t}\frac{\tau ^{2}}{2}d\tau -\frac{1}{4}\int _{0}^{t}\left ( \int _{0}^{t_{1}}y\left ( \tau _{2}\right ) d\tau _{2}\right ) d\tau \\ y\left ( t\right ) -y\left ( 0\right ) & =2t-\frac{t^{3}}{6}-\frac{1}{4}\int _{0}^{t}\left ( t-\tau _{2}\right ) y\left ( \tau _{2}\right ) d\tau _{2}\\ y\left ( t\right ) & =2t-\frac{t^{3}}{6}-\frac{1}{4}\int _{0}^{t}\left ( t-\tau \right ) y\left ( \tau \right ) d\tau \end{align*}
Check the above. Can I do this below? Taking derivative w.r.t. \(t\)\begin{align*} \frac{d^{2}y\left ( t\right ) }{dt^{2}} & =-t-\frac{1}{4}\left ( \int _{0}^{t}\frac{d}{dt}y\left ( \tau \right ) d\tau +\frac{dt}{dt}y\left ( t\right ) -0\right ) \\ & =-t-\frac{1}{4}y\left ( t\right ) \end{align*}
Hence\[ y^{\prime \prime }+\frac{1}{4}y\left ( t\right ) =-t \] The roots are \(\frac{i}{2},-\frac{i}{2}\), hence \[ y_{h}=B\cos \frac{t}{2}+C\sin \frac{t}{2}\] For particular solution, let \[ y_{p}=c_{1}t+c_{2}\] and substituting in the ODE gives \(\frac{1}{4}c_{1}=-1\) or \(c_{1}=-4\), so \[ y_{p}=-4t \] Hence the solution is\[ y=B\cos \frac{t}{2}+C\sin \frac{t}{2}-4t \] When \(x=0\)\[ 0=B \] Hence the solution becomes\begin{equation} y\left ( t\right ) =C\sin \frac{t}{2}-4t \tag{1} \end{equation} To find the constant \(C\), from the integral equation, at \(t=0\)\begin{align*} \frac{dy\left ( 0\right ) }{dt} & =2-\frac{0}{2}-\frac{1}{4}\int _{0}^{0}y\left ( \tau \right ) d\tau \\ & =2 \end{align*}
And from Eq. (1) \(y^{\prime }\left ( t\right ) =C\frac{1}{2}\cos \frac{t}{2}-4\), hence at \(t=0\)\begin{align*} 2 & =C\frac{1}{2}-4\\ C & =12 \end{align*}
Hence the solution in Eq. (1) becomes\[ y\left ( t\right ) =12\sin \frac{t}{2}-4t \]
Solve the integral equation using any method
\[ y\left ( x\right ) =f\left ( x\right ) -A\int _{0}^{x}xte^{\lambda \left ( x-t\right ) }y\left ( t\right ) dt \]
Solution:
The above is a Volterra integral equation, inhomogeneous since \(f\left ( x\right ) \) exist, and second kind since the function we solving for is under the integral as well. We start by removing \(x\) dependencies from inside the integral to the outside\[ y\left ( x\right ) =f\left ( x\right ) -Axe^{\lambda x}\int _{0}^{x}te^{-\lambda t}y\left ( t\right ) dt \] Now divide by \(xe^{\lambda x}\) (notice that \(x\) can not be zero, hence initial conditions must start at some other value).\[ \frac{y\left ( x\right ) }{xe^{\lambda x}}=\frac{f\left ( x\right ) }{xe^{\lambda x}}-A\int _{0}^{x}te^{-\lambda t}y\left ( t\right ) dt \] Let \(\phi \left ( x\right ) =\frac{y\left ( x\right ) }{xe^{\lambda x}}\) and \(F\left ( x\right ) =\frac{f\left ( x\right ) }{xe^{\lambda x}}\) the above becomes\[ \phi \left ( x\right ) =F\left ( x\right ) -A\int _{0}^{x}te^{-\lambda t}y\left ( t\right ) dt \] Now inside the integral, multiply by \(\frac{te^{\lambda t}}{te^{\lambda t}}\) in order to obtain the same form as on LHS\begin{align*} \phi \left ( x\right ) & =F\left ( x\right ) -A\int _{0}^{x}te^{-\lambda t}te^{\lambda t}\left ( \frac{y\left ( t\right ) }{te^{\lambda t}}\right ) dt\\ & =F\left ( x\right ) -A\int _{0}^{x}t^{2}\phi \left ( t\right ) dt \end{align*}
