# Help simplifying expression

I have the following set of equations:

1)

$$\text{b_1} \cos (\text{\beta_1} u)-\text{b_1} \cosh (\text{\beta_1} u)-\text{b2} \cos (\text{\beta_1} \theta y)-\text{d_2} \cosh (\text{\beta_1} \theta y)\sin (\text{\beta_1} u)-\sinh (\text{\beta_1} u)=0$$

2)

$$-\text{b_1} \sin (\text{\beta_1} u)-\text{b_1} \sinh (\text{\beta_1} u)+\text{b_2} \theta \sin (\text{\beta 1} \theta y)-\text{d_2} \theta \sinh (\text{\beta_1} \theta y)+\cos (\text{\beta_1} u)-\cosh (\text{\beta_1} u)=0$$

3)

$$-\text{b_1} \cos (\text{\beta_1} u)-\text{b_1} \cosh (\text{\beta_1} u)+\alpha ^4 \text{b_2} \theta ^2 \cos (\text{\beta_1} \theta y)-\alpha ^4 \text{d_2} \theta ^2 \cosh (\text{\beta_1} \theta y)-\sin (\text{\beta_1} u)-\sinh (\text{\beta_1} u)=0$$

... and I would like to find the constants $$b_1$$, $$b_2$$ and $$d_2$$.

$$b_1 =\frac{\sin \left(\beta _1 u\right)-\sinh \left(\beta _1 u\right)}{\cosh \left(\beta _1 u\right)-\cos \left(\beta _1 u\right)}$$

$$b_2 =\frac{2 \cos \left(\beta _1 u\right) \left(\cos \left(\beta _1 u\right) \cosh \left(\beta _1 u\right)-1\right)}{\theta \left(\cosh \left(\beta _1 u\right)-\cos \left(\beta _1 u\right)\right) \left(\cos \left(\beta _1 \theta y\right) \sinh \left(\beta _1 \theta y\right)+\sin \left(\beta _1 \theta y\right) \cosh \left(\beta _1 \theta y\right)\right)}$$

$$d_2 = -b_2 \frac{\cos \left(\beta _1 \theta y\right)}{\cosh \left(\beta _1 \theta y\right)}$$

I want to find those expressions myself, but so far I only have that (using Mathematica):

Click on it, to see it full-scale...

Any help would be very much appreciated !

Writing the system as $$Ab_1+Bb_2+Cd_2+D=0\tag1$$ $$Eb_1+Fb_2+Gd_2+A=0\tag2$$ $$Hb_1+Ib_2+Jd_2+E=0\tag3$$ should make it easy to solve where

$$\qquad\begin{cases}A=\cos (\text{\beta_1} u)-\cosh (\text{\beta_1} u)& \\\\B=-\cos (\text{\beta_1} \theta y)& \\\\C=-\cosh (\text{\beta_1} \theta y)\sin (\text{\beta_1} u)& \\\\D=-\sinh (\text{\beta_1} u)& \\\\E=-\sin (\text{\beta_1} u)-\sinh (\text{\beta_1} u)\end{cases}$$ $$\qquad\begin{cases}F=\theta \sin (\text{\beta 1} \theta y)& \\\\G=-\theta \sinh (\text{\beta_1} \theta y)& \\\\H=-\cos (\text{\beta_1} u)-\cosh (\text{\beta_1} u)& \\\\I=\alpha ^4 \theta ^2 \cos (\text{\beta_1} \theta y)& \\\\J=-\alpha ^4 \theta ^2 \cosh (\text{\beta_1} \theta y)\end{cases}$$

$$E\times (1)-A\times (2)$$ gives $$Kb_2+Ld_2+M=0\tag4$$

$$H\times (1)-A\times (3)$$ gives $$Nb_2+Pd_2+Q=0\tag5$$ where $$K=BE-AF,\quad L=CE-AG,\quad M=DE-A^2$$ $$N=BH-AI,\quad P=CH-AJ,\quad Q=DH-AE$$

$$P\times (4)-L\times (5)$$ gives $$(PK-NL)b_2+PM-QL=0,$$ i.e. $$b_2=\frac{QL-PM}{PK-NL}$$

From $$(4)$$, $$d_2=\frac{MN-KQ}{PK-NL}$$

Finally, from $$(1)$$, we get $$b_1=\frac{-B(QL-PM)-C(MN-KQ)-D(PK-NL)}{A(PK-NL)}$$

• Thank you so much ! This is a very helpful answer ! :) – james Oct 6 '18 at 9:47