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$y^{\prime \prime}+4 y=1$

$y(\pi)=0, y^{\prime}(\pi)=0$

IVP

$y^{\prime \prime}+4 y=0 \quad r^{2}+4=0 \quad r^{2}=-4 \quad r=\pm 2i$

$\lambda=0, \mu=2$

$y=c_{1} e^{\lambda t} \cos \mu t+c_{2} e^{\lambda t} \sin \mu t$

$y_{h}=c_{1} \cos 2 x+c_{2} \sin 2 x$

$y_{1}=\cos 2 x$

$y_{1}^{\prime}=-2 \sin 2 x$

$y_{2}=\sin 2 x$

$y_{2}^{\prime}=2 \cos 2 x$

$w=\left|\begin{array}{ll}{y_{1}} & {y_{2}} \\ {y_{1}^\prime} & {y_{2}^\prime}\end{array}\right|=\left|\begin{array}{c}{\cos 2 t}& {\sin 2 t} \\ {-2 \sin 2 t} & { 2 \cos 2 t}\end{array}\right|$

$=2 \cos ^{2} 2 t+2 \sin ^{2} 2 t$

$=2\left(\cos ^{2} 2 t+\sin ^{2} 2 t\right)$

$w(t)=2$

$G(x, t)=\frac{y_{1}(t) y_{2}(x)-y_{1}(x) y_{2}(t)}{w}$

$G(x, t)=\frac{\cos 2 t \sin 2 x-\cos 2 x \sin 2 t}{2}$

$\sin (x-y)=\sin x \cos y-\sin y \cos x$

$G(x, t)=\frac{1}{2} \sin (2 x-2 t)$

$y_{p}=\int_{\pi}^{x} \frac{1}{2} \sin (2 x-2 t) d t$

$\frac{1}{2} \sin (2 x-2 t)$

$\frac{d}{d x} \cos (2 x-2 t)=-2 \sin (2 x -2 t)$

$\frac{1}{2} \sin (2 x-2 t)$

$\frac{d}{d t} \cos (2 x-2 t)=+2 \sin (2 x-2 t)$

$=\frac{1}{4} \int_{\pi}^{x} 2 \sin (2 x-2 t) d t$

$=\left.\frac{1}{4} \cos (2 x-2 t)\right]_{\pi} ^{x}$

$=\frac{1}{4}[1-\cos (2 x-2 \pi)]$

$\cos -x=\cos x$

$\cos (2 x-2 \pi)=\cos (2 \pi-2 x)$

$\cos (2 \pi-x)=\cos x$

$y_{p}=\frac{1}{4}-\frac{1}{4} \cos 2 x$

$y_{h}=c_{1} \cos 2 x+c_{2} \sin 2 x$

$y_{p}=\frac{1}{4}(1-\cos 2 x)$

$y=c_{1} \cos 2 x+c_{2} \sin 2 x+\frac{1}{4}(1-\cos 2 x)$

$y(\pi)=0$

$0=c_{1}(1)+c_{2}(0)+\frac{1}{4}(1-1)$

$\Rightarrow C_{1}=0$

$y=c_{2} \sin 2 x+\frac{1}{4}(1-\cos 2 x)$

$y^{\prime}=2 c_{2} \cos 2 x+\frac{1}{4}(0+2 \sin 2 x)$

$0=2 c_{2}(1)+\frac{1}{4}(0) \Rightarrow C_{2}=0$

$y=\frac{1}{4}(1-\cos 2 x)$

$y^{\prime \prime}=x^{2}, y(0)=0, y(1)=0 \quad a=0, b=1$

$y^{\prime \prime}=0 \quad y^{\prime}=A \quad y=A x+B$

$0=A(0)+B \quad B=0 \quad A$

$y_{1}=A x \quad y_{1}=x$

$y(1)=0 \quad 0=A(1)+B \quad {A}=-{B}$

$y_{2}=-x+1$

$y_{h}=c_{1} x+c_{2}(1-x)$

$y_{1}^{\prime}=1, y_{2}^{\prime}=-1$

$w(t)=\left|\begin{array}{ll}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right|$

$=\left|\begin{array}{cc}{t} & {1-t} \\ {1} & {-1}\end{array}\right|$

$w=-t-1+t$

$w=-1$

$G(x, t)=$

$\frac{y_{1}(t) y_{2}(x)}{w} \quad 0 \leq t \leq x$

$G(x, t)=$

$\frac{y_{1}(x) y_{2}(t)}{w} x \leq t \leq 1$

$G(x, t)=\left[\begin{array}{cc}{-t(1-x)} & {0 \leq t \leq x} \\ {-x(1-t)} & {x \leq t \leq 1}\end{array}\right.$

$y_{p}=\int_{0}^{x}-t(1-x) t^{2} d t+\int_{x}^{1}-x(1-t) t^{2} d t$

$=\left[-\frac{t^{4}}{4}(1-x)\right]_{0}^{x}+\left[\frac{-x t^{3}}{3}+\frac{x t^{4}}{4}\right]_{x}^{1}$

$=\frac{-x^{4}}{4}(1-x)+\left[\frac{-x}{3}+\frac{x}{4}\right]-\left[\frac{-x^{4}}{3}+\frac{x^{5}}{4}\right]$

$=\frac{-x^{4}}{4}+\frac{x^{5}}{4}-\frac{1}{12} x+\frac{x^{4}}{3}-\frac{x^{5}}{4}$

$=\frac{1}{12} x^{4}-\frac{1}{12} x \Rightarrow y_{p}=\frac{1}{12}\left(x^{4}-x\right)$

$y=y_{h}+y_{p}$