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I need help with this one, is by the method of Undetermined Coefficients.

Maybe you can explain to me... 

4y''+9y=15

SBonilla  May 9, 2018
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 #1
avatar+19376 
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I need help with this one, is by the method of Undetermined Coefficients.

Maybe you can explain to me... 

4y''+9y=15

\(ay''+by=c\)

 

2 th-order differential equation: \(4y''+9y=15\)

 

1. characteristic equation:

\(\begin{array}{|rcll|} \hline 4\lambda^2 + 9 &=& 0 \\ 4\lambda^2 &=& -9 \\ \lambda^2 &=& -\frac{9}{4} \\ \lambda^2 &=& \frac{9}{4} \cdot(-1) \\ \lambda &=& \pm \frac{3}{2} i \\ \mathbf{\lambda} & \mathbf{=} & \mathbf{ \underbrace{0}_{=\alpha} \pm \underbrace{\frac{3}{2}}_{=\beta} i } \\ \hline \end{array}\)

 

2. Homogeneous Equations \(y_h =\ ?\)

\(\large{ \begin{array}{|rcll|} \hline y_h & =& e^{\alpha\cdot x}\Big(c_1\sin(\beta x)+c_2\cos(\beta x) \Big) \quad & | \quad \lambda \text{ complex!} \\ &=& e^{0\cdot x}\Big(c_1\sin(\frac{3}{2} x)+c_2\cos(\frac{3}{2} x) \Big) \\ \mathbf{y_h} &\mathbf{=}& \mathbf{ c_1\sin(\frac{3}{2} x)+c_2\cos(\frac{3}{2} x) }\\ \hline \end{array} } \)

 

3. Particular Solutions \(y_p = \ ?\)

\(\begin{array}{|rcll|} \hline y_p = c \qquad y_p^{'} = 0 \qquad y_p^{''} = 0 \\\\ 4y_p^{''}+9y_p &=& 15 \\ 4\cdot 0 +9 \cdot c &=& 15 \\ 9 \cdot c &=& 15 \\ c &=& \frac{15}{9} \\ \mathbf{c=y_p} &\mathbf{=}& \mathbf{\frac{5}{3}} \\ \hline \end{array} \)

 

4. \(y(x) = \ ?\)

\(\begin{array}{|rcll|} \hline y(x) &=& y_h + y_p \\ \mathbf{y(x)} &\mathbf{=}& \mathbf{ c_1\sin(\frac{3}{2} x)+c_2\cos(\frac{3}{2} x) + \frac{5}{3} } \\ \hline \end{array}\)

 

laugh

heureka  May 9, 2018
 #2
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Thank you ! laugh

SBonilla  May 9, 2018

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