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# 4y''+9y=15 Please, someone help me!

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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

May 9, 2018

#1
+21869
+2

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

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

May 9, 2018
#2
+10
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Thank you !

SBonilla  May 9, 2018