Zestaw I

1. Oblicz

∬x²y dx dy, gdzie x² + y² ≤ 9; $\int_{- \sqrt{9 - x^{2}}}^{\sqrt{9 - x^{2}}}{\int_{- 3}^{3}{x^{2}\text{ydxdy}} =}\int_{- \sqrt{9 - x^{2}}}^{\sqrt{9 - x^{2}}}{\frac{x^{2}y^{2}}{2}\text{dx}\left. \ \right|_{- 3}^{3} =}\int_{- \sqrt{9 - x^{2}}}^{\sqrt{9 - x^{2}}}{\left( \frac{x^{2}*9}{2} - \frac{x^{2}*9}{2} \right)dx = 0}$

2. Oblicz za pomocą całki podwójnej pole obszaru ograniczonego krzywymi

y = 0,  y = −x² + x + 2; ∬_Ddxdy; =b² − 4ac = 9; $x_{1} = \frac{- b \pm \sqrt{}}{2a} = \frac{- 4}{- 2} = 2,\ \ \ \ \ \ x_{2} = \frac{2}{- 2} = - 1$; $\int_{- 1}^{2}{\int_{0}^{- x^{2} + x + 2}{dxdy = \int_{- 1}^{2}{y\left. \ \right|_{0}^{- x^{2} + x + 2}\text{dx} = \int_{- 1}^{2}{- \left( x^{2} + x + 2 \right)dx = \left( - \frac{x^{3}}{3} + \frac{x^{2}}{2} + 2x \right)\left. \ \right|_{- 1}^{2}}}}} = - \frac{2^{3}}{3} + \frac{2^{2}}{2} + 2*2 - \left( - \frac{\left( - 1 \right)^{3}}{3} + \frac{\left( - 1 \right)^{2}}{2} + 2*\left( - 1 \right) \right) = - \frac{8}{2} + 2 + 4 - \frac{1}{3} - \frac{1}{2} + 2 = - 3 + 8 - \frac{1}{2} = 4\frac{1}{2}$

3. Oblicz za pomocą całki podwójnej objętość bryły ograniczonej płaszczyzną z = 0,    oraz powierzchnią  V : x² + y² + z² = 4,   z ≥ 0; $z = \sqrt{4 - x^{2} - y^{2}}$; $\iint_{D}^{}{\sqrt{4 - x^{2} - y^{2}}\text{dxdy}}$; x = r cosφ ,    y = r sinφ,     I = r; 0 ≤ r ≤ 2,     0 ≤ φ ≤ 2π; $\int_{0}^{2\pi}{\int_{0}^{2}{\sqrt{4 - r^{2}\operatorname{}\varphi - r^{2}\operatorname{}\varphi}*rdrd\varphi =}}\int_{0}^{2\pi}{\int_{0}^{2}{\sqrt{4 - r^{2}\operatorname{}\varphi + \operatorname{}\varphi)}*rdrd\varphi =}}\int_{0}^{2\pi}{\int_{0}^{2}{\sqrt{4 - r^{2}}*rdrd\varphi = \left| \begin{matrix} 4 - r^{2} = t \\ \begin{matrix} dr*\left( - 2 \right)r = dt \\ rdr = - \frac{\text{dt}}{2} \\ \end{matrix} \\ \end{matrix} \right|}\ } = - \int_{0}^{2\pi}{\int_{0}^{2}{\frac{\sqrt{t}}{2}*dtd\varphi =}} - \int_{0}^{2\pi}{\ \left( \frac{t^{\frac{3}{2}}}{3} \right)\left. \ \right|_{0}^{2}}\ d\varphi = - \int_{0}^{2\pi}{\ \left( \frac{{(4 - r^{2})}^{\frac{3}{2}}}{3} \right)\left. \ \right|_{0}^{2}}\text{\ dφ} = - \int_{0}^{2\pi}{\ \left( \frac{{(4 - 4)}^{\frac{3}{2}}}{3} - \frac{{(4 - 0)}^{\frac{3}{2}}}{3} \right)}\ d\varphi = \int_{0}^{2\pi}\frac{\sqrt{64}}{3}\ d\varphi = \frac{8}{3}\int_{0}^{2\pi}\text{dφ} = \ \frac{8}{3}\varphi\left. \ \right|_{0}^{2\pi} = \frac{16}{3}\pi$

4. Oblicz

∭_Ve^{−(x2+y2+z2)}dxdydz ; gdzie V :   x² + y² + z² ≤ 4; $\iiint_{V}^{}{\frac{e^{- \left( x^{2} + y^{2} + z^{2} \right)}}{\sqrt{x^{2} + y^{2} + z^{2}}}\text{dxdydz\ }}$

z = rcosθ,    r^′ = rsinθ,   x = r^′cosφ = rsinθcosφ

y = r^′sinφ = rsinθsinφ,     I = r²sinθ; 0 ≤ r ≤ 2,   0 ≤ θ ≤ π,    0 ≤ φ ≤ 2π; $\frac{e^{- \left( r^{2}\operatorname{}\theta\operatorname{}\varphi + r^{2}\operatorname{}\theta\operatorname{}\varphi + r^{2}\operatorname{}\theta \right)}}{\sqrt{r^{2}\operatorname{}\theta\operatorname{}\varphi + r^{2}\operatorname{}\theta\operatorname{}\varphi + r^{2}\operatorname{}\theta}} = \frac{e^{- \left( r^{2}\operatorname{}\theta\operatorname{}\varphi + \operatorname{}\varphi) + r^{2}\operatorname{}\theta \right)}}{\sqrt{r^{2}\operatorname{}\theta\operatorname{}\varphi + \operatorname{}\varphi) + r^{2}\operatorname{}\theta}} = \frac{e^{- \left( r^{2}\operatorname{}\theta + r^{2}\operatorname{}\theta \right)}}{\sqrt{r^{2}\operatorname{}\theta + r^{2}\operatorname{}\theta}} = \frac{e^{- \left( r^{2}(\operatorname{}\theta + \operatorname{}\theta) \right)}}{\sqrt{r^{2}(\operatorname{}\theta + \operatorname{}\theta)}} = \frac{e^{- r^{2}}}{\sqrt{r^{2}}} = \frac{e^{- r^{2}}}{r}$; $\int_{0}^{\pi}{\int_{0}^{2\pi}{\int_{0}^{2}{\frac{e^{- r^{2}}}{r}r^{2}}}}\sin\theta drd\varphi d\theta = \left| \begin{matrix} - r^{2} = t \\ \begin{matrix} dr*\left( - 2 \right)r = dt \\ rdr = - \frac{\text{dt}}{2} \\ \end{matrix} \\ \end{matrix} \right| = - \frac{1}{2}\int_{0}^{\pi}{\int_{0}^{2\pi}{\int_{0}^{2}e^{t}}}\sin\theta\text{dtdφdθ} = - \frac{1}{2}\int_{0}^{\pi}{\int_{0}^{2\pi}{e^{t}\left. \ \right|_{0}^{2}}}\sin\theta d\varphi d\theta = - \frac{1}{2}\int_{0}^{\pi}{\int_{0}^{2\pi}{e^{- r^{2}}\left. \ \right|_{0}^{2}}}\sin\theta\text{dφdθ} = - \frac{1}{2}\int_{0}^{\pi}{\int_{0}^{2\pi}{{(e}^{- 4} - e^{0})}}\sin\theta\text{dφdθ} = - \frac{1}{2}\left( \frac{1}{e^{4}}\ - 1 \right)\int_{0}^{\pi}{\int_{0}^{2\pi}{\sin\theta\text{dφdθ}}} = - \frac{1}{2}\left( \frac{1}{e^{4}} - 1 \right)\int_{0}^{\pi}{\varphi\left. \ \right|_{0}^{2\pi}\sin\theta\text{dθ}} = - \frac{1}{2}\left( \frac{1}{e^{4}} - 1 \right)\int_{0}^{\pi}{\varphi\left. \ \right|_{0}^{2\pi}\sin\theta\text{dθ}} = - \frac{1}{2}\left( \frac{1}{e^{4}} - 1 \right)2\pi\int_{0}^{\pi}{\sin\theta\text{dθ}} = - \pi\left( \frac{1}{e^{4}} - 1 \right)\left( - \cos\theta\left. \ \right|_{0}^{\pi} \right) = - \pi\left( \frac{1}{e^{4}} - 1 \right)( - \cos\theta)\left. \ \right|_{0}^{\pi} = = - \pi\left( \frac{1}{e^{4}} - 1 \right)( - \cos\pi + \cos{0)} = - \pi\left( \frac{1}{e^{4}} - 1 \right)\left( 1 + 1 \right) = 2\pi\left( 1 - \frac{1}{e^{4}} \right)$

