Forum serwisu

Łącze do rozwiązania krok po kroku:
\(\sin^{3}\left(x\right)+\sin\left(x\right)\,\cos^{2}\left(x\right)-2\,\cos^{3}\left(x\right)=0\)

Dodano kalkulator granic funkcji: \displaystyle\lim_{x\to{\infty}}{\frac{1+\sin\left(x\right)}{x}} \displaystyle\lim_{x\to{\infty}}{\frac{\ln\left(1+3^{x}\right)}{\ln\left(1+2^{x}\right)}} \displaystyle\lim_{x\to{\infty}}{\sin\left(\sqrt{x+1}\right)-\sin\left(\sqrt{x}\right)} \displaystyle\lim_{x\to...

2) \(y'+y=3\,e^{7\,x}\;\rightarrow\;y=\dfrac{3\,e^{7\,x}}{8}+\dfrac{C}{e^{x}}\)
3) \(\frac{\mathrm{d}y}{\mathrm{d}x}=2\,y\)
4) \(-x'''+2\,x''+x'-2\,x=0\;\rightarrow\;x=C\,e^{2\,t}+C_{1}\,e^{t}+\dfrac{C_{2}}{e^{t}}\)

Metodą niezdefiniowanych współczynników:
\(x^{2}\,y''-2\,x\,y'+2\,y=x^{5}\,\ln\left(x\right)\;\rightarrow\;y=\dfrac{x^{5}\,\ln\left(x\right)}{12}-\dfrac{7\,x^{5}}{144}+C\,x^{2}+C_{1}\,x\)

Oprócz górnej odpowiedzi: \left\{\begin{matrix}x'=-3\,z-y+2\,x\\y'=-z-y+2\,x\\z'=-2\,z-y+x\end{matrix}\right.\;\rightarrow\;\left\{\begin{matrix}x=\dfrac{C\,\sin\left(t\right)+C_{1}\,\cos\left(t\right)}{e^{t}}+C\,e^{t}\\y=\dfrac{C_{1}\,\sin\left(t\right)-C\,\cos\left(t\right)}{e^{t}}+C\,e^{t}\\z=\df...

Dodano kalkulator numeryczny: https://mathdf.com/calc/pl/ https://cdn1.bbcode0.com/uploads/2021/1/29/070b7b475d2b8cd4e1b983db7631a941-full.png https://cdn1.bbcode0.com/uploads/2021/1/29/d65287f5ff19d388c27617ed6d79bbba-full.png https://cdn1.bbcode0.com/uploads/2021/1/29/9997d19b9edbfdfadce0d40f664cd...

X=\left(\begin{matrix}-1&-3&0\end{matrix}\right)\cdot {\left(\begin{matrix}-1&0&0\\0&-1&-1\\1&1&2\end{matrix}\right)^{-1}}=\left(\begin{matrix}4&6&3\end{matrix}\right) krok po kroku Kontrola: \left(\begin{matrix}4&6&3\end{matrix}\right)\cdot \left(\be...

\(x\,y\,y'=y^{2}+x^{2}\;\rightarrow\;y^{2}=2\,x^{2}\,\left(\ln\left(x\right)+C\right)\)

krok po kroku

\left(\dfrac{\sin\left(2\,x\right)+1}{\cos\left(2\,x\right)}\right)'_{x}=\dfrac{2\,\sin\left(2\,x\right)\,\left(\sin\left(2\,x\right)+1\right)+2\,\cos^{2}\left(2\,x\right)}{\cos^{2}\left(2\,x\right)}=\dfrac{2\,\sin^{2}\left(2\,x\right)+2\,\sin\left(2\,x\right)}{\cos^{2}\left(2\,x\right)}+2 krok po ...

janusz55,
Po prostu szukam metod rozwiązywania całek i testuję je, nic wielkiego

Całka
\(\int{\dfrac{1}{\sqrt{u-2}\,u^{2}}}{\;\mathrm{d}u}=\dfrac{\sqrt{u-2}}{2\,u}+\dfrac{\operatorname{arctg}\left(\frac{\sqrt{u-2}}{\sqrt{2}}\right)}{2\,\sqrt{2}}+C\)
krok po kroku

janusz55,
Tylko jeśli używasz zastępowania
\(u=\left(x-1\right)^2,\;\,\mathrm{d}u=2\left(x-1\right)\mathrm{d}x,\;\,\mathrm{d}x=\dfrac{1}{2\left(x-1\right)}\,\mathrm{d}u\)
okaże się tak
\(\displaystyle\frac{1}{2}\int\frac{1}{u^2\sqrt{u-2}}\;\mathrm{d}u\)

\int{\dfrac{1}{\left(x-1\right)^{3}\,\sqrt{x^{2}-2\,x-1}}}{\;\mathrm{d}x}=\dfrac{\sqrt{x^{2}-2\,x-1}}{4\,\left(x-1\right)^{2}}-\dfrac{\arcsin\left(\frac{\sqrt{2}}{x-1}\right)}{4\,\sqrt{2}}+C krok po kroku: https://mathdf.com/int/pl/?expr=1%2F((x-1)%5E3*sqrt(x%5E2-2x-1)&arg=x lub tak https://mat...

Lub \mathrm{A}^{\mathrm{T}}\,\mathrm{A}= {\left(\begin{matrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\a_{41}&a_{42}&a_{43}\end{matrix}\right)^{\mathrm{T}}}\cdot \left(\begin{matrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a...

W ten sposób? \mathrm{A}\cdot \mathrm{A}^{\mathrm{T}}= \left(\begin{matrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\\a_{41}&a_{42}&a_{43}\end{matrix}\right)\cdot {\left(\begin{matrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a...

Zaktualizowano: Gdyby {\small\quad y=y\left(x\right),\;z=z\left(x\right)} wtedy: \left(y\,z\right)'_{x}=y\,z'+y'\,z \left(y\,z'+y'\,z\right)'_{x}=y\,z''+2\,y'\,z'+y''\,z \left(y\,z''+2\,y'\,z'+y''\,z\right)'_{x}=y\,z'''+3\,y'\,z''+3\,y''\,z'+y'''\,z lub \left(t\,\cos\left(y'\right)\right)'_{t}{\smal...