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Równanie różniczkowe drugiego stopnia

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#1 SzymonK

SzymonK

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Napisano 19.04.2022 - 17:17

Rozwiązać równania różniczkowe z zadanymi warunkami początkowymi.
 
x^2(1-ln(x))y''+xy'-y=0 ,y(1)=2,y'(1)=1
 
 x^2y''-2xy'+2y=0, y(1)=3,y'(1)=1

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Afroman

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Napisano 25.09.2011 - 17:55

#2 Mariusz M

Mariusz M

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Napisano 26.02.2023 - 09:59

1) obniżanie rzędu równania liniowego

 

</p>\\<p>x^2\left(1-\ln{\left(x\right)}\right)\cdot y''\left(x\right) + x\cdot y'\left(x\right) -y\left(x\right) = 0 \qquad y\left(1\right) = 2 \qquad y'\left(1\right) = 1<br>\\y_{1}\left(x\right) = x\\</p>\\<p>x^2\left(1-\ln{\left(x\right)}\right)\cdot 0 + x\cdot 1 -x = 0\\</p>\\<p>0 + x - x = 0\\</p>\\<p>0 = 0\\</p>\\<p>y\left(x\right) = x\int{u\left(x\right)\mbox{d}x}\\</p>\\<p>y'\left(x\right) = \int{u\left(x\right)\mbox{d}x} + xu\left(x\right)\\</p>\\<p>y''\left(x\right) = u\left(x\right) + u\left(x\right) + xu'\left(x\right)\\</p>\\<p>y''\left(x\right) = 2u\left(x\right) + xu'\left(x\right)\\</p>\\<p>x^2\left(1-\ln{\left(x\right)}\right)\cdot\left(2u\left(x\right) + xu'\left(x\right)\right)+x\left(\int{u\left(x\right)\mbox{d}x} + xu\left(x\right)\right) - x\int{u\left(x\right)\mbox{d}x} = 0\\</p>\\<p>x^3\left(1-\ln{\left(x\right)}\right)u'\left(x\right)+2x^2\left(1-\ln{\left(x\right)}\right)u\left(x\right)+x^2u\left(x\right)+x\int{u\left(x\right)\mbox{d}x} - x\int{u\left(x\right)\mbox{d}x} = 0\\</p>\\<p>x^3\left(1-\ln{\left(x\right)}\right)u'\left(x\right)+x^2\left(3-2\ln{\left(x\right)}\right)u\left(x\right) = 0\\</p>\\<p>x\left(1-\ln{\left(x\right)}\right)u'\left(x\right)+\left(3-2\ln{\left(x\right)}\right)u\left(x\right) = 0\\</p>\\<p>x\left(1-\ln{\left(x\right)}\right)u'\left(x\right) = \left(2\ln{\left(x\right)} - 3\right)u\left(x\right) \\</p>\\<p>\frac{u'\left(x\right)}{u\left(x\right)} = \frac{2\ln{\left(x\right)} - 3}{x\left(1-\ln{\left(x\right)}\right)}\\</p>\\<p>t = 1-\ln{\left(x\right)}\\</p>\\<p>\mbox{d}t = -\frac{\mbox{d}x}{x}\\</p>\\<p>-2t-1 = 2\ln{\left(x\right)}-3\\</p>\\<p>\int{\frac{2\ln{\left(x\right)} - 3}{x\left(1-\ln{\left(x\right)}\right)}\mbox{d}x}= \int{\frac{2t+1}{t}\mbox{d}t}\\</p>\\<p>\int{\frac{2\ln{\left(x\right)} - 3}{x\left(1-\ln{\left(x\right)}\right)}\mbox{d}x} = 2t + \ln{\left(t\right)}\\</p>\\<p>\frac{u'\left(x\right)}{u\left(x\right)} = - 2\ln{\left(x\right)} + \ln{\left(1-\ln{\left(x\right)}\right)} + C\\</p>\\<p>\ln{\left(u\left(x\right)\right)} = \ln{\left(\frac{1-\ln{\left(x\right)}}{x^2}\right)} +C\\</p>\\<p>u\left(x\right) = \frac{1-\ln{\left(x\right)}}{x^2}\\</p>\\<p>y = x\left(C_{1}\int{\frac{1-\ln{\left(x\right)}}{x^2}\mbox{d}x}\right)\\</p>\\<p>y = C_{1}x\left(-\frac{1}{x}+\frac{\ln{\left(x\right)}}{x} - \int{\frac{1}{x^2}\mbox{d}x}\right)\\</p>\\<p>y = C_{1}x\left(-\frac{1}{x}+\frac{\ln{\left(x\right)}}{x} + \frac{1}{x} + C_{2}\right)\\</p>\\<p>y = C_{1}x\left(\frac{\ln{\left(x\right)}}{x} + C_{2}\right)\\</p>\\<p>y = C_{1}\ln{\left(x\right)} + C_{2}x\\</p>\\<p>y' = \frac{C_{1}}{x}+C_{2}\\</p>\\<p>C_{2} = 2\\</p>\\<p>C_{1}+C_{2}=1\\</p>\\<p>C_{1} = -1\\</p>\\<p>y = 2x - \ln{\left(x\right)}</p>\\<p>