5. Oblicz za pomocą całki potrójnej objętość bryły zawartej pomiędzy powierzchniami; z² = x² + y²,     z = 0,    z = 4; $z = \sqrt{x^{2} + y^{2}},\ \ \ \ \ x^{2} + y^{2} = 4^{2}$; $V = \iiint_{V}^{}\text{dxdydz} = \iint_{\Omega}^{}{\left( \int_{\sqrt{x^{2} + y^{2}}}^{4}\text{dz} \right)dxdy = \iint_{\Omega}^{}{\left( z\left. \ \right|_{\sqrt{x^{2} + y^{2}}}^{4} \right)\text{dxdy} =}}\iint_{\Omega}^{}{\left( 4 - \sqrt{x^{2} + y^{2}} \right)dxdy =};\ x = r\cos\varphi,\ \ \ \ y = r\sin\varphi,\ \ \ \ I = r$; 0 ≤ r ≤ 4,      0 ≤ φ ≤ 2π $= \int_{0}^{2\pi}{\int_{0}^{4}{(4 - \sqrt{r^{2}\operatorname{}\varphi + r^{2}\operatorname{}\varphi)}}\text{\ rdrdφ} = \int_{0}^{2\pi}{\int_{0}^{4}{(4 - \sqrt{r^{2}\operatorname{}\varphi + \operatorname{}\varphi)}}\ )\ rdrd\varphi =}}\int_{0}^{2\pi}{\int_{0}^{4}{(4 - r}\ )\ rdrd\varphi = \int_{0}^{2\pi}{\int_{0}^{4}{(4r - r^{2}}\ )\ drd\varphi =}}\int_{0}^{2\pi}{\left( \frac{4r^{2}}{2} - \frac{r^{3}}{3} \right)\left. \ \right|_{0}^{4}\text{dφ} =}\int_{0}^{2\pi}{\left( 2*4^{2} - \frac{4^{3}}{3} \right)\text{dφ} =}\int_{0}^{2\pi}{\left( 32 - \frac{64}{3} \right)\text{dφ} = \frac{32}{3}\int_{0}^{2\pi}{d\varphi =}}\frac{32}{3}\varphi\left. \ \right|_{0}^{2\pi} = \frac{32}{3}*2\pi = \frac{64}{3}\pi$

4.Oblicz całkę krzywoliniową nieskierowaną

∫_Γxyds Γϵ x² + y² = 1,    r = 1; $x = cos\varphi,\ \ y = sin\varphi\ \ \ \ \ \ 0 \leq \ \varphi \leq \frac{\pi}{2}$; $\int_{}^{}{f\left( x,y \right)ds = \int_{}^{}{f\left( x\left( t \right),y\left( t \right) \right)*\sqrt{x^{'2} + {y^{'}}^{2}\ }\text{dt}}};\ \int_{0}^{\frac{\pi}{2}}{cos\varphi sin\varphi d\varphi = \operatorname{}{\left. \ \right|_{0}^{\frac{\pi}{2}} = \frac{1}{2} - 0 = \frac{1}{2}}}$; $\int_{}^{}{cos\varphi sin\varphi d\varphi = \left| \begin{matrix} t = sin\varphi \\ dt = cos\varphi d\varphi \\ \end{matrix} \right|} = \int_{}^{}{tdt = \frac{t^{2}}{2} = \operatorname{}\frac{\varphi}{2}}$; x^′ = −sinφ,     y^′ = cosφ; x^′2 + y^′2 = φ + φ = 1

5.Oblicz całkę krzywoliniową skierowaną

Ponieważ $\frac{\partial D}{\partial y} = - \sin y + \cos x\text{\ oraz\ }\frac{\partial Q}{\partial x} = \cos x - \sin y\ $, są różne to całka nie zależy od drogi czyli ∫_F=∫_Ω+∫_Ω ; Dla łuku AC x=0, dx=0 ; $y \in < \frac{\pi}{4};\pi >$

$$\int_{\frac{\pi}{4}}^{\pi}{2dy} = 2y\left. \ \right|_{\frac{\pi}{4}}^{\pi} = 2\left( \pi - \frac{\pi}{4} \right) = 2*\frac{3}{4}\pi = \frac{3}{2}\pi$$

Dla łuku CB y = π,  dy = 0 ; $x \in < 0,\frac{\pi}{2} >$ ;

$$\int_{0}^{\frac{\pi}{2}}{\left( \cos\pi + \pi\cos x \right)dx = - 1\left( \frac{\pi}{2} - 0 \right) + \pi\sin x\left. \ \right|_{0}^{\frac{\pi}{2}} = - \frac{\pi}{2} + \pi = \frac{\pi}{2}}$$