Taking derivative\begin{align*} \phi ^{\prime }\left ( x\right ) & =F^{\prime }\left ( x\right ) -A\left ( \int _{0}^{x}\frac{d}{dx}t^{2}\phi \left ( t\right ) dt+\frac{dx}{dx}x^{2}\phi \left ( x\right ) -0\right ) \\ & =F^{\prime }\left ( x\right ) -Ax^{2}\phi \left ( x\right ) \\ \phi ^{\prime }\left ( x\right ) +Ax^{2}\phi \left ( x\right ) & =F^{\prime }\left ( x\right ) \end{align*}
Integrating factor is \(e^{\int Ax^{2}dx}=e^{A\frac{x^{3}}{3}}\), hence\[ d\left ( e^{A\frac{x^{3}}{3}}\phi \left ( x\right ) \right ) =e^{A\frac{x^{3}}{3}}F^{\prime }\left ( x\right ) \] Integrate both sides from \(0\) to \(x\)\begin{align*} \int _{0}^{x}d\left ( e^{A\frac{z^{3}}{3}}\phi \left ( z\right ) \right ) & =\int _{0}^{x}e^{A\frac{z^{3}}{3}}\frac{dF}{dz}dz\\ e^{A\frac{x^{3}}{3}}\phi \left ( x\right ) -\phi \left ( 0\right ) & =\left [ e^{A\frac{z^{3}}{3}}F\right ] _{0}^{x}-\int _{0}^{x}Fd\left ( e^{A\frac{z^{3}}{3}}\right ) \\ e^{A\frac{x^{3}}{3}}\phi \left ( x\right ) -\phi \left ( 0\right ) & =\left ( e^{A\frac{x^{3}}{3}}F\left ( x\right ) -F\left ( 0\right ) \right ) -\int _{0}^{x}FAz^{2}e^{A\frac{z^{3}}{3}}dz\\ e^{A\frac{x^{3}}{3}}\phi \left ( x\right ) & =\phi \left ( 0\right ) +e^{A\frac{x^{3}}{3}}F\left ( x\right ) -F\left ( 0\right ) -\int _{0}^{x}FAz^{2}e^{A\frac{z^{3}}{3}}dz\\ \phi \left ( x\right ) & =e^{-A\frac{x^{3}}{3}}\phi \left ( 0\right ) +F\left ( x\right ) -e^{-A\frac{x^{3}}{3}}F\left ( 0\right ) -Ae^{-A\frac{x^{3}}{3}}\int _{0}^{x}Fz^{2}e^{A\frac{z^{3}}{3}}dz \end{align*}
But \(\phi \left ( x\right ) =\frac{y\left ( x\right ) }{xe^{\lambda x}}\) and \(F\left ( x\right ) =\frac{f\left ( x\right ) }{xe^{\lambda x}}\), therefore the above becomes (what to do with the division by zero?)\begin{align*} \frac{y\left ( x\right ) }{xe^{\lambda x}} & =e^{-A\frac{x^{3}}{3}}\frac{y\left ( 0\right ) }{\lim _{x\rightarrow 0}xe^{\lambda x}}+\frac{f\left ( x\right ) }{xe^{\lambda x}}-e^{-A\frac{x^{3}}{3}}\frac{f\left ( 0\right ) }{\lim _{x\rightarrow 0}xe^{\lambda x}}-Ae^{-A\frac{x^{3}}{3}}\int _{a}^{x}\frac{f\left ( z\right ) }{ze^{\lambda z}}z^{2}e^{A\frac{z^{3}}{3}}dz\\ y\left ( x\right ) & =xe^{\left ( -A\frac{x^{3}}{3}+\lambda x\right ) }\frac{y\left ( 0\right ) }{\lim _{x\rightarrow 0}xe^{\lambda x}}+f\left ( x\right ) -xe^{\left ( -A\frac{x^{3}}{3}+\lambda x\right ) }\frac{f\left ( 0\right ) }{\lim _{x\rightarrow 0}xe^{\lambda x}}-xAe^{\left ( -A\frac{x^{3}}{3}+\lambda x\right ) }\int _{a}^{x}\frac{f\left ( z\right ) }{e^{\lambda z}}ze^{A\frac{z^{3}}{3}}dz \end{align*}
Assuming zero initial conditions for now, which means \(f\left ( 0\right ) =0\) and \(y\left ( 0\right ) =0\), then in the limit the above reduces to\[ y\left ( x\right ) =f\left ( x\right ) -\frac{xAe^{\lambda x}}{e^{\frac{Ax^{3}}{3}}}\int _{a}^{x}zf\left ( z\right ) \frac{e^{\frac{Az^{3}}{3}}}{e^{\lambda z}}dz \]