 

2) równanie Eulera - zamiana zmiennej niezależnej sprowadzi równanie do równania o stałych współczynnikach

 

 

</p>\\<p>x^2y''-2xy'+2y = 0 \qquad y\left(1\right) = 3 \qquad y'\left(1\right) = 1</p>\\<p>x=e^{t}\\</p>\\<p>\frac{\mbox{d}x}{\mbox{d}t} = e^{t}\\</p>\\<p>\frac{\mbox{d}y}{\mbox{d}x} = \frac{\mbox{d}y}{\mbox{d}t}\cdot\frac{\mbox{d}t}{\mbox{d}x}\\</p>\\<p>\frac{\mbox{d}y}{\mbox{d}x} = \frac{\mbox{d}y}{\mbox{d}t}e^{-t}\\</p>\\<p>x\frac{\mbox{d}y}{\mbox{d}x} = e^{t}\frac{\mbox{d}y}{\mbox{d}t}e^{-t}\\</p>\\<p>x\frac{\mbox{d}y}{\mbox{d}x} = \frac{\mbox{d}y}{\mbox{d}t}\\</p>\\<p>\frac{\mbox{d}^2y}{\mbox{d}x^2} = \frac{\mbox{d}}{\mbox{d}x}\left(\frac{\mbox{d}y}{\mbox{d}x}\right)\\</p>\\<p>\frac{\mbox{d}^2y}{\mbox{d}x^2} = \frac{\mbox{d}}{\mbox{d}t}\left(\frac{\mbox{d}y}{\mbox{d}t}\cdot\frac{\mbox{d}t}{\mbox{d}x}\right)\frac{\mbox{d}t}{\mbox{d}x}\\</p>\\<p>\frac{\mbox{d}^2y}{\mbox{d}x^2} = \frac{\mbox{d}}{\mbox{d}t}\left(\frac{\mbox{d}y}{\mbox{d}t}\cdot e^{-t}\right)e^{-t}\\</p>\\<p>\frac{\mbox{d}^2y}{\mbox{d}x^2} = \left(\frac{\mbox{d}^2y}{\mbox{d}t^2}\cdot e^{-t} - \frac{\mbox{d}y}{\mbox{d}t} \cdot e^{-t}\right)e^{-t}\\</p>\\<p>\frac{\mbox{d}^2y}{\mbox{d}x^2} = \left(\frac{\mbox{d}^2y}{\mbox{d}t^2} -\frac{\mbox{d}y}{\mbox{d}t} \right)e^{-2t}\\</p>\\<p>x^2\frac{\mbox{d}^2y}{\mbox{d}x^2} = e^{2t}\left(\frac{\mbox{d}^2y}{\mbox{d}t^2} - \frac{\mbox{d}y}{\mbox{d}t}\right)e^{-2t}\\</p>\\<p>x^2\frac{\mbox{d}^2y}{\mbox{d}x^2} =\left(\frac{\mbox{d}^2y}{\mbox{d}t^2} -\frac{\mbox{d}y}{\mbox{d}t} \right)\\</p>\\<p>\left(\frac{\mbox{d}^2y}{\mbox{d}t^2} - \frac{\mbox{d}y}{\mbox{d}t}\right) - 2\frac{\mbox{d}y}{\mbox{d}t}+2y = 0\\</p>\\<p>\frac{\mbox{d}^2y}{\mbox{d}t^2} - 3 \frac{\mbox{d}y}{\mbox{d}t}+ 2y = 0\\</p>\\<p></p>\\<p>y = e^{\lambda t}\\</p>\\<p>\lambda^2e^{\lambda t} - 3\lambda e^{\lambda t} + 2e^{\lambda t} = 0\\</p>\\<p>\left(\lambda^2 - 3\lambda + 2\right)e^{\lambda t} = 0\\</p>\\<p>\lambda^2 - 3\lambda + 2 = 0\\</p>\\<p>\left(\lambda - 1\right)\left(\lambda - 2\right) = 0\\</p>\\<p>\lambda_{1} = 1\\</p>\\<p>\lambda_{2} = 2\\</p>\\<p>y\left(t\right) = C_{1}e^{t}+C_{2}e^{2t}\\</p>\\<p>y\left(x\right) = C_{1}x + C_{2}x^2\\</p>\\<p>y'\left(x\right) = C_{1} + 2C_{2}x\\</p>\\<p>\begin{cases}</p>\\<p>C_{1}+C_{2} = 3\\</p>\\<p>C_{1}+2C_{2} = 1\\</p>\\<p>\end{cases}\\</p>\\<p>\begin{cases}</p>\\<p>C_{1}+C_{2} = 3\\</p>\\<p>C_{2} = -2\\</p>\\<p>\end{cases}\\</p>\\<p>\begin{cases}</p>\\<p>C_{1} = 5\\</p>\\<p>C_{2} = -2\\</p>\\<p>\end{cases}\\</p>\\<p></p>\\<p>y\left(x\right) = 5x -2x^2\\</p>\\<p></p>\\<p>


Użytkownik Mariusz M edytował ten post 26.02.2023 - 10:00

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