Czyli: ∫_Γ(cosy+ycosx)dx + (sinx−xsiny+2)dy = 2π

6. Oblicz

$\iiint_{V}^{}{(e^{x} + \frac{1}{y} + \frac{1}{z^{2}}})dxdydz$; gdzie    V = {(x,y,z)   :   1 ≤ x ≤ 2,   2 ≤ y ≤ 3,   − 3 ≤ z ≤ −1}; $\int_{- 3}^{- 1}{(\int_{2}^{3}{(\int_{1}^{2}{\left( e^{x} + \frac{1}{y} + \frac{1}{z^{2}} \right)dx)dy)dz =}}}\int_{- 3}^{- 1}{(\int_{2}^{3}{\left( e^{x} + \frac{x}{y} + \frac{x}{z^{2}} \right)\left. \ \right|_{1}^{2}dy)dz}} = \int_{- 3}^{- 1}{(\int_{2}^{3}{\left( e^{2} + \frac{2}{y} + \frac{2}{z^{2}} - (e^{1} + \frac{1}{y} + \frac{1}{z^{2}}) \right)dy)dz}} = \int_{- 3}^{- 1}{(\int_{2}^{3}{\left( e^{2} - e + \frac{1}{y} + \frac{1}{z^{2}} \right)dy)dz}} = \int_{- 3}^{- 1}{\left( e^{2}y - ey + \ln\left| y \right| + \frac{1}{z^{2}}y \right)\left. \ \right|_{2}^{3}}\text{dz} = \int_{- 3}^{- 1}\left( {3e}^{2} - 3e + \ln 3 + \frac{3}{z^{2}} - ({2e}^{2} - 2e + \ln 2 + \frac{2}{z^{2}}) \right)\text{dz} = \int_{- 3}^{- 1}\left( e^{2} - e + \ln\frac{3}{2} + \frac{1}{z^{2}} \right)dz = \left( \text{ze}^{2} - ze + z\ln\frac{3}{2} - \frac{1}{z} \right)\left. \ \right|_{- 3}^{- 1} = {- e}^{2} + e - \ln\frac{3}{2} + 1 - \left( {- 3e}^{2} - 3e - 3\ln\frac{3}{2} + \frac{1}{3} \right) = {2e}^{2} - 2e + 2\ln\frac{3}{2} - \frac{2}{3}$

ZESTAW II

1.Wyznacz dywergencje pola $\overrightarrow{F} = \left\lbrack 5^{\text{xz}},\operatorname{}{\left( x^{2} + 2y \right),\ }\text{tg}\frac{x}{z} \right\rbrack$; $\nabla\overrightarrow{F} = \frac{\text{δP}}{\text{δx}} + \frac{\text{δQ}}{\text{δy}} + \frac{\text{δR}}{\text{δz}}$; a^x′ = a^xlna; $\frac{\text{δP}}{\text{δx}} = 5^{\text{xz}}ln5*z$; $\frac{\text{δQ}}{\text{δy}} = 2\sin\left( x^{2} + 2y \right)\cos\left( x^{2} + 2y \right)2$; $\frac{\text{δR}}{\text{δy}} = 1/(\operatorname{}{\frac{x}{2})}*\left( - \frac{x}{z^{2}} \right)$; $\nabla\overrightarrow{F} = 5^{\text{xz}}ln5*z + 2\sin\left( x^{2} + 2y \right)\cos\left( x^{2} + 2y \right)2 + 1/(\operatorname{}{\frac{x}{2})}*\left( - \frac{x}{z^{2}} \right)$

2.Pokaż dla funkcji dwukrotnie różniczkowanej rot(grad U)=$\overrightarrow{\mathbf{0}}$

U = U(x, y, z); ∇ × (∇U) = 0; rot(gradU) = 0; $\nabla U = \left\lbrack \frac{\partial U}{\partial x},\frac{\partial U}{\partial y},\frac{\partial U}{\partial z} \right\rbrack$; $\nabla \times \left( \nabla u \right) = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial U}{\partial x} & \frac{\partial U}{\partial y} & \frac{\partial U}{\partial z} \\ \end{matrix} \right| = \left( \frac{\delta}{\text{δy}}\left( \frac{\text{δU}}{\text{δz}} \right) - \frac{\delta}{\text{δz}}\left( \frac{\text{δU}}{\text{δy}} \right) \right)\hat{i} + \left( \frac{\delta}{\text{δx}}\left( \frac{\text{δU}}{\text{δz}} \right) - \frac{\delta}{\text{δz}}\left( \frac{\text{δU}}{\text{δx}} \right) \right)\hat{j} + \left( \frac{\delta}{\text{δx}}\left( \frac{\text{δU}}{\text{δy}} \right) - \frac{\text{δU}}{\text{δy}}\left( \frac{\text{δU}}{\text{δx}} \right) \right)\hat{k} = 0$

3.Oblicz długość łuku

x(t) = 3t,     y(t) = 3t²,    z(t) = 2t³;    tϵ[0,1]; $\int_{t0}^{t1}\sqrt{\left( x^{'} \right)^{2} + \left( y^{'} \right)^{2} + \left( z^{'} \right)^{2}}\text{dt}$; x^′ = 3,   y^′ = 6t,    z^′ = 6t²; $\int_{0}^{1}\sqrt{9 + 36t^{2} + 36t^{4}}dt = 3\int_{0}^{1}\sqrt{1 + 4t^{2} + 4t^{4}}\text{dt} = 3\int_{0}^{1}{(2t^{2} + 1)}dt = 3\left( 2*\frac{t^{3}}{3} + t \right)\left. \ \right|_{0}^{1} = 5$

6.Oblicz całkę krzywoliniową skierowaną:

∫_Γ(x²+y²)dx + xydy; Γ : x = t,   y = e^t tϵ[0,1] ; $\int_{0}^{1}{\left\lbrack \left( t^{2} + e^{2t} \right)*1 + \left( te^{t} \right)*e^{t} \right\rbrack\text{dt} = \int_{0}^{1}{t^{2}dt + \int_{0}^{1}{e^{2t}dt + \int_{0}^{1}{te^{2t}\text{dt}}}}} = \left| \begin{matrix} u = t & v^{'} = e^{2t} \\ u^{'} = 1 & v = \frac{1}{2}e^{2t} \\ \end{matrix} \right| = \frac{t^{3}}{3}\left. \ \right|_{0}^{1} + \frac{1}{2}e^{2t}\left. \ \right|_{0}^{1} + t\frac{1}{2}e^{2t}\left. \ \right|_{0}^{1} - \frac{1}{2}\int_{0}^{1}{e^{2t}\text{dt}} = \left( \frac{t^{3}}{3} + \frac{1}{2}e^{2t} + \frac{1}{2}te^{2t} - \frac{1}{4}e^{2t} \right)\left. \ \right|_{0}^{1} = \frac{1}{3} + \frac{1}{2}e^{2} - \frac{1}{2} + \frac{1}{2}e^{2} - \frac{1}{4}e^{2} + \frac{1}{4} = \frac{3}{4}e^{2} + \frac{1}{12}$

7.Oblicz masę płata powierzchniowego S: $z = \frac{1}{2}\left( x^{2} + y^{2} \right) = f\left( x \right)\ ,\ $

 z ≤ 2 o   gestosci ρ(x,y,z) = 5 $\frac{1}{2}\left( x^{2} + y^{2} \right) \leq 2\ \ \ \ \ \ \ \ x^{2} + y^{2} \leq 4$

