91Ó°ÊÓ

\(f(x)=x^{2}, \quad 0

$$\frac{\partial u}{\partial t}=5 \frac{\partial^{2} u}{\partial x^{2}}, 0<\mathcal{x}<1,t>0,$$ $$u(0, t)=u(1, t)=0, \quad t>0,$$ $$u(x, 0)=(1-x) x^{2},0<\mathcal{x}<1$$

\(\frac{\partial^{2} u}{\partial t^{2}}=\frac{\partial^{2} u}{\partial x^{2}}\) 0\(<\)x\(<\)1 t\(>\)0 $$ u(0, t)=u(1, t)=0, \quad t>0$$ $$ u(x, 0) $$$=x(1-x)$$0\(<\)x\(<\)1\( $$ \frac{\partial u}{\partial t}(x, 0) $$$=\sin 7 \pi x$$0\)<\(x\)<\(1\)

\begin{equation}y ^ { \prime \prime } - y = 0 ; \quad 0 < x < 1\end{equation} \begin{equation}y ( 0 ) = 0 , \quad y ( 1 ) = - 4\end{equation}

$$f(x)=x^{3}+\sin 2 x$$

Find a formal solution to the given boundary value problem. $$ \begin{array}{l} \frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, \quad 0

$$f(x)=\sin ^{2} x$$

\(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, \quad 0< x <\pi, \quad 0< y <\pi,\) \(\frac{\partial u}{\partial x}(0, y)=\frac{\partial u}{\partial x}(\pi, y)=0, \quad 0 \leq y \leq \pi,\) \(u(x, 0)=\cos x-2 \cos 4 x, \quad 0 \leq x \leq \pi,\) \(u(x, \pi)=0, \quad 0 \leq x \leq \pi\)

$$ \frac{\partial^{2} u}{\partial t^{2}} $$$=16 \frac{\partial^{2} u}{\partial x^{2}}$$ 0<\(x\)<\pi, \quad t>0\( $$ u(0, t)=u(\pi, t)=0, \quad t>0$$ $$ u(x, 0) $$$=\sin ^{2} x, $$0<\)x\(<\pi\) $$ \frac{\partial u}{\partial t}(x, 0) $$$1-\cos x$$0<\(x\)<\pi$

$$\frac{\partial u}{\partial t}=\frac{\partial^{2} u}{\partial x^{2}}, 0<\mathcal{x}<\pi,t>0,$$ $$u(0, t)=u(1, t)=0, \quad t>0,$$ $$u(x, 0)=(1-x) x^{2},0<\mathcal{x}<\pi$$