$\frac{\partial f}{\partial x} = x,\ \ \ \ \ \frac{\partial f}{\partial y} = y$; $\iint_{\Omega}^{}{5(\sqrt{x^{2} + y^{2} + 1}}dxdy = \left| \begin{matrix} x = rcos\varphi & 0 \leq \varphi \leq 2\pi \\ y = rsin\varphi & 0 \leq r \leq 2 \\ \end{matrix} \right| = 5\int_{0}^{2\pi}{\int_{0}^{2}{\sqrt{r^{2}\operatorname{}{\varphi + r^{2}\operatorname{}{\varphi + 1}}}rdrd\varphi = 5}}\int_{0}^{2\pi}{\int_{0}^{2}{\sqrt{r^{2} + 1}rdrd\varphi = \left| \begin{matrix} r^{2} + 1 = t \\ 2rdr = dt \\ rdr = \frac{\text{dt}}{2} \\ \end{matrix} \right|}}$=; $\int_{0}^{2}{\frac{1}{2}\sqrt{t}}dt = \frac{1}{2}t^{\frac{3}{2}}*\frac{2}{3} = \frac{1}{3}t^{\frac{3}{2}} = \frac{1}{3}\left( r^{2} + 1 \right)^{\frac{3}{2}}\left. \ \right|_{0}^{2} = \frac{1}{3}5^{\frac{3}{2}} - \frac{1}{3}$ ; $= 5\int_{0}^{2\pi}{\frac{1}{3}{(5}^{\frac{3}{2}} - 1)d\varphi = \frac{5}{3}(5^{\frac{3}{2}} - 1)\int_{0}^{2\pi}{d\varphi = \frac{5}{3}\left( 5^{\frac{3}{2}} - 1 \right)\varphi\left. \ \right|_{0}^{2\pi} = \frac{5}{3}\left( 5^{\frac{3}{2}} - 1 \right)2\pi}};\ $

8.Oblicz całą powierzchnię zorientowaną

∫_Sxdydz + ydzdx + zdxdy    S = x² + y² + z² = 16 ; $\iint_{S}^{}\overset{\overline{}}{F}\overset{\overline{}}{\text{ds}} = \iiint_{V}^{}{\nabla\overrightarrow{F}dxdydz =}\iiint_{V}^{}{\left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \right)\text{dxdydz}}$; $\overrightarrow{F}\left( x,y,z \right) = P\left( x,y,z \right)\hat{i} + Q\left( x,y,z, \right)\hat{j} + R(x,y,z)\hat{k}$; $\text{dir}\overrightarrow{F} = \nabla\overrightarrow{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$; $\nabla = \frac{\partial}{\partial x}\hat{i} + \frac{\partial}{\partial y}\hat{j} + \frac{\partial}{\partial z}\hat{k}$; $\iint_{S}^{}\overset{\overline{}}{F}\overset{\overline{}}{\text{ds}} = Pdydz + Qdxdz + Rdxdy$; ∬_Sxdydz + ydzdx + zdxdy

$\frac{\partial P}{\partial x} = 1,\ \ \frac{\partial Q}{\partial y} = 1\ \ \frac{\partial R}{\partial z} = 1$; $\iiint_{V}^{}{3( - dxdydz) =} - 3\iiint_{V}^{}{dxdydz =} - {\frac{4}{3}\pi 4^{3} = - 4}^{4}\pi$

9.Oblicz całkę powierzchniową zorientowaną

∫_Sxdydz + zdxdy,     S : x = cosu,   y = sinu,    z = v,    ; u ∈ [0,2π],   vϵ[−1, 1]; ∬_Sxdydz + zdxdy; 0 ≤ u ≤ 2π,    − 1 ≤ v ≤ 1; $\int_{0}^{2\pi}{\int_{- 1}^{1}{(cosu*\left| \begin{matrix} \text{cosu} & 0 \\ 0 & 1 \\ \end{matrix} \right|}} + 0 + v*\left| \begin{matrix} - sinu & 0 \\ \text{cosu} & 0 \\ \end{matrix} \right|)dudv = \ \int_{0}^{2\pi}{(\int_{- 1}^{1}{\operatorname{}{udv)du}}}$; ∫₋₁¹dv = v |₋₂¹ = 2; $\operatorname{}{u = \frac{1 + cos2u}{2}}$; $\int_{0}^{2\pi}{(\int_{- 1}^{1}{\operatorname{}{udv)du}}} = 2\int_{0}^{2\pi}{\operatorname{}{\text{udu} = \int_{0}^{2\pi}{\left( 1 + cos2u \right)du = u + \frac{1}{2}2u\left. \ \right|_{0}^{2\pi} = 2\pi}}}$

ZESTAW III

1. $\mathbf{\ }\left\{ \begin{matrix} \mathbf{x}^{\mathbf{'}}\mathbf{= x - y + 2}\mathbf{z} \\ \mathbf{y}^{\mathbf{'}}\mathbf{= - x + y - 2}\mathbf{z} \\ \mathbf{z}^{\mathbf{'}}\mathbf{= 2}\mathbf{x - 2}\mathbf{y} \\ \end{matrix} \right.\ \mathbf{\text{\ \ }}$; $A = \begin{bmatrix} 1 & - 1 & 2 \\ - 1 & 1 & - 2 \\ 2 & - 2 & 0 \\ \end{bmatrix}$; $\det{|A - \lambda I|} = \left| \begin{matrix} 1 - \lambda & - 1 & 2 \\ - 1 & 1 - \lambda & - 2 \\ 2 & - 2 & - \lambda \\ \end{matrix} \right| = \left( 1 - \lambda \right)^{2}\left( - \lambda \right) + 4 + 4 - \left( 4\left( 1 - \lambda \right) + 4\left( 1 - \lambda \right) - \lambda \right) = - \lambda^{3} + 2\lambda^{2} + 8\lambda = - \lambda(\lambda^{2} - 2\lambda - 8)$; $\Delta = b^{2} - 4ac = 4 + 32 = 36,\ \ \ \ \ \sqrt{\Delta} = 6$ ;$\lambda_{1} = \frac{- b - \sqrt{\Delta}}{2a} = \frac{2 + 6}{2} = 4$; $\lambda_{2} = \frac{2 - 6}{2} = - 2$; λ₁ = 4,    λ₂ = −2,    λ₃ = 0;

dla λ = 0; $V_{0} = \left\{ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} \right|\begin{bmatrix} 1 & - 1 & 2 \\ - 1 & 1 & - 2 \\ 2 & - 2 & 0 \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \left. \ \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} \right\}$ ; $\left\{ \begin{matrix} x - y + 2z = 0 \\ - x + y - 2z = 0 \\ 2x - 2y = 0 \\ \end{matrix} \right.\ = > \begin{matrix} x = y \\ z = 0 \\ \end{matrix}$; V₀ = V₀⁽¹⁾; dla λ = −2; $V_{- 2} = \left\{ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} \right|\begin{bmatrix} 3 & - 1 & 2 \\ - 1 & 3 & - 2 \\ 2 & - 2 & 2 \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \left. \ \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} \right\}$ ; $\left\{ \begin{matrix} 3x - y + 2z = 0 \\ - x + 3y - 2z = 0 \\ 2x - 2y + 2z = 0 \\ \end{matrix} \right.\ = > z = s,\ \ \ \begin{matrix} s = 2y = > y = \frac{s}{2} \\ x = - \frac{s}{2} \\ \end{matrix}$; $V_{- 2} = \left\{ \begin{bmatrix} - \frac{s}{2} \\ \frac{s}{2} \\ s \\ \end{bmatrix} = s\begin{bmatrix} - \frac{1}{2} \\ \frac{1}{2} \\ 1 \\ \end{bmatrix},\ s \in R \right\}$; $\text{dla\ }\lambda = 4;\ V_{4} = \left\{ \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} \right|\begin{bmatrix} - 3 & - 1 & 2 \\ - 1 & - 3 & - 2 \\ 2 & - 2 & - 4 \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \left. \ \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix} \right\}$ ; $\left\{ \begin{matrix} - 3x - y + 2z = 0 \\ - x - 3y - 2z = 0 \\ 2x - 2y - 4z = 0 \\ \end{matrix} \right.\ = > z = s,\ \ \ \begin{matrix} y = - s \\ x = s \\ \end{matrix}$; $V_{4} = \left\{ \begin{bmatrix} s \\ - s \\ s \\ \end{bmatrix} = s\begin{bmatrix} 1 \\ - 1 \\ 1 \\ \end{bmatrix},\ s \in R \right\}$ ; $X = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ z\left( t \right) \\ \end{bmatrix} = C_{1}\begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix}e^{0t} + C_{2}\begin{bmatrix} - \frac{1}{2} \\ \frac{1}{2} \\ 1 \\ \end{bmatrix}e^{- 2t} + C_{3}\begin{bmatrix} 1 \\ - 1 \\ 1 \\ \end{bmatrix}e^{4t} = C_{1}\begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix} + C_{2}\begin{bmatrix} - \frac{1}{2} \\ \frac{1}{2} \\ 1 \\ \end{bmatrix}e^{- 2t} + C_{3}\begin{bmatrix} 1 \\ - 1 \\ 1 \\ \end{bmatrix}e^{4t}$

$\left\{ \begin{matrix} - x + 2y = 0 \\ x - 2y = 0 \\ \end{matrix} = > x = s,\ \ \ y = \frac{1}{2}s \right.\ $; $V_{- 4} = \left\{ \begin{bmatrix} s \\ \frac{1}{2}s \\ \end{bmatrix} = s\begin{bmatrix} 1 \\ \frac{1}{2} \\ \end{bmatrix},\ s \in R \right\}$; $X_{0} = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ \end{bmatrix} = C_{1}\begin{bmatrix} 1 \\ - 1 \\ \end{bmatrix}e^{- 7t} + C_{2}\begin{bmatrix} 1 \\ \frac{1}{2} \\ \end{bmatrix}e^{- 4t}$ ; Drugi przypadek dla x’=F(t) ; $x_{1}\left( t \right) = \begin{bmatrix} A_{1}e^{t} + B_{1}e^{2t} \\ A_{2}e^{t} + B_{2}e^{2t} \\ \end{bmatrix}$; ${x'}_{1}\left( t \right) = \begin{bmatrix} A_{1}e^{t} + {2B}_{1}e^{2t} \\ A_{2}e^{t} + {2B}_{2}e^{2t} \\ \end{bmatrix}$ ; A₁e^t + 2B₁e^2t = −5(A₁e^t+B₁e^2t) + 2(A₂e^t + B₂e^2t)+e^t ; A₁ + 2B₁e^t = −5A₁ − 5B₁e^t + 2A₂ + 2B₂e^t + 1 ; (7B₁−2B₂)e^t + 6A₁ − 2A₂ − 1 = 0 ; $\left\{ \begin{matrix} 7B_{1} - 2B_{2} = 0 \\ 6A_{1} - {2A}_{2} - 1 = 0 \\ \end{matrix} \right.\ $ ; A₂e^t + 2B₂e^2t = A₁e^t + B₁e^2t − 6(A₂e^t + B₂e^2t)+e^2t ; A₂ + 2B₂e^t = A₁ + B₁e^t − 6A₂ − 6B₂e^t + e^t ; (8B₂−B₁−1)e^t + 7A₂ − A₁ = 0 ; $\left\{ \begin{matrix} - 8B_{2} - B_{1} - 1 = 0 \\ {7A}_{2} - A_{1} = 0 \\ \end{matrix} \right.\ $ ; $A_{1} = \frac{7}{40},\ \ \ A_{2} = \frac{1}{40},\ \ \ B_{1} = - \frac{1}{27},\ \ \ B_{2} = - \frac{7}{54}$; $X = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ \end{bmatrix} = C_{1}\begin{bmatrix} 1 \\ - 1 \\ \end{bmatrix}e^{- 7t} + C_{2}\begin{bmatrix} 1 \\ \frac{1}{2} \\ \end{bmatrix}e^{- 4t} + \begin{bmatrix} \frac{7}{40}e^{t} - \frac{1}{27}e^{2t} \\ \frac{1}{40}e^{t} - \frac{7}{54}e^{2t} \\ \end{bmatrix}$

2. $\left\{ \begin{matrix} \mathbf{x}^{\mathbf{'}}\mathbf{= - 5}\mathbf{x - y +}\mathbf{e}^{\mathbf{t}} \\ \mathbf{y}^{\mathbf{'}}\mathbf{= - 3}\mathbf{y +}\mathbf{e}^{\mathbf{2}\mathbf{t}} \\ \end{matrix} \right.\ $; x^′ = Ax + F(t),  najpierw x^′ = Ax; $A = \begin{bmatrix} - 5 & - 1 \\ 0 & - 3 \\ \end{bmatrix}$; $\det\left| A - \text{λI} \right| = \left| \begin{matrix} - 5 - \lambda & - 1 \\ 0 & - 3 - \lambda \\ \end{matrix} \right| = \left( - 5 - \lambda \right)\left( - 3 - \lambda \right) - 0 = \lambda^{2} + 8\lambda + 15$; $\Delta = b^{2} - 4ac = 64 - 60 = 4,\ \ \ \ \ \sqrt{\Delta} = 2$; $\lambda_{1} = \frac{- b - \sqrt{\Delta}}{2a} = \frac{- 8 + 2}{2} = - 5$; $\lambda_{2} = \frac{- 8 + 2}{2} = - 3$; dla λ = −5; $V_{- 5} = \left\{ \begin{bmatrix} x \\ y \\ \end{bmatrix} \right|\begin{bmatrix} 0 & - 1 \\ 0 & 2 \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ \end{bmatrix} = \left. \ \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} \right\}\ $; $\left\{ \begin{matrix} - y = 0 \\ 2y = 0 \\ \end{matrix} = > x = s,\ \ \ y = 0 \right.\ $

$V_{- 5} = \left\{ \begin{bmatrix} s \\ 0 \\ \end{bmatrix} = s\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix},\ s \in R \right\}$; dla λ = −3; $V_{- 3} = \left\{ \begin{bmatrix} x \\ y \\ \end{bmatrix} \right|\begin{bmatrix} - 2 & - 1 \\ 0 & 0 \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ \end{bmatrix} = \left. \ \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} \right\}$ ; −2x − y = 0 = >x = s,    y = −2s ; $V_{- 5} = \left\{ \begin{bmatrix} s \\ - 2s \\ \end{bmatrix} = s\begin{bmatrix} 1 \\ - 2 \\ \end{bmatrix},\ s \in R \right\}$; $X_{0} = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ \end{bmatrix} = C_{1}\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}e^{- 5t} + C_{2}\begin{bmatrix} 1 \\ - 2 \\ \end{bmatrix}e^{- 3t}$; Drugi przypadek dla x’=F(t); $x_{1}\left( t \right) = \begin{bmatrix} A_{1}e^{t} + B_{1}e^{2t} \\ A_{2}e^{t} + B_{2}e^{2t} \\ \end{bmatrix}$ ; ${x'}_{1}\left( t \right) = \begin{bmatrix} A_{1}e^{t} + {2B}_{1}e^{2t} \\ A_{2}e^{t} + {2B}_{2}e^{2t} \\ \end{bmatrix}$; A₁e^t + 2B₁e^2t = −5(A₁e^t+B₁e^2t) − A₂e^t − B₂e^2t + e^t ; A₁ + 2B₁e^t = −5A₁ − 5B₁e^t − A₂ − B₂e^t + 1; (7B₁+B₂)e^t + 6A₁ + A₂ − 1 = 0; $\left\{ \begin{matrix} 7B_{1} + B_{2} = 0 \\ 6A_{1} + A_{2} - 1 = 0 \\ \end{matrix} \right.\ $ ; A₂e^t + 2B₂e^2t = −3(A₂e^t+B₂e^2t) + e^2t ; A₂ + 2B₂e^t = −3A₂ − 3B₂e^t + e^t ;(5B₂−1)e^t + 4A₂ = 0; $\left\{ \begin{matrix} 5B_{2} - 1 = 0 \\ {4A}_{2} = 0 \\ \end{matrix} \right.\ $; $A_{1} = \frac{1}{6},\ \ \ A_{2} = 0,\ \ \ B_{1} = - \frac{1}{35},\ \ \ B_{2} = \frac{1}{5}$ ; $X = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ \end{bmatrix} = C_{1}\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}e^{- 5t} + C_{2}\begin{bmatrix} 1 \\ - 2 \\ \end{bmatrix}e^{- 3t} + \begin{bmatrix} \frac{1}{6}e^{t} - \frac{1}{35}e^{2t} \\ \frac{1}{5}e^{2t} \\ \end{bmatrix}$ ;

3. $\left\{ \begin{matrix} \mathbf{x}^{\mathbf{'}}\mathbf{= - 5}\mathbf{x + 2}\mathbf{y +}\mathbf{e}^{\mathbf{t}} \\ \mathbf{y}^{\mathbf{'}}\mathbf{= x - 6}\mathbf{y +}\mathbf{e}^{\mathbf{2}\mathbf{t}} \\ \end{matrix} \right.\ $; $A = \begin{bmatrix} - 5 & 2 \\ 1 & - 6 \\ \end{bmatrix}$; $\det\left| A - \text{λI} \right| = \left| \begin{matrix} - 5 - \lambda & 2 \\ 1 & - 6 - \lambda \\ \end{matrix} \right| = 30 + 5\lambda + 6\lambda + \lambda^{2} - 2 = \lambda^{2} + 11\lambda + 28$; $\Delta = b^{2} - 4ac = 121 - 112 = 9,\ \ \ \ \ \sqrt{\Delta} = 3$; $\lambda_{1} = \frac{- b - \sqrt{\Delta}}{2a} = \frac{- 11 + 3}{2} = - 7$; $\lambda_{2} = \frac{- 11 + 3}{2} = - 4$; dla λ = −7; $V_{- 7} = \left\{ \begin{bmatrix} x \\ y \\ \end{bmatrix} \right|\begin{bmatrix} 2 & 2 \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ \end{bmatrix} = \left. \ \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} \right\}$ ; $\left\{ \begin{matrix} 2x + 2y = 0 \\ x + y = 0 \\ \end{matrix} = > x = s,\ \ \ y = - s \right.\ $ ; $V_{- 7} = \left\{ \begin{bmatrix} s \\ - s \\ \end{bmatrix} = s\begin{bmatrix} 1 \\ - 1 \\ \end{bmatrix},\ s \in R \right\}$; dla λ = −4; $V_{- 4} = \left\{ \begin{bmatrix} x \\ y \\ \end{bmatrix} \right|\begin{bmatrix} - 1 & 2 \\ 1 & - 2 \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ \end{bmatrix} = \left. \ \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} \right\}$

4. $\left\{ \begin{matrix} \mathbf{x}^{\mathbf{'}}\mathbf{= - 5}\mathbf{x - 2}\mathbf{y} \\ \mathbf{y}^{\mathbf{'}}\mathbf{= x - 7}\mathbf{y} \\ \end{matrix} \right.\ $; $A = \begin{bmatrix} - 5 & - 2 \\ 1 & - 7 \\ \end{bmatrix}$ ; $\det\left| A - \text{λI} \right| = \left| \begin{matrix} - 5 - \lambda & - 2 \\ 1 & - 7 - \lambda \\ \end{matrix} \right| = 35 + 5\lambda + 7\lambda + \lambda^{2} + 2 = \lambda^{2} + 12\lambda + 37$ ; $\Delta = b^{2} - 4ac = 144 - 148 = - 4,\ \ \ \ \ \sqrt{\Delta} = 2i$ ; $\lambda_{1} = \frac{- b - \sqrt{\Delta}}{2a} = \frac{- 12 - 2i}{2} = - 6 - i$ ; $\lambda_{2} = \frac{- 12 + 2i}{2} = - 6 + i$; λ₁ = −6 − i możemy pominąć bez żadnych negatywnych skutków na wynik ostateczny

dla λ = −6 + i ; $V_{- 6 + i} = \left\{ \begin{bmatrix} x \\ y \\ \end{bmatrix} \right|\begin{bmatrix} 1 - i & - 2 \\ 1 & - 1 - i \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ \end{bmatrix} = \left. \ \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} \right\}$ ; $\left\{ \begin{matrix} \left( 1 - i \right)x - 2y = 0 \\ x + ( - 1 - i)y = 0 \\ \end{matrix} = > y = s,\ \ \ x = \frac{2s}{1 - i}*\left( \frac{1 + i}{1 + i} \right) \right.\ = \frac{2s\left( 1 + i \right)}{2} = s\left( 1 + i \right)$ ; $V_{- 6 + i} = \left\{ \begin{bmatrix} s\left( 1 + i \right) \\ s \\ \end{bmatrix} = s\begin{bmatrix} 1 + i \\ 1 \\ \end{bmatrix},\ s \in R \right\}$ ; $X = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ \end{bmatrix} = C_{1}\begin{bmatrix} 1 + i \\ 1 \\ \end{bmatrix}e^{\left( - 6 + i \right)t} = C_{1}\begin{bmatrix} 1 + i \\ 1 \\ \end{bmatrix}e^{- 6t + it} = C_{1}\begin{bmatrix} 1 + i \\ 1 \\ \end{bmatrix}e^{- 6t}e^{\text{it}}$ ; wzor Eulera e^it = cost + isint ; $X = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ \end{bmatrix} = C_{1}\begin{bmatrix} 1 + i \\ 1 \\ \end{bmatrix}\left( \cos t + i\sin t \right)e^{- 6t} = C_{1}\begin{bmatrix} \cos t + i\sin t + i\cos t + i^{2}\sin t \\ \cos t + i\sin t \\ \end{bmatrix}e^{- 6t} = C_{1}\begin{bmatrix} \cos t - \sin t + i(\sin t + \cos t) \\ \cos t + i\sin t \\ \end{bmatrix}e^{- 6t} = \left( C_{1}\begin{bmatrix} \cos t - \sin t \\ \cos t \\ \end{bmatrix} + C_{2}\begin{bmatrix} \sin t + \cos t \\ \sin t \\ \end{bmatrix} \right)e^{- 6t}$ ;

5. $\mathbf{\ }\left\{ \begin{matrix} \mathbf{x}^{\mathbf{'}}\mathbf{= 2}\mathbf{x - 5}\mathbf{y + 4}\mathbf{t} \\ \mathbf{y}^{\mathbf{'}}\mathbf{= x - 2}\mathbf{y + 1} \\ \end{matrix} \right.\ $; x^′ = Ax + F(t),  najpierw x^′ = Ax; $A = \begin{bmatrix} 2 & - 5 \\ 1 & - 2 \\ \end{bmatrix}$

$\det\left| A - \text{λI} \right| = \left| \begin{matrix} 2 - \lambda & - 5 \\ 1 & - 2 - \lambda \\ \end{matrix} \right| = - 4 - 2\lambda + 2\lambda + \lambda^{2} + 5 = \lambda^{2} + 1$ ; λ² + 1 = 0 = >λ₁ = i,    λ₂ = −i ; λ₂ = −i,   możemy pominąć bez żadnych negatywnych skutków na wynik ostateczny ; dla λ = i ; $V_{i} = \left\{ \begin{bmatrix} x \\ y \\ \end{bmatrix} \right|\begin{bmatrix} 2 - i & - 5 \\ 1 & - 2 - i \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ \end{bmatrix} = \left. \ \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} \right\}$ ; $\left\{ \begin{matrix} \left( 2 - i \right)x - 5y = 0 \\ x + \left( - 2 - i \right)y = 0 \\ \end{matrix} = > x = s,\ \ \ y = \frac{s}{2 + i}(\frac{2 - i}{2 - i}) \right.\ = \frac{s\left( 2 - i \right)}{5}$ ; $V_{i} = \left\{ \begin{bmatrix} s \\ \frac{s\left( 2 - i \right)}{5} \\ \end{bmatrix} = s\begin{bmatrix} 1 \\ \frac{2 - i}{5} \\ \end{bmatrix},\ s \in R \right\}$ ; $X_{0} = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ \end{bmatrix} = C_{1}\begin{bmatrix} 1 \\ \frac{2 - i}{5} \\ \end{bmatrix}e^{\text{it}} = C_{1}\begin{bmatrix} 1 \\ \frac{2 - i}{5} \\ \end{bmatrix}\left( \cos t + i\sin t \right) = C_{1}\begin{bmatrix} \cos t + i\sin t \\ \frac{2\cos t + 2i\sin t - i\cos t - i^{2}\sin t}{5} \\ \end{bmatrix} = C_{1}\begin{bmatrix} \cos t + i\sin t \\ \frac{2\cos t + \sin t + i(2\sin t - \cos t)}{5} \\ \end{bmatrix} = C_{1}\begin{bmatrix} \cos t \\ \frac{2}{5}\cos t + \frac{1}{5}\sin t \\ \end{bmatrix} + C_{2}\begin{bmatrix} \sin t \\ \frac{2}{5}\sin t - \frac{1}{5}\cos t\ \\ \end{bmatrix}$ ; Drugi przypadek dla x’=F(t) ; $x_{1}\left( t \right) = \begin{bmatrix} A_{1}t + B_{1} \\ A_{2}t + B_{2} \\ \end{bmatrix}$ ; ${x'}_{1}\left( t \right) = \begin{bmatrix} A_{1} \\ A_{2} \\ \end{bmatrix}$ ; A₁ = −2(A₁t+B₁) − 5(A₂t + B₂)+4t ; (2A₁−A₂+4)t − A₁ + 2B₁ − 5B₂ = 0 ; $\left\{ \begin{matrix} 2A_{1} - A_{2} + 4 = 0 \\ A_{1} + 2B_{1} - 5B_{2} = 0 \\ \end{matrix} \right.\ $ ; A₂ = A₁t + B₁ − 2(A₂t+B₂) + 1 ; (A₁−A₂)t − 2B₂ + B₁ − A₂ + 1 = 0 ; $\left\{ \begin{matrix} A_{1} - A_{2} = 0 \\ 2B_{2} + B_{1} - A_{2} + 1 = 0 \\ \end{matrix} \right.\ $

$A_{1} = \frac{4}{3},\ \ \ A_{2} = \frac{4}{3},\ \ \ B_{1} = \frac{13}{3},\ \ \ B_{2} = 2$ ;

$X = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ \end{bmatrix} = C_{1}\begin{bmatrix} \cos t \\ \frac{2}{5}\cos t + \frac{1}{5}\sin t \\ \end{bmatrix} + C_{2}\begin{bmatrix} \sin t \\ \frac{2}{5}\sin t - \frac{1}{5}\cos t\ \\ \end{bmatrix} + \begin{bmatrix} \frac{4}{3}t + \frac{13}{3} \\ \frac{4}{3}t + 2 \\ \end{bmatrix}$

$${X = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ \end{bmatrix} = C}_{1}\begin{bmatrix} 1 \\ - i \\ \end{bmatrix}\operatorname{(cos}{( - 2t)} + i\sin{( - 2t))} = C_{1}\begin{bmatrix} \cos{( - 2t)} + i\sin{( - 2t)} \\ \operatorname{-i(cos}{( - 2t)} + i\sin{( - 2t))} \\ \end{bmatrix} = C_{1}\begin{bmatrix} \cos{( - 2t)} + i\sin{( - 2t)} \\ \sin{( - 2t)} - i\cos{( - 2t)} \\ \end{bmatrix} = C_{1}\begin{bmatrix} \cos{( - 2t)} \\ \sin{( - 2t)} \\ \end{bmatrix} + C_{2}\begin{bmatrix} \sin{( - 2t)} \\ \operatorname{-cos}{( - 2t)} \\ \end{bmatrix}$$

6. $\left\{ \begin{matrix} \mathbf{x}^{\mathbf{'}}\mathbf{= x - 4}\mathbf{y} \\ \mathbf{y}^{\mathbf{'}}\mathbf{= 2}\mathbf{x - 3}\mathbf{y} \\ \end{matrix}\mathbf{\ \ \ ,\ \ \ gdzie\ x}\left( \mathbf{0} \right)\mathbf{= y}\left( \mathbf{0} \right)\mathbf{= 0} \right.\ $ ; $A = \begin{bmatrix} 1 & - 4 \\ 2 & - 3 \\ \end{bmatrix}$; $\det\left| A - \text{λI} \right| = \left| \begin{matrix} 1 - \lambda & - 4 \\ 2 & - 3 - \lambda \\ \end{matrix} \right| = - 3 - \lambda + 3\lambda + \lambda^{2} + 8 = \lambda^{2} + 2\lambda + 5$ ; λ² + 2λ + 5 = 0 ; $\Delta = b^{2} - 4ac = 4 - 20 = - 16,\ \ \ \ \ \sqrt{\Delta} = 4i$ ; $\lambda_{1} = \frac{- b + \sqrt{\Delta}}{2a} = \frac{- 2 + 4i}{2} = - 1 + 2i$ ; $\lambda_{2} = \frac{- 2 - 4i}{2} = - 1 - 2i$ ; λ₂ = −1 − 2i, możemy pominąć bez żadnych negatywnych skutków na wynik ostateczny ; dla λ = −1 + 2i ; $V_{- 1 - 2i} = \left\{ \begin{bmatrix} x \\ y \\ \end{bmatrix} \right|\begin{bmatrix} 2 - 2i & - 4 \\ 2 & - 2 - 2i \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ \end{bmatrix} = \left. \ \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} \right\}$ ; $\left\{ \begin{matrix} \left( 2 - 2i \right)x - 4y = 0 \\ 2x + \left( 2 - 2i \right)y = 0 \\ \end{matrix}\ \ \ x = s = > y = \frac{1 - i}{2}s \right.\ $ ; $V_{- 1 - 2i} = \left\{ \begin{bmatrix} s \\ \frac{1 - i}{2}s \\ \end{bmatrix} = s\begin{bmatrix} 1 \\ \frac{1 - i}{2} \\ \end{bmatrix},\ s \in R \right\}$ ; $X = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ \end{bmatrix} = C_{1}\begin{bmatrix} 1 \\ \frac{1 - i}{2} \\ \end{bmatrix}e^{( - 1 + 2i)t}$ ; wzor Eulera e^2it = cos(2t) + isin(2t) ; ${X = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ \end{bmatrix} = C}_{1}\begin{bmatrix} 1 \\ \frac{1 - i}{2} \\ \end{bmatrix}\operatorname{(cos}\left( 2t \right) + i\sin{(2t))}e^{- t} = C_{1}\begin{bmatrix} \cos\left( 2t \right) + i\sin\left( 2t \right) \\ \frac{1}{2}\cos{2t} + \frac{1}{2}i\sin{2t} - \frac{1}{2}i\cos{2t} - \frac{1}{2}i^{2}\sin{2t} \\ \end{bmatrix}e^{- t} = C_{1}\begin{bmatrix} \cos\left( 2t \right) + i\sin\left( 2t \right) \\ \frac{1}{2}\cos{2t} + \frac{1}{2}\sin{2t} + i\left( \frac{1}{2}\sin{2t} - \frac{1}{2}\cos{2t} \right) \\ \end{bmatrix}e^{- t} = \left( C_{1}\begin{bmatrix} \cos\left( 2t \right) \\ \frac{1}{2}\cos{2t} + \frac{1}{2}\sin{2t} \\ \end{bmatrix} + C_{2}\begin{bmatrix} \sin\left( 2t \right) \\ \frac{1}{2}\sin{2t} - \frac{1}{2}\cos{2t} \\ \end{bmatrix} \right)e^{- t}$ ; x(0) = y(0) = 0 ; $\begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} = \left( C_{1}\begin{bmatrix} 1 \\ \frac{1}{2} \\ \end{bmatrix} + C_{2}\begin{bmatrix} 0 \\ - \frac{1}{2} \\ \end{bmatrix} \right)e^{- t}$ ; C₁ = 0,    C₂ = 0

7. $\left\{ \begin{matrix} \mathbf{x}^{\mathbf{'}}\mathbf{= 2}\mathbf{y} \\ \mathbf{y}^{\mathbf{'}}\mathbf{= - 2}\mathbf{x} \\ \end{matrix} \right.\ $; $A = \begin{bmatrix} 0 & 2 \\ - 2 & 0 \\ \end{bmatrix}$ ; $\det\left| A - \text{λI} \right| = \left| \begin{matrix} - \lambda & 2 \\ - 2 & - \lambda \\ \end{matrix} \right| = \lambda^{2} + 4$ ; $\lambda^{2} = - 4 = > \ \lambda = \sqrt{- 4}$ ; λ₁ = −2i ; λ₂ = 2i ; λ₂ = 2i, możemy pominąć bez żadnych negatywnych skutków na wynik ostateczny ; dla λ = −2i ; $V_{- 2i} = \left\{ \begin{bmatrix} x \\ y \\ \end{bmatrix} \right|\begin{bmatrix} 2i & 2 \\ - 2 & 2i \\ \end{bmatrix}\begin{bmatrix} x \\ y \\ \end{bmatrix} = \left. \ \begin{bmatrix} 0 \\ 0 \\ \end{bmatrix} \right\}$ ; $\left\{ \begin{matrix} 2ix + 2y = 0 \\ - 2x + 2iy = 0 \\ \end{matrix}\ \ \ x = s = > y = - is\ \right.\ $; $V_{- 2i} = \left\{ \begin{bmatrix} s \\ - is \\ \end{bmatrix} = s\begin{bmatrix} 1 \\ - i \\ \end{bmatrix},\ s \in R \right\}$ ; $X = \begin{bmatrix} x\left( t \right) \\ y\left( t \right) \\ \end{bmatrix} = C_{1}\begin{bmatrix} 1 \\ - i \\ \end{bmatrix}e^{- 2it}$ ; wzor Eulera e^−2it = cos( − 2t) + isin( − 2t